## 1 Introduction and preliminaries

Since its appearance in 2005, the terminology lineability and spaceability has attracted the attention of many researchers and, just recently, the American Mathematical Society introduced this terminology in its 2020 Mathematical Subject Classification under the references 15A03 and 46B87. In a nutshell, this notion consists on finding (when possible) large algebraic structures within non-linear subsets of a topological vector space.

Some early examples of results within this theory are due to V. I. Gurariy (1935–2005), who proved that the set of continuous nowhere differentiable functions on $$\mathbb {R}$$ contains (except for $$\{0\}$$) infinite dimensional linear spaces. Furthermore, in 2005 [1], he proved that the set of differentiable nowhere monotone functions also contains (except for $$\{0\}$$) infinite dimensional linear spaces. After these seminal works, a lot has been done linking many areas of mathematics, such as Set Theory [8, 9, 11], Real and Complex Analysis [10, 18], Linear and Multilinear Algebra [5], Linear Dynamics [15], or Statistics [12]. Let us recall some terminology we shall need throughout this work (which can be found in [3, 4, 22, 24]). Assume that X is a vector space and $$\alpha$$ is a cardinal number. Then a subset $$A\subset X$$ is said to be:

• lineable if there is an infinite dimensional vector space M such that $$M {\setminus } \{0\}\subset A$$.

• $$\alpha$$-lineable if there exists a vector space M with $$\dim (M)=\alpha$$ and $$M{\setminus } \{0\}\subset A$$. In the case $$\dim (M)=\dim (X)$$, the set A is called maximal lineable.

If, in addition, X is a topological vector space, then the subset A is said to be:

• spaceable in X whenever there is a closed infinite-dimensional vector subspace M of X such that $$M{\setminus } \{0\}\subset A$$. In the case $$\dim (M)=\dim (X)$$, the set A is called maximal spaceable.

• $$\alpha$$-latticeable if there exists a Riesz space M such that $$M\backslash \{0\}\subset A$$ and M is an $$\alpha$$-dimensional vector space. If, in addition, M is closed then A is said to be $$\alpha$$-spaceable latticeable. Even more, it is said to be maximal whenever $$\dim (M)=\dim (X)$$.

• M is said to be $$\mu$$-dense-lineable if $$M\cup \{0\}$$ contains a dense vector space of dimension $$\mu$$.

This paper is arranged as follows. Section 2 focuses on studying the lineability properties of certain subsets of continuous functions on bounded intervals. Within this class we consider, among several others: (i.) $$C^\infty (]0,1])$$, i.e., the class of all elements of C([0, 1]) which are infinitely differentiable on ]0, 1], (ii.) $$\mathcal {D}_0$$, i.e., the class of all functions $$h \in C([0,1])$$ which are differentiable $$\lambda$$-almost everywhere with derivative 0 but are not Lipschitz continuous, or (iii.) $$\mathcal {H}$$, that is, the family of all functions $$h \in C([0,1])$$ which are $$\alpha$$-Hölder continuous for every $$\alpha \in ]0,1[$$ but not Lipschitz continuous.

Next, if we fix an integer $$m\ge 0$$, a real number $$1\le p<\infty$$ and a vector space X, Sect. 3 studies the problem of finding lineable/latticeable subsets of the Sobolev Space $$W^{m,p}\left( ]0,1[\right)$$. We shall provide a brief background on theory of Sobolev Spaces over one dimensional bounded intervals in order to have this section self-contained. Section 4 considers the unbounded counterpart of Sect. 3 for Sobolev Spaces. The notation used throughout the paper shall be rather usual.

## 2 Lineability and spaceability in C([0, 1])

We consider the Banach space $$(l^\infty ,\Vert \cdot \Vert _\infty )$$ of all bounded sequences in $$\mathbb {R}$$ as well as the Banach space $$({\text {Bd}}([0,1]),d_\infty )$$ of all bounded functions, and the Banach space $$(C([0,1],d_\infty )$$ of all real-valued continuous functions on [0, 1], endowed with the uniform distance $$d_\infty$$, respectively (we write $$d_\infty$$ instead of $$\Vert \cdot \Vert _\infty$$ to avoid the usage of one symbol for two different objects). In the sequel we will also view each of these Banach spaces as Banach algebras and lattices (with the usual coordinate-wise/pointwise operations). The same holds for the closed subspace $$c_0$$ of $$l^\infty$$ containing all sequences in $$l^\infty$$ converging to 0.

Throughout this section $$f:[0,1] \rightarrow [0,1]$$ will denote a function for which there exists $$x_0 \in ]0,1[$$ with $$f(x_0)=1$$. The only exception is the proof of Theorem 2.10 in which f is $$[-1,1]$$-valued. Writing $$\mathbf {s}=(s_1,s_2,\ldots ) \in l^\infty$$ and $$I_n=]\frac{1}{2^n},\frac{1}{2^{n-1}}[$$ for every $$n \in \mathbb {N}$$), define the operator $$\Phi _f: l^\infty \rightarrow {\text {Bd}}([0,1])$$ by

\begin{aligned} \Phi _f(\mathbf {s})(x) = \left\{ \begin{array}{rl} &{}s_n f(2^nx-1) \text {if } x \in I_n \text { for some } n \in \mathbb {N}, \\ &{}0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text {otherwise}. \end{array} \right. \end{aligned}
(2.1)

The subsequent lemma gathers the most important properties of $$\Phi _f$$ and is straightforward to prove:

### Lemma 2.1

$$\Phi _f$$ is a linear isometric embedding of $$(l^\infty ,\Vert \cdot \Vert _\infty )$$ in $$({\text {Bd}}([0,1]),d_\infty )$$ and at the same time an algebra- and lattice isomorphism. Moreover, $$\Phi _f(l^\infty )$$ is a closed subspace (algebra, lattice) of $$({\text {Bd}}([0,1]),d_\infty )$$. The same holds for $$\Phi _f$$ if $$(l^\infty ,\Vert \cdot \Vert _\infty )$$ is replaced by $$(c_0,\Vert \cdot \Vert _\infty )$$. Additionally, for $$\mathbf {s} \in c_0$$ the function $$\Phi _f(\mathbf {s})$$ is continuous at 0.

### Proof

The fact that $$\Phi _f$$ is a structure-preserving isometry on $$(l^\infty ,\Vert \cdot \Vert _\infty )$$ (and hence also on $$(c_0,\Vert \cdot \Vert _\infty )$$) is straightforward to verify (the fact that $$f(x_0) = 1$$ is crucial to guarantee this since, otherwise, we would not have an isometry, but a contraction). Considering that $$(l^\infty ,\Vert \cdot \Vert _\infty )$$ is complete and that $$\Phi _f$$ is an isometry it follows that $$\Phi _f(l^\infty )$$ is complete and consequently closed in $$({\text {Bd}}([0,1]),d_\infty )$$. The same reasoning applies to $$(c_0,\Vert \cdot \Vert _\infty )$$.

Finally, continuity of $$\Phi _f(\mathbf {s})$$ at 0 for $$\mathbf {s} \in c_0$$ is a straightforward consequence of the fact that $$\vert \Phi _f(\mathbf {s})(x) \mathbf {1}_{I_n}(x) \vert \le \vert s_n \vert$$ holds for every $$n \in \mathbb {N}$$ and for every $$x \in [0,1]$$. $$\square$$

Selecting f adequately yields various results on spaceability/lineability/latticeability of seemingly ‘small’ subfamilies of $$({\text {Bd}}([0,1]),d_\infty )$$ and $$(C([0,1]),d_\infty )$$. In order to simplify notation we will let $$\mathcal {N}$$ denote the class of non-measurable, bounded functions on [0, 1]. Recall that the notation $$C^\infty (]0,1])$$, $$\mathcal {D}_0$$, and $$\mathcal {H}$$ was already presented in the previous section.

### Theorem 2.2

The set $$C^\infty (]0,1])$$ is maximal spaceable and latticeable in $$(C([0,1]),d_\infty )$$.

### Proof

Let $$g:\mathbb {R} \rightarrow [0,1]$$ be defined by $$g(t)=e^{-\frac{1}{(t(1-t))^2}}$$ and set $$f=\frac{1}{g(\frac{1}{2})} \, g$$. Then f maps [0, 1] into [0, 1] and fulfills $$f(0)=f(1)=0$$ as well as $$f^{(n)}(0)=f^{(n)}(1)=0$$ for every $$n \in \mathbb {N}$$ where $$f^{(n)}$$ denotes the derivative of order $$n \in \mathbb {N}$$. For every $$\mathbf {s}\in l^\infty$$ it therefore follows that $$\Phi _f(\mathbf {s})$$ is infinitely differentiable on (0, 1]. According to Lemma 2.1$$\Phi _f(\mathbf {s})$$ is continuous at 0 for every $$\mathbf {s} \in c_0$$, which, again using Lemma 2.1, altogether yields that $$\Phi _f(c_0)$$ is a closed subspace and sublattice of $$(C([0,1]),d_\infty )$$. This completes the proof. $$\square$$

The next theorem has already been established and proved in [20]—our approach using $$\Phi _f$$, however, allows for an alternative very short and simple proof.

### Theorem 2.3

The set $$\mathcal {D}_0$$ is maximal spaceable and latticeable in $$(C([0,1]),d_\infty )$$.

### Proof

Let $$F:[0,1] \rightarrow [0,1]$$ denote a non-decreasing continuous function with $$F(0)=0,F(1)=1$$ fulfilling $$F'(x)=0$$ for $$\lambda$$-almost every $$x \in [0,1]$$. We could, for instance, choose F as the famous Cantor function (a.k.a. evil’s staircase, e.g., [14]) or Minkowski’s questions mark function (see, e.g., [17, 20] and the references therein) or work with fractal interpolation (see [26]). Obviously each such F corresponds to a probability measure which is singular w.r.t. $$\lambda$$, and is not Lipschitz continuous. Choose such a function F and define the function $$f: [0,1] \rightarrow [0,1]$$ by

\begin{aligned} f(x)=F(2x) \mathbf {1}_{[0,\frac{1}{2}]}(x) + F(2-2x) \mathbf {1}_{(\frac{1}{2},1]}(x). \end{aligned}
(2.2)

Then $$f \in C([0,1])$$, f is differentiable $$\lambda$$-everywhere with derivative 0, and f is not Lipschitz continuous. As direct consequence of the construction of $$\Phi _f$$ the same is true for $$\Phi _f(\mathbf {s})$$ if we consider $$\mathbf {s} \in c_0$$. Figure 1 depicts two examples of singular functions F, as well as the corresponding functions f and $$\Phi _f(\mathbf {s})$$ for some $$\mathbf {s} \in c_0$$. According to Lemma 2.1$$\Phi _f(c_0)$$ is a closed subspace and sublattice of $$(C([0,1]),d_\infty )$$, so the proof is complete. $$\square$$

### Theorem 2.4

The set $$\mathcal {H}$$ is maximal spaceable and latticeable in $$(C([0,1]),d_\infty )$$.

### Proof

Letting $$T:[0,1] \rightarrow [0,1]$$ denote Takagi function (see [25]), defined by

\begin{aligned} T(x)=\sum _{n=0}^\infty \frac{1}{2^n} d(2^n\,x,\mathbb {Z}), \end{aligned}

whereby $$d(y,\mathbb {Z}):=\min \{d(y,z):z \in \mathbb {Z}\}$$ and setting $$f:=\frac{3}{2}\,T$$ yields a function $$f:[0,1] \rightarrow [0,1]$$ with $$\max _{x \in [0,1]}f(x)=1$$, which is $$\alpha$$-Hölder continuous for every $$\alpha \in (0,1)$$ but not Lipschitz continuous (again see [25]). Considering $$\Phi _f$$ and proceeding as in the last two proofs yields the desired result. $$\square$$

### Theorem 2.5

The set $$\mathcal {N}$$ is maximal spaceable and latticeable in $$({\text {Bd}}([0,1]),d_\infty )$$.

### Proof

Letting $$N \subseteq ]0,1[$$ denote a non-measurable set, setting $$f=\mathbf {1}_N$$ as well as $$\Phi _f$$ and proceeding as before directly yields the desired result. $$\square$$

Considering the following slight modification of $$\Phi _f$$ allows for an alternative simple proof of the fact that the set of all functions $$f \in C([0,1])$$ whose graph has Hausdorff- and Box-Counting dimension equal to some fixed $$s \in ]1,2]$$ is $$\mathfrak {c}$$-lineable and latticeable in C([0, 1]) (see [6]).

Using the same notation as for $$\Phi _f$$ define the operator $$\Psi _f: l^\infty \rightarrow {\text {Bd}}([0,1])$$ by

\begin{aligned} \Psi _f(\mathbf {s})(x) = \left\{ \begin{array}{ll} &{}\frac{s_n}{2^n} f(2^nx-1) \text {if } x \in I_n \text { for some } n \in \mathbb {N}\\ &{}0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text {otherwise }. \end{array} \right. \end{aligned}
(2.3)

Then $$\Psi _f$$ is well-defined, obviously linear, injective, and Lipschitz continuous with Lipschitz constant $$L=1$$ but no isometry. Before focusing on the aforementioned result we prove the following simple lemma which will be used afterwards.

### Lemma 2.6

Suppose that $$f: [0,1] \rightarrow [0,1]$$ fulfills that its graph $$\Gamma (f)$$ has Hausdorff and Box–Counting dimension equal to $$\alpha \in (1,2]$$ and let $$\Psi _f$$ be defined according to Eq. (2.3). Then for every $$\mathbf {s} \in l^\infty$$ the same is true for $$\Psi _f(\mathbf {s})$$, i.e.,

\begin{aligned} dim_H(\Gamma (\Psi _f(\mathbf {s}))) = dim_B(\Gamma (\Psi _f(\mathbf {s}))) = \alpha \end{aligned}

holds.

### Proof

Fix $$\mathbf {s} \ne \mathbf {0}$$ and set $$M:= \max \{\Vert \mathbf {s} \Vert _\infty ,1\}$$. Some bi-Lipschitz argument in combination with countable stability of the Hausdorff dimension implies $$dim_H(\Gamma (\Psi _f(\mathbf {s}))) = \alpha$$. It therefore suffices to show that the upper Box-Counting dimensionFootnote 1 (see [16]) fulfills $$\overline{dim_B}(\Gamma (\Psi _f(\mathbf {s}))) \le \alpha$$ which can be done as follows: According to [16] in the calculation of the box-counting dimension it suffices to work with $$\delta _k=\frac{M}{2^k}$$ meshes and $$k \in \mathbb {N}$$. Since the set

\begin{aligned} \Gamma (\Psi _f(\mathbf {s})) \cap \, \left[ 0,\frac{1}{2^{k+1}}\right] \times \mathbb {R} \end{aligned}

can be covered by one square of side length $$\delta _k$$, the minimum number $$N_{\delta _k} \left( \Gamma (\Psi _f(\mathbf {s})) \right)$$ of squares of side length $$\delta _k$$ needed to cover $$\Gamma (\Psi _f(\mathbf {s}))$$ fulfills

\begin{aligned} N_{\delta _k} \left( \Gamma (\Psi _f(\mathbf {s})) \right)\le & {} 1 + \sum _{i=1}^{k+1} N_{\delta _k}( \Gamma (\Psi _f(\mathbf {s})) \cap I_i \times \mathbb {R}) \\\le & {} 1 + (k+1) N_{\delta _k} \left( \Gamma (f \cdot M) \right) , \end{aligned}

which yields

\begin{aligned} \limsup _{k \rightarrow \infty } \frac{\ln \left( N_{\delta _k} \left( \Gamma (\Psi _f(\mathbf {s})) \right) \right) }{-\ln (\delta _k)}\le & {} \limsup _{k \rightarrow \infty } \frac{\ln \left( (k+2) N_{\delta _k} \left( \Gamma (f \cdot M) \right) \right) }{-\ln (\delta _k)} \\= & {} \limsup _{k \rightarrow \infty } \frac{\ln (k+1)}{-\ln (\delta _k)} \, + \, \limsup _{k \rightarrow \infty } \frac{\ln \left( N_{\delta _k} \left( \Gamma (f \cdot M) \right) \right) }{-\ln (\delta _k)} \\= & {} 0 + \overline{dim_B}(\Gamma (f\cdot M)) = \overline{dim_B}(\Gamma (f))= \dim _B(\Gamma (f)) = \alpha \end{aligned}

$$\square$$

Lemma 2.6 directly yields the following result already proved in a different manner in [6]. Notice that (contrary to the results on the previous pages) we do not get spaceability since $$\Psi _f$$ is not an isometry and we can not simply conclude that the subspace $$\Psi _f(l^\infty )$$ is closed.

### Theorem 2.7

The family of all functions $$f \in C([0,1])$$ whose graph has Hausdorff- and Box-Counting dimension equal to some fixed $$s \in ]1,2]$$ is $$\mathfrak {c}$$ lineable and latticeable in $$(C([0,1]),d_\infty )$$.

Focusing exclusively on the Hausdorff dimension, working with $$\Phi _f: c_0 \rightarrow {\text {Bd}}([0,1])$$ for some continuous $$f:[0,1] \rightarrow [0,1]$$ whose graph $$\Gamma (f)$$ fulfills $$dim_H(\Gamma (f))=s \in [1,2]$$, and again using some bi-Lipschitz argument together with countable stability of the Hausdorff dimension even yields spaceability:

### Theorem 2.8

The family of all functions $$f \in C([0,1])$$ whose graph has Hausdorff dimension equal to some fixed $$s \in ]1,2]$$ is spaceable and latticeable in $$(C([0,1]),d_\infty )$$.

Considering yet another small modification of $$\Phi _f$$ allows for quick alternative proofs for some of the results going back to [18, 19]. In fact, setting

\begin{aligned} \hat{\Psi }_f(\mathbf {s})(x) = \left\{ \begin{array}{ll} \frac{s_n}{4^n} f(2^nx-1) &{} \text {if } x \in I_n \text { for some } n \in \mathbb {N}\\ 0 &{} \text {otherwise }, \end{array} \right. \end{aligned}
(2.4)

analogously to $$\Psi _f$$ the new operator $$\hat{\Psi }_f$$ is well-defined, linear, injective, and Lipschitz continuous with Lipschitz constant $$L=1$$ (but not an isometry). Using $$\hat{\Psi }_f$$ we can show the subsequent result (Theorem 2.3 combined with Corollary 2.1 in [18])—thereby $$\mathcal {D}_{dis}$$ denotes the set of all functions $$h \in C([0,1])$$ which are differentiable on [0, 1] (at 0 and 1 we consider the one-sided derivates) with a derivative that is discontinuous at every point of a set with positive $$\lambda$$-measure, and $$\mathcal {D}_{\lnot R}$$ the family of all functions $$h \in C([0,1])$$ which are differentiable on [0, 1] with a derivative that is bounded but not Riemann integrable.

### Theorem 2.9

$$\mathcal {D}_{dis}$$ and $$\mathcal {D}_{\lnot R}$$ are $$\mathfrak {c}$$-lineable and latticeable in $$(C([0,1]),d_\infty )$$.

### Proof

(i) The assertion concerning $$\mathcal {D}_{dis}$$ can be proved as follows: Let $$f_h$$ denote one of the functions constructed in the proof of Theorem 2.3. in [18] and proceed as follows. Defining $$g:[0,1] \rightarrow [0,\infty )$$ by

\begin{aligned} g(x)= x^2(1-x)^2\left( f_h(x)- \min \{f_h(z): z \in [0,1]\} \right) \end{aligned}

and setting $$f(x):= \frac{g(x)}{\max \{g(z): z \in [0,1]\}}$$ yields a non-negative, continuous function $$f: [0,1] \rightarrow [0,1]$$ with $$f(0)=f(1)=0$$, which attains its maximum 1 in (0, 1), which is differentiable on [0, 1] and fulfills that its derivative $$f'$$ is discontinuous on a fat Cantor set C. As a direct result, the function $$\hat{\Psi }_f(\mathbf {s})$$ is obviously differentiable on (0, 1]. Considering that the very definition of $$\hat{\Psi }_f$$ implies that $$\hat{\Psi }_f(\mathbf {s})$$ is also differentiable with (right-hand) derivative 0 at 0 it follows that $$\hat{\Psi }_f(\mathbf {s})$$ is differentiable on [0, 1]. Moreover, if $$s_n \ne 0$$ for some $$n \in \mathbb {N}$$, then $$(\hat{\Psi }_f(\mathbf {s}))'$$ is discontinuous on some fat Cantor set contained in $$I_n$$, implying $$\hat{\Psi }_f(\mathbf {s}) \in \mathcal {D}_{dis}$$.

(ii) The assertion concerning $$\mathcal {D}_{\lnot R}$$ follows in the same fashion. $$\square$$

We now turn to the family $$\mathcal {M}_s \subseteq C([0,1])$$ of all functions f fulfilling that the sets $$\underline{U}_h, \overline{U}_h$$, defined by

\begin{aligned} \underline{U}_h=\{x \in [0,1]: h(x)= \min _{z \in [0,1]} h(z)\}, \quad \overline{U}_h=\{x \in [0,1]: h(x)= \max _{z \in [0,1]} h(z)\} \end{aligned}

both have Hausdorff dimension $$s \in ]0,1[$$ and, again using $$\Phi _f$$ for some appropriately chosen f, show that $$\mathcal {M}_s$$ is spaceable in C([0, 1]). Notice that in this context we do not obtain latticeability since the range of the constructed function f is [0, 1].

### Theorem 2.10

$$\mathcal {M}_s$$ is spaceable in $$(C([0,1]),d_\infty )$$ for every $$s \in [0,1]$$.

### Proof

Since the assertion is trivial for $$s \in \{0,1\}$$ it suffices to consider $$s \in (0,1)$$. Fix $$\beta \in ]0,\frac{1}{2}[$$ and define the contractions $$w_1,w_2: [0,1] \rightarrow [0,1]$$ by $$w_1(x)= \beta x$$ and $$w_2(x)= \beta x + 1- \beta$$. Considering the Iterated Function System (IFS, for short) $$\{w_1,w_2\}$$ and using the standard properties of IFSs (see [16] and [21]) it follows that there exists a unique non-empty compact subset $$C_\beta ^*$$ of [0, 1] fulfilling

\begin{aligned} C_\beta ^*= w_1(C_\beta ^*) \cup w_2(C_\beta ^*). \end{aligned}

To simplify notation we will write $$\mathcal {W}(K) = w_1(K) \cup w_2(K)$$ for every non-empty compact subset K of [0, 1] and refer to $$\mathcal {W}$$ as Hutchinson operator induced by the IFS. Again following [16] and [21] and considering that $$w_1,w_2$$ are similarities the set $$C_\beta ^*$$ is self-similar and its Hausdorff dimension $$\text {dim}_H(C_\beta ^*)$$ is the unique solution s of the equation $$2\beta ^s=1$$, i.e., $$\text {dim}_H(C_\beta ^*)=\frac{-\log (2)}{-\log (\beta )} \in (0,1)$$.

As next step we construct a continuous function $$g:[0,1] \rightarrow [0,1]$$ fulfilling $$\overline{U}_g=C_\beta ^*$$ in several steps. Set $$g_1(x)=1$$ for every $$x \in [0,1]$$ and define the function $$g_2: [0,1] \rightarrow [0,1]$$ by

\begin{aligned} g_2(x) = \left\{ \begin{array}{ll} g_1(x) &{} \text {if } x \in \mathcal {W}([0,1]), \\ \frac{1}{\frac{1}{2}-\beta }\,(x - \frac{1}{2}) &{} \text {if } x \in [\beta ,\frac{1}{2}], \text { and}\\ \frac{1}{\frac{1}{2}-\beta }\,(\frac{1}{2} - x) &{} \text {if } x \in [\frac{1}{2},1-\beta ]. \end{array} \right. \end{aligned}
(2.5)

The basic idea for the construction of $$g_2$$ is to replace $$g_1$$ on the (open) interval $$[0,1]{\setminus } \mathcal {W}([0,1])$$ by a reflected tent map. Modifying $$g_2$$ on each of the intervals constituting $$[0,1] {\setminus } \mathcal {W}^2([0,1])$$ in the same manner yields the function $$g_3$$. Proceeding analogously yields a sequence $$(g_n)_{n \in \mathbb {N}}$$ of continuous functions that converges uniformly to a continuous function g, which obviously fulfills $$\overline{U}_g=C_\beta ^*$$. The first and the second panel in the first row of Fig. 2 depict $$g_2$$ and $$g_7$$. Shrinking [0, 1] to $$[\beta ,1-\beta ]$$ and extending linearly on both sides to 0 yields the continuous function $$g^*$$ fulfilling $$g^*(0)=0=g^*(1)$$ (third upper panel in Fig. 2). Finally defining $$f:[0,1] \rightarrow [-1,1]$$ by (see fourth panel in the first row of Fig. 2)

\begin{aligned} f(x) = \left\{ \begin{array}{rl} g^*(x) &{} \text {if } x \in [0,\frac{1}{2}], \\ -g^*(x) &{} \text {if } x \in [\frac{1}{2},1]. \end{array} \right. \end{aligned}
(2.6)

it follows that f is continuous, fulfills $$f(0)=0=f(1)$$, and attains its maximum 1 and its minimum $$-1$$ both in sets of Hausdorff dimension $$\frac{-\log (2)}{-\log (\beta )}$$. Considering the induced operator $$\Phi _f: c_0 \rightarrow {\text {Bd}}([0,1])$$ it follows that $$\Phi _f(\mathbf {s}) \in C([0,1])$$ for every $$\mathbf {s} \in c_0$$ and the assertion of the theorem follows for $$s=\frac{-\log (2)}{-\log (\beta )}$$. Since $$\{\frac{-\log (2)}{-\log (\beta )}: \beta \in ]0,\frac{1}{2}[\}=]0,1[$$ this completes the proof. The lower panel in Fig. 2 depicts $$\Phi _f(\mathbf {s})$$ for $$\beta =\frac{3}{8}$$ and $$\mathbf {s}=(s_1,s_2,s_3,\ldots )$$ with $$s_i=\frac{1}{i} (-1)^{i+1}$$ $$\square$$

### Remark 2.11

An alternative simple way for constructing the (Lipschitz) continuous function $$g^*$$ used in the proof of Theorem 2.10 would be as follows: Letting $$\mathcal {F}$$ denote the family of functions in C([0, 1]) fulfilling $$f(0)=f(1)=1$$ it follows that $$\mathcal {F}$$ is closed in $$(C([0,1]),d_\infty )$$. Defining the operator $$T_\beta : \mathcal {F} \rightarrow \mathcal {F}$$ by

\begin{aligned} T_\beta (h)(x) = \left\{ \begin{array}{ll} \displaystyle \beta h(\frac{x}{\beta }) &{} \text {if } x \in [0,\beta ], \\ \displaystyle \frac{1}{\frac{1}{2}-\beta }\,(x - \frac{1}{2}) &{} \text {if } x \in [\beta ,\frac{1}{2}],\\ \displaystyle \frac{1}{\frac{1}{2}-\beta }\,(\frac{1}{2} - x) &{} \text {if } x \in [\frac{1}{2},1-\beta ], \text { and}\\ \displaystyle \beta \left( \frac{x-(1-\beta )}{\beta } \right) + 1-\beta &{} \text {if } x \in [1-\beta ,1]. \end{array} \right. \end{aligned}
(2.7)

it is straightforward to verify that $$T_\beta$$ is well-defined and a contraction on $$(\mathcal {F},d_\infty )$$, so Banach’s Fixed Point Theorem implies the existence of a unique, globally attractive fixed point, which is easily verified to coincide with $$g^*$$.

We conclude this section by carrying over the main results from C([0, 1]) established so far to

\begin{aligned} L_p([0,1],\mathcal {B}([0,1]),\lambda )=:L^p([0,1]) \end{aligned}

and start with the $$L^p$$-version $$\Phi _f^p$$ of the operator $$\Phi _f$$. For $$p \in [1,\infty )$$, $$f \in L^p([0,1])$$ and $$\mathbf {s} \in l^p$$ define $$\Phi _f^p: l^p \rightarrow L^p([0,1])$$ by

\begin{aligned} \Phi _f^p(\mathbf {s})(x) = \left\{ \begin{array}{rl} s_n 2^{\frac{n}{p}} f(2^nx-1) &{} \text {if } x \in I_n \text { for some } n \in \mathbb {N},\\ 0 &{} \text {otherwise}, \end{array} \right. \end{aligned}
(2.8)

where the right-hand side is to be interpreted as equivalence class in $$L^p([0,1])$$. In the sequel we will write $$\Vert \cdot \Vert _p$$ both for the norm on $$l^p$$ and the norm on $$L^p([0,1])$$ since no confusion will arise. Considering that for every $$f \in L^p([0,1])$$ with $$\Vert f \Vert _p=1$$ we have

\begin{aligned} \Vert \Phi _f^p(\mathbf {s}) \Vert _p^p= & {} \int _{[0,1]} \vert \Phi _f^p(\mathbf {s})(x) \vert ^p\, d\lambda (x)= \sum _{n=1}^\infty \vert s_n \vert ^p 2^{n} \int _{I_n} \vert f(2^nx-1)\vert ^p \, d\lambda (x)\\= & {} \sum _{n=1}^\infty \vert s_n \vert ^p = \Vert \mathbf {s} \Vert _p^p \end{aligned}

it follows that for each such f the operator $$\Phi _f^p: l^p \rightarrow L^p([0,1])$$ is a linear isometry and a lattice isomorphism. As a direct consequence, $$\Phi _f^p(l^p)$$ is a closed subspace of $$L^p([0,1])$$. Notice, however, that although obviously $$l^p \subseteq c_0$$ holds, in general $$\Phi _f^p(\mathbf {s})$$ is not necessarily continuous at 0 (and the same holds true for each representative of $$\Phi _f^p(\mathbf {s})$$). Working with the operator $$\Phi _f^p$$ and preceding as before yields the following complementing results on lineability/spaceability of subsets of $$L^p([0,1])$$:

### Theorem 2.12

The following assertions hold for every fixed $$p \in [1,\infty )$$:

1. 1.

The set of equivalence classes in $$L^p([0,1])$$ which contain some representative contained in $$C^\infty (]0,1])$$ is spaceable and latticeable in $$L^p([0,1])$$.

2. 2.

The set of equivalence classes in $$L^p([0,1])$$ which contain some representative that is differentiable $$\lambda$$-almost everywhere with derivative 0 but not Lipschitz continuous is spaceable and latticeable in $$L^p([0,1])$$.

3. 3.

The set of equivalence classes in $$L^p([0,1])$$ which contain some representative that is $$\alpha$$-Hölder continuous for every $$\alpha \in ]0,1[$$ but not Lipschitz continuous is spaceable and latticeable in $$L^p([0,1])$$.

4. 4.

The set of equivalence classes in $$L^p([0,1])$$ which contain some representative whose graph has Hausdorff- and Box–Counting dimension equal to some $$s \in ]1,2]$$ is $$\mathfrak {c}$$-lineable and latticeable in $$L^p([0,1])$$.

5. 5.

The set of equivalence classes in $$L^p([0,1])$$ which contain some representative whose graph has Hausdorff dimension equal to some $$s \in ]1,2]$$ is spaceable and latticeable in $$L^p([0,1])$$.

## 3 Lineability, spaceability and latticeability in Sobolev spaces over bounded intervals

In this section we consider the problem of finding lineable/latticeable subsets of Sobolev Spaces over one-dimensional bounded intervals. Thus, fixed an integer $$m\ge 0$$ and a real number $$1\le p<\infty$$ we consider, as vector space X, the Sobolev Space $$W^{m,p}\left( ]0,1[\right)$$. Before recalling basic facts about Sobolev spaces over intervals we start with some preparations which will be used subsequently.

Let $$(t_n)_{n \in \mathbb {N}}$$ denote the Thue–Morse sequence (also known as Prouhet–Thue–Morse sequence) defined to be zero if the sum of the digits in the binary expansion of n is even and $$t_n=1$$ otherwise. Let us recall that $$t_n$$ is a binary digits sequence for which the sequence $$d_n = t_0t_1\cdots t_{2^{n}-1}$$ satisfies that $$d_{n+1}$$ is the concatenation of $$d_n$$ and is Boolean complementary, i.e. $$d_0 = 0, d_1= 01, d_2 = 0110, d_3 = 01101001, d_4 = 0110100110010110, \ldots$$

Since binary sequences are identified with $$\mathbb {Z}_{2}$$ numbers, the number $$t = \sum _{n=0}^{\infty }t_{n}2^{n}$$ associated to the sequence $$t_{n}$$ is the unique fixed point of the contraction

\begin{aligned} x=\sum _{n=0}^{\infty }x_{n}2^{n} \rightarrow \sum _{n=0}^{\infty }\left( x_{n}+(1-x_n)2\right) 2^{2n}. \end{aligned}

For each fixed integer $$m\ge 1$$, we will use the following functions in $$\mathbb {R}^{[0,1]}$$:

\begin{aligned} s_{m,0}(x):= {\left\{ \begin{array}{ll} 1-2t_{n}, &{} \text {if }x\in ] \frac{n}{2^{m}},\frac{n+1}{2^{m}}] \text { with }n\in \left\{ 1,\ldots ,2^{m}-1\right\} , \\ 1, &{} \text {if }x\in \left[ 0,\frac{1}{2^{m}}\right] . \end{array}\right. } \end{aligned}

In addition, for each $$j\in \{1, \ldots , m\}$$, we recursively define the function $$s_{m,j}\in \mathbb {R}^{[0,1]}$$ by

\begin{aligned} s_{m,j}(x):=\int _{\left[ 0,x\right] }s_{m,j-1}(t)dt. \end{aligned}

In the following lemma we collect some elementary properties of the functions $$s_{m,j}$$.

### Lemma 3.1

For every integer $$m\ge 1$$ the function $$s_{m,m}$$ is non-negative and

1. (a)

$$s_{m, j}$$ is bounded for every $$j \in \left\{ 0, 1,\ldots ,m\right\}$$,

2. (b)

$$s_{m,j}(0) = s_{m,j}(1)=0$$ for every $$j\in \left\{ 1,\ldots ,m\right\}$$.

We now recall some basic definitions and main properties of Sobolev spaces in one dimension, see [7] for an extended study using weak derivatives (for a distributional point of view see also [23]). We recall that, given $$m\in \mathbb {N}$$ and $$1\le p \le \infty$$ the Sobolev space $$W^{m,p}\left( ]a,b[\right)$$ can be defined as follows

\begin{aligned} W^{m,p}\left( ]a,b[\right) =\left\{ \begin{array}{l} u\in L^p(]a,b[): \text {for each } j\in \{1,\ldots , m\}\text { there exists } g_j\in L^p(]a,b[)\text { with} \\ \displaystyle \int _{]a,b[} u(x) \varphi ^{j}(x) dx =(-1)^{j} \displaystyle \int _{]a,b[} g_j(x) \varphi (x) dx, \, \forall \varphi \in C_{c}^{\infty }(]a,b[) \end{array} \right\} , \end{aligned}

where $$C_{c}^{\infty }(]a,b[)$$ denotes the space of compactly supported, infinitely differentiable functions on ]ab[.

Also, the function $$g_j$$ involved in the previous definition is the well known weak derivative of j-order of the function u and is denoted as usual by $$g_j\equiv u^{j)}$$.

The standard norm in the Sobolev space $$W^{m,p}\left( ]a,b[\right)$$ is given by

\begin{aligned} \Vert u\Vert _{m,p} = \Vert u\Vert _p + \sum _{j=1}^m \Vert u^{j)}\Vert _p, \end{aligned}

where $$\Vert \cdot \Vert _p$$ denotes the usual $$L^p$$-norm. The space $$W^{m,p}\left( ]a,b[\right)$$ endowed with the norm $$\Vert \cdot \Vert _{m,p}$$ satisfies the following basic properties:

### Lemma 3.2

Let $$m\in \mathbb {N}$$, $$1\le p\le \infty$$ and $$]a,b[\subset \mathbb {R}$$. Then:

1. 1.

$$W^{m,p}\left( ]a,b[\right)$$ is a Banach space.

2. 2.

$$W^{m,p}\left( ]a,b[\right)$$ is reflexive for $$1< p < \infty$$.

3. 3.

$$W^{m,p}\left( ]a,b[\right)$$ is separable for $$1 \le p < \infty$$.

4. 4.

For every $$1\le p<\infty$$, $$C_c^{\infty }(\mathbb {R})$$ is dense $$W^{m,p}\left( ]a,b[\right)$$ with respect to the norm $$\Vert \cdot \Vert _{m,p}$$.

In the particular case of ]ab[ being bounded we have, in addition, that:

5. 5.

For every $$1\le p<\infty$$, the polynomials are dense in $$W^{m,p}\left( ]a,b[\right)$$.Footnote 2

6. 6.

$$W^{m,p}\left( ]a,b[\right)$$ is continuously embedded in $$C^{m-1}([a,b])$$. Moreover the embedding is compact for $$1<p \le \infty$$.

The clousure of $$C_c^\infty (]a,b[)$$ in the space $$W^{m,p}\left( ]a,b[\right)$$, for $$1\le p<\infty$$ is denoted by $$W_0^{m,p}\left( ]a,b[\right)$$ and satisfies the following properties:

### Lemma 3.3

Let $$m\in \mathbb {N}$$, $$1\le p<\infty$$ and $$]a,b[\subset \mathbb {R}$$. Then:

1. 1.

$$W_0^{m,p}\left( ]a,b[\right)$$ is a separable Banach space.

2. 2.

$$W_0^{m,p}\left( ]a,b[\right)$$ is reflexive for $$1 < p$$.

3. 3.

$$u\in W^{m,p}\left( ]a,b[\right) \cap C^{m-1}\left( \overline{]a,b[}\right)$$ belongs to $$W_0^{m,p}\left( ]a,b[\right)$$ if and only if $$u=u'=\cdots =u^{m-1)}=0$$ on the boundary of ]ab[.

4. 4.

When ]ab[ is bounded, $$\Vert u^{m)}\Vert _p$$ for $$u\in W_0^{m,p}\left( ]a,b[\right)$$ is a norm equivalent to $$\Vert u\Vert _{m,p}$$.

The main result in this section is the following.

### Theorem 3.4

For every $$1\le p<+\infty$$ the set $$W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$ is spaceable latticeable.

### Proof

Fixed m and p, we consider the function $$f_{n}\in \mathbb {R}^{[0,1]}$$ defined by

\begin{aligned} f_{n}(x):= {\left\{ \begin{array}{ll} s_{m,m}\left( 2^{n}x-1\right) \cdot 2^{-\left( m-\frac{1}{p}\right) n}, &{} \text {if }x\in ] \frac{1}{2^{n}},\frac{1}{2^{n-1}}[, \\ 0, &{} \text {otherwise.} \end{array}\right. } \end{aligned}

We observe that $$f_{n}$$ has derivatives up to order m and $$f_{n}^{j)}$$ is continuous for every $$j\in \{0, 1, \ldots , m-1\}$$ and $$f_{n}^{j)}\left( \frac{1}{2^{n}}\right) =f_{n}^{j)}\left( \frac{1}{2^{n-1}}\right) =0$$. Therefore, for every $$j=1,2,\ldots ,m$$

\begin{aligned} \int _{\left[ \frac{1}{2^{n}},\frac{1}{2^{n-1}}\right] }f_{n}^{j)}(x)\varphi (x)dx=\left( -1\right) ^{j}\int _{\left[ \frac{1}{2^{n}},\frac{1}{2^{n-1}} \right] }f_{n}(x)\varphi ^{j)}(x)dx, \end{aligned}

holds for all sufficiently regular $$\varphi$$, implying

\begin{aligned} \int _{] 0,1[ }f_{n}^{j)}(x)\varphi (x)dx=\left( -1\right) ^{j}\int _{] 0,1[ }f_{n}(x)\varphi ^{j)}(x)dx, \quad \text {for every } \varphi \in C_{c}^{\infty }(]0,1[). \end{aligned}

Now we consider, for every $$d\in \mathbb {N}$$, an infinite subset $$A_d\subset \mathbb {N}$$ such that $$\mathbb {N}=\bigcup _{d\in \mathbb {N}} A_d$$ and any two of these subsets are disjoint. Let us denote by d(n) the element in the position n of $$A_d$$ with the usual order. Moreover, we define the function $$\gamma _{d}\in \mathbb {R}^{[0,1]}$$ by

\begin{aligned} \gamma _{d}(x):= {\left\{ \begin{array}{ll} f_{d(n)}(x)a_{n}, &{} \text {if }x\in ] \frac{1}{2^{d(n)}},\frac{1}{ 2^{d(n)-1}}[, \\ 0, &{} \text {otherwise,} \end{array}\right. } \end{aligned}

where $$a_{n}:=\left( n\ln ^{2}(n+1)\right) ^{-1/p}$$.

We claim that $$\gamma _{d}\in W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$. In order to prove the claim we first show that $$\gamma _{d}$$ admits weak derivatives up to order m. Indeed, given a function $$\varphi$$ regular enough with $${{\,\mathrm{supp}\,}}\varphi \cap ] \frac{1}{2^{d(n)}},\frac{1}{2^{d(n)-1}}[ \ne \emptyset$$, we have, using the properties of $$f_{d(n)}$$, that

\begin{aligned} \int _{\left[ \frac{1}{2^{d(n)}},\frac{1}{2^{d(n)-1}}\right] }\gamma _{d}^{j)}(x)\varphi (x)dx=\left( -1\right) ^{j}\int _{\left[ \frac{1}{2^{d(n)} },\frac{1}{2^{d(n)-1}}\right] }\gamma _{d}(x)\varphi ^{j)}(x)dx. \end{aligned}

In particular, given $$\varphi \in C_{c}^{\infty }(]0,1[)$$ we deduce that

\begin{aligned} \int _{]0,1[}\gamma _{d}^{j)}(x)\varphi (x)dx=\left( -1\right) ^{j}\int _{]0,1[}\gamma _{d}(x)\varphi ^{j)}(x)dx. \end{aligned}

Since $$\gamma _{d}$$ and its derivatives up to order $$m-1$$ are bounded functions we have that $$\gamma _{d}^{j)} \in L^{p}(]0,1[)$$ for every $$j\in \{0, 1, \ldots , m-1\}$$. With respect to $$\gamma _{d}^{m)}$$ we know that

\begin{aligned} \gamma _{d}^{m)}(x) = s_{m,0}(2^{d(n)}x-1) 2^{d(n)/p}a_{n}, \quad \text {for every } x\in ] \frac{1}{2^{d(n)}},\frac{1}{2^{d(n)-1}}[. \end{aligned}

In particular $$\gamma _{d}\in W^{m,p}\left( ]0,1[\right)$$ since

\begin{aligned} \int _{]0,1[} \left| \gamma _{d}^{m)}(x)\right| ^p = \sum _{n=1}^{\infty } 2^{d(n)}a_{n}^p \frac{1}{2^{d(n)}}=\sum _{n=1}^{\infty }\left( n\ln ^{2}(n+1)\right) ^{-1}< +\infty . \end{aligned}

However, $$\gamma _{d}^{m)}$$ does not belong to $$L^{q}(]0,1[)$$ when $$q>p$$ since

\begin{aligned} \int _{]0,1[} \left| \gamma _{d}^{m)}(x)\right| ^q = \sum _{n=1}\frac{1 }{2^{d(n)}}\left( \frac{2^{d(n)}}{n\ln ^{2}(n+1)}\right) ^{q/p}= +\infty . \end{aligned}

This proves the claim, i.e. $$\gamma _{d}\in W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$.

Let us write $$h_{d}:=\gamma _{d}/\left\| \gamma _{d}\right\| _{m,p}$$ and consider the set $$H\subset W^{m,p}\left( ]0,1[\right)$$ given by

\begin{aligned} H=\left\{ h\in W^{m,p}\left( ]0,1[\right) : h= \sum _{d=1}^{\infty }c_{d}h_{d} \right\} . \end{aligned}

By construction, the coefficients $$c_d$$ associated to any function $$h\in H$$ are uniquely determined and H is a nontrivial vector space. We prove now that $$H{\setminus }\{0\} \subset W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$, H is closed in $$W^{m,p}\left( ]0,1[\right)$$ and that it is a lattice.

In order to prove that H is closed we assume that $$g_r\in H$$ and that $$\lim _{r\rightarrow \infty } \Vert g_r - g\Vert _{m,p} =0$$ for some $$g\in W^{m,p}\left( ]0,1[\right)$$. We may assume, if necessary, that $$g_r$$ and g are continuous according to the Sobolev embedding (see (6), Lemma 3.2). In particular, since $$\left\| g_{r}-g\right\| _{p} \le \Vert g_r-g\Vert _{m,p}$$, we have that

\begin{aligned} \lim _{r\rightarrow \infty } \int _{] 2^{-d(n)},2^{-d(n)+1}[ }\left| g_{r}(x)-g(x)\right| ^{p}dx =0, \quad \text {for every } d,n\in \mathbb {N} . \end{aligned}

For each $$d\in \mathbb {N}$$ there exists a unique coefficient $$c_{d_r}\in \mathbb {R}$$ such that, given $$n\in \mathbb {N}$$ we have that

\begin{aligned} g_{r}(x)=c_{d_{r}}\frac{f_{d(n)}(x)}{\left\| \gamma _{d}\right\| _{m,p} }a_{n}, \quad \text { for every } x\in ] 2^{-d(n)},2^{-d(n)+1}[. \end{aligned}

Therefore it follows that

\begin{aligned} \lim _{r\rightarrow \infty } \int _{] 2^{-d(n)},2^{-d(n)+1}[ }\left| c_{d_{r}}\frac{f_{d(n)}(x)}{\left\| \gamma _{d}\right\| _{m,p}} a_{n}-g(x)\right| ^{p}dx=0. \end{aligned}

Now we claim that, fixed d, the sequence $$\left\{ c_{d_{r}}\right\}$$ is bounded. Otherwise we may assume - up to a subsequence - that $$c_{d_{r}}$$ is increasing and unbounded (observe that we may replace g by $$-g$$). Using the definition of $$f_{d(n)}$$ we may choose $$\omega > 0$$ such that $$S_\omega = f_{d(n)}^{-1}([\omega ,+\infty [)$$ is a subset of $$] 2^{-d(n)},2^{-d(n)+1}[$$ with positive measure. Moreover, g is bounded and $$f_{d(n)}(x)a_{n}\ge 0$$ in the interval $$] 2^{-d(n)},2^{-d(n)+1} [$$ and, as a particular case, in $$S_\omega$$. Thus, given $$M>0$$ there exists $$r_{M}>0$$ such that $$\left| c_{d_{r}}\frac{f_{d(n)}(x)}{ \left\| \gamma _{d}\right\| _{m,p}}a_{n}-g(x)\right| ^{p}>M$$ for every $$x \in S_\omega$$ and every $$r>r_{M}$$. This implies, using that the measure of $$S_\omega$$ is positive, that

\begin{aligned} 0= & {} \lim _{r\rightarrow \infty } \int _{] 2^{-d(n)},2^{-d(n)+1}[ }\left| c_{d_{r}}\frac{f_{d(n)}(x)}{\left\| \gamma _{d}\right\| _{m,p}} a_{n}-g(x)\right| ^{p}dx\\\ge & {} \underset{r\rightarrow \infty }{\lim } \int _{S_\omega }\left| c_{d_{r}}\frac{f_{d(n)}(x)}{\left\| \gamma _{d}\right\| _{m,p}}a_{n}-g(x)\right| ^{p}dx = +\infty . \end{aligned}

This is a contradiction and this completes the proof that $$\left\{ c_{d_{r}}\right\}$$ is bounded. In addition, we may assume that - up to a subsequence - $$\left\{ c_{d_{r}}\right\}$$ converges to some $$c_{d}\in \mathbb {R}$$. In particular

\begin{aligned} \lim _{r\rightarrow \infty } \int _{] 2^{-d(n)},2^{-d(n)+1}[ }\left| c_{d_{r}}\frac{f_{d(n)}(x)}{\left\| \gamma _{d}\right\| _{m,p}} a_{n}-c_{d}\frac{f_{d(n)}(x)}{\left\| \gamma _{d}\right\| _{m,p}} a_{n}\right| ^{p}dx =0. \end{aligned}

This implies that the restriction of the function g to the interval $$] 2^{-d(n)},2^{-d(n)+1}[$$ is equal to $$c_{d}\frac{f_{d(n)}(x)}{ \left\| \gamma _{d}\right\| _{m,p}}a_{n}$$ for every $$n\in \mathbb {N}$$ (observe that $$c_{d_r}$$ and $$c_d$$ do not depend on n). Therefore g restricted to the set

\begin{aligned} \bigcup _{n\in \mathbb {N}} ] 2^{-d(n)},2^{-d(n)+1} [ \end{aligned}

is equal to $$c_{d}h_{d}$$ or equivalently $$g=\sum _{d=1}^{\infty }c_{d}h_{d}$$ and we have proved that H is closed.

Observe also that, by construction, given $$h\in H{\setminus }\{0\}$$ then $$h=\sum _{d=1}^{\infty }c_{d}h_{d}$$ with some $$c_{d}\ne 0$$, thus, if $$h\in W^{m,q}\left( ]0,1[\right)$$ for some $$q>p$$ then the corresponding function $$h_{d}$$ belongs to $$W^{m,q}\left( ]0,1[\right)$$ which is a contradiction, i.e. $$H{\setminus }\{0\} \subset W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$.

Moreover H is infinite dimensional, actually we may define the injective map $$T : l_{\infty } \rightarrow H$$ given by $$T\left( c\right) = \sum _{d=1}^{\infty }c_{d} h_{d}$$ for every $$c=\{c_d\}\in l^\infty$$. We observe that

\begin{aligned} \Vert T(c)\Vert _{m,p}= & {} \sum _{j=0}^{m}\left( \int _{]0,1[}\left| \sum _{r=1}^{\infty }c_{r}h_{r}^{j)}(x)\right| ^{p}\right) ^{1/p}dx =\sum _{j=0}^{m}\left( \int _{]0,1[}\sum _{r=1}^{\infty }\left| c_{r}h_{r}^{j)}(x)\right| ^{p}dx\right) ^{1/p} \\\le & {} \left\| c\right\| _{\infty }\sum _{j=0}^{m}\left( \int _{]0,1[}\sum _{r=1}^{\infty }\left| h_{r}^{j)}(x)\right| ^{p}dx\right) ^{1/p}=\left\| c\right\| _{\infty }, \end{aligned}

which implies that T is well defined and even more, that $$l^\infty$$ is continuously embedded in H.

In particular, H contain $$\mathfrak {c}$$ independent vectors.

We finally show that H is a lattice. Indeed, given $$z,g\in H$$ with $$z(x)=\sum _{d=1}^{\infty }c_{d}h_{d}(x)$$ and $$g(x)=\sum _{d=1}^{\infty }g_{d}h_{d}(x)$$ then $$z\vee g$$ is given by $$\left( z\vee g\right) (x)=\sum _{d=1}^{\infty }\left( c_{d}\vee g_{d}\right) h_{d}(x)$$. Observe that, for $$\varphi$$ regular enough we have

\begin{aligned} \int _{\left[ \frac{1}{2^{d(n)}},\frac{1}{2^{d(n)-1}}\right] }\left( c_{d}\vee g_{d}\right) h_{d}^{j)}(x)\varphi (x)dx=\left( -1\right) ^{j}\int _{ \left[ \frac{1}{2^{d(n)}},\frac{1}{2^{d(n)-1}}\right] }\left( c_{d}\vee g_{d}\right) h_{d}^{j)}(x)\varphi ^{j)}(x)dx. \end{aligned}

In particular,

\begin{aligned} \int _{]0,1[}\left( z\vee g\right) ^{j)}(x)\varphi (x)dx=\left( -1\right) ^{j}\int _{]0,1[}\left( z\vee g\right) (x)\varphi ^{j)}(x)dx,\quad \text {for every }\varphi \in C_{c}^{\infty }\left( ]0,1[\right) . \end{aligned}

Moreover, since

\begin{aligned} \left| \left( z\vee g\right) ^{j)}\right| ^{p}(x)=\sum _{d=1}^{\infty }\left| c_{d}\vee g_{d}\right| ^{p}\left| h_{d}^{j)}(x)\right| ^{p}\le \sum _{d=1}^{\infty }\left( \left| c_{d}\right| ^{p}+\left| g_{d}\right| ^{p}\right) \left| h_{d}^{j)}(x)\right| ^{p}, \end{aligned}

we have that $$z\vee g\in W^{m,p}\left( ]0,1[\right)$$ and $$\Vert z\vee g\Vert _{m,p}^{p}\le \Vert z\Vert _{m,p}^{p}+\Vert g\Vert _{m, p}^{p}$$. Therefore $$z\vee g\in H$$.

Finally the maximality is deduced from the fact that $$W^{m,p}\left( ]0,1[\right)$$ is continuously embedded in C(]0, 1[). Thus the space $$W^{m,p}\left( ]0,1[\right)$$ has cardinality $$\mathfrak {c}$$. $$\square$$

### Remark 3.5

Observe that, since the functions $$h_d$$ in the proof above vanish at the boundary of ]0, 1[, from Lemma 3.3 it is deduced that in fact we have proved that, for every $$1\le p<+\infty$$ the set

\begin{aligned} W_0^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W_0^{m,q}\left( ]0,1[\right) \end{aligned}

is maximal spaceable latticeable.

We conclude this section proving maximal dense-lineability by using the sufficient condition in [2] (see Theorem 3.7 bellow). Let us also recall the following definition from that paper.

### Definition 3.6

Let AB be subsets of a vector space X. We say that A is stronger than B if $$A+B\subseteq A$$.

### Theorem 3.7

[2] Let X be a separable Banach space, and consider two subsets AB of X such that A is lineable and B dense-lineable. If A is stronger than B, then A is dense-lineable.

Now we can prove the maximal dense-lineability of $$W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$.

### Corollary 3.8

For every $$1\le p<+\infty$$ the set $$W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$ is maximal dense-lineable.

### Proof

The proof follows immediately from Theorem 3.7 with $$X=W^{m,p}\left( ]0,1[ \right)$$, the set A is given by $$W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$ which, according to Theorem 3.4, is in particular lineable, and finally the set B is the vector space of polynomials in ]0, 1[ which, by Lemma 3.2, is dense in $$W^{m,p}\left( ]0,1[ \right)$$. Thus we only have to prove that A is stronger than B. i.e., $$A+B\subseteq A$$. In order to prove that we observe that $$A+B\subseteq W^{m,p}\left( ]0,1[\right)$$ and given $$f\in A$$ and $$p_0\in B$$ we have that $$(f+p_0)\not \in \bigcup _{q>p}W^{m,q} \left( ]0,1[\right)$$. Otherwise, for some $$q_0>p$$, $$(f+p_0) \in W^{m,q_0} \left( ]0,1[\right)$$ and, since $$p_0\in W^{m,q_0} \left( ]0,1[\right)$$, we have that $$f=(f+p_0)-p_0 \in W^{m,q_0} \left( ]0,1[\right)$$ which contradicts that $$f\in A$$. $$\square$$

## 4 Lineability, spaceability and latticeability in Sobolev spaces over unbounded intervals

In this section we focus on the problem of finding latticeable subsets of Sobolev Spaces over one dimensional unbounded intervals, namely, the Sobolev Space $$W^{m,p}\left( ]1, +\infty [\right)$$.

The analogues to the functions $$s_{m,j}$$ are given by $$k_{m,j} \in \mathbb {R}^{ \left[ 0,2^{m}\right] }$$ with

\begin{aligned} k_{m,0}(x):= {\left\{ \begin{array}{ll} 1-2t_{n}, &{} \text {if }x\in ] n,n+1] \text { with }n\in \left\{ 1,\ldots ,2^{m}-1\right\} , \\ 1, &{} \text {if }x\in \left[ 0,1\right] , \end{array}\right. } \end{aligned}

and, for $$j\in \{1, \ldots , m\}$$,

\begin{aligned} k_{m,j}(x):=\int _{\left[ 0,x\right] }k_{m,j-1}(t)dt. \end{aligned}

The properties of functions $$k_{m,j}$$ are analogous to those of $$s_{m,j}$$ and are collected in the following lemma.

### Lemma 4.1

For every integer $$m\ge 1$$ the function $$k_{m,m}$$ is non-negative and

1. (a)

$$k_{m, j}$$ is bounded for every $$j \in \left\{ 0, 1,\ldots ,m\right\}$$,

2. (b)

$$k_{m,j}(0) = k_{m,j}(2^m)=0$$ for every $$j\in \left\{ 1,\ldots ,m\right\}$$.

The main result about latticeability on Sobolev Spaces over unbounded intervals is the following.

### Theorem 4.2

For every $$1< p<+\infty$$ the set $$W^{m,p}\left( ]1,+\infty [\right) \backslash \bigcup _{q<p}W^{m,q}\left( ]1,+\infty [\right)$$ is spaceable latticeable.

### Proof

Let us consider, for every $$e\in \mathbb {N}$$, an infinite subset $$B_{e}\subset \mathbb {N}$$ such that $$\mathbb {N}=\bigcup _{e\in \mathbb {N} }B_{e}$$ and any two of these subsets are disjoint. Let us denote by e(n) the element in the position n of $$B_{e}$$ with the usual order. The proof follows as in Theorem 3.4 replacing functions $$\gamma _{d}$$ by

\begin{aligned} \rho _{e}(x):= {\left\{ \begin{array}{ll} k_{m,m}(2^{m+1-e(n)}x-2^{m}){2^{-\frac{e(n)}{p}}}a_{n}, &{} \text { if }x\in ] 2^{e(n)-1},2^{e(n)}[ , \\ 0, &{} \text {otherwise,} \end{array}\right. } \end{aligned}

where, as before, $$a_{n}:=\left( n\ln ^{2}(n+1)\right) ^{-1/p}$$.

Write $$\eta _{e}:=\rho _{e}/\left\| \rho _{e}\right\| _{m,p}$$ and consider now the set $$B \subset W^{m,p}\left( ]1, +\infty [\right)$$ given by

\begin{aligned} B=\left\{ \rho \in W^{m,p}\left( ]1,+\infty [\right) : \rho = \sum _{e=1}^{\infty }c_{e}\eta _{e} \right\} . \end{aligned}

Arguing as in the previous section the coefficients $$c_e$$ associated to any function $$\rho \in B$$ are uniquely determined and B is a vector space with $$B{\setminus }\{0\} \subset W^{m,p}\left( ]1, +\infty [\right) \backslash \bigcup _{q<p}W^{m,q}\left( ]1,+\infty [\right)$$. Let us show that B is closed in $$W^{m,p}\left( ]1,+\infty [\right)$$. Indeed, let us assume that $$h_r\in B$$ and that $$\lim _{r\rightarrow \infty } \Vert h_r - h\Vert _{m,p} =0$$ for some $$h\in W^{m,p}\left( ]1,+\infty [\right)$$.

We may assume, if necessary, that $$h_r$$ and h are continuous according to the Sobolev embedding. In particular, since $$\left\| h_{r}-h\right\| _{p} \le \Vert h_r-h\Vert _{m,p}$$, we have that

\begin{aligned} \lim _{r\rightarrow \infty } \int _{] 2^{e(n)-1},2^{e(n)}[ }\left| h_{r}(x)-h(x)\right| ^{p}dx =0, \quad \text {for every } e,n\in \mathbb {N} . \end{aligned}

For each $$e\in \mathbb {N}$$ there exists a unique coefficient $$c_{e_r}\in \mathbb {R}$$ such that, given $$n\in \mathbb {N}$$ we have that

\begin{aligned} h_{r}(x)=c_{e_{r}}\frac{k_{m,m}(2^{m+1-e(n)}x-2^{m})2^{-\frac{e(n)}{p}}}{\left\| \rho _{e}\right\| _{m,p} }a_{n}, \quad \text { for every } x\in ] 2^{e(n)-1},2^{e(n)}[. \end{aligned}

Therefore we have that, for every $$e,n\in \mathbb {N}$$,

\begin{aligned} 0&=\lim _{r\rightarrow \infty } \int _{] 2^{e(n)-1},2^{e(n)} [ }\left| c_{e_{r}}\frac{k_{m,m}(2^{m+1-e(n)}x-2^{m})2^{-\frac{e(n)}{p}}}{\left\| \rho _{e}\right\| _{m,p}} a_{n}-h(x)\right| ^{p}dx\\&=\lim _{r\rightarrow \infty } \int _{]0,2^{m} [ }\left| c_{e_{r}}\frac{k_{m,m}(t)2^{-\frac{e(n)}{p}}}{\left\| \rho _{e}\right\| _{m,p}} a_{n}-h((2^m+t)2^{e(n)-m-1})\right| ^{p} 2^{e(n)-m-1}dt\\&= \frac{1}{2^{m+1}}\lim _{r\rightarrow \infty } \int _{]0,2^{m} [ }\left| c_{e_{r}}\frac{k_{m,m}(t)}{\left\| \rho _{e}\right\| _{m,p}} a_{n}-h(2^{e(n)-1}+t 2^{e(n)-m-1})2^{\frac{e(n)}{p}}\right| ^{p} dt. \end{aligned}

Now we claim that, fixed $$e\in \mathbb {N}$$, the sequence $$\left\{ c_{e_{r}}\right\}$$ is bounded. We will use that for every fixed $$n\in \mathbb {N}$$ we have

\begin{aligned} \lim _{r\rightarrow \infty } \int _{]0,2^{m} [ }\left| c_{e_{r}}\frac{k_{m,m}(t)}{\left\| \rho _{e}\right\| _{m,p}} a_{n}-h(2^{e(n)-1}+t 2^{e(n)-m-1})2^{\frac{e(n)}{p}}\right| ^{p} dt=0. \end{aligned}

Assume on the contrary that - up to a subsequence - $$c_{e_{r}}$$ is increasing and unbounded (observe that we may replace h by $$-h$$). Since $$k_{m,m}$$ is non-negative and non-trivial we may choose $$\omega > 0$$ such that $$S_\omega = k_{m,m}^{-1}([\omega ,+\infty [)$$ is a positive measure subset of $$] 0,2^{m}[$$. Moreover, since h is bounded on compact sets,

\begin{aligned} |h(2^{e(n)-1}+t 2^{e(n)-m-1})2^{\frac{e(n)}{p}}|\le C_{n,e} \end{aligned}

for some positive constant $$C_{n,e}\in \mathbb {R}$$ not depending on r. Thus, given $$M>0$$ there exists $$r_0\equiv r_0(M,e,n)>0$$ such that, for every $$t\in S_\omega$$,

\begin{aligned} \left| c_{e_{r}}\frac{k_{m,m}(t)}{\left\| \rho _{e}\right\| _{m,p}}a_{n}-h(2^{e(n)-1}+t 2^{e(n)-m-1})2^{\frac{e(n)}{p}}\right| ^{p}> \left| c_{e_{r}}\frac{\omega }{\left\| \rho _{e}\right\| _{m,p}}a_{n}-C_{n,e}\right| ^{p} > M, \, \forall r\ge r_0. \end{aligned}

This implies, considering that the measure of $$S_\omega$$ is positive, that

\begin{aligned} 0 =&\lim _{r\rightarrow \infty } \int _{]0,2^{m} [ }\left| c_{e_{r}}\frac{k_{m,m}(t)}{\left\| \rho _{e}\right\| _{m,p}} a_{n}-h(2^{e(n)-1}+t 2^{e(n)-m-1})2^{\frac{e(n)}{p}}\right| ^{p} dt \\ \ge&\underset{r\rightarrow \infty }{\lim } \int _{S_\omega }\left| c_{e_{r}}\frac{\omega }{\left\| \rho _{e}\right\| _{m,p}}a_{n}-C_{n,e} \right| ^{p}dx = +\infty . \end{aligned}

This is a contradiction and thus we have the claim proved, i.e. $$\left\{ c_{e_{r}}\right\}$$ is bounded. In addition, we may assume that - up to a subsequence - $$\left\{ c_{e_{r}}\right\}$$ converges to some $$c_{e}\in \mathbb {R}$$. In particular

\begin{aligned} \lim _{r\rightarrow \infty } \int _{]0,2^{m} [ }\left| c_{e_{r}}\frac{k_{m,m}(t)}{\left\| \rho _{e}\right\| _{m,p}} a_{n}-c_{e}\frac{k_{m,m}(t)}{\left\| \rho _{e}\right\| _{m,p}} a_{n}\right| ^{p} dt =0. \end{aligned}

This implies that the restriction of the function h to the interval $$] 2^{e(n)-1},2^{e(n)}[$$ is equal to

\begin{aligned} c_{e}\frac{k_{m,m}(2^{m+1-e(n)}x-2^{m})2^{-\frac{e(n)}{p}}}{\left\| \rho _{e}\right\| _{m,p} }a_{n} \end{aligned}

for every $$n\in \mathbb {N}$$ (observe that $$c_{e_r}$$ and $$c_e$$ do not depend on n). Therefore h restricted to the set

\begin{aligned} \bigcup _{n\in \mathbb {N}} ] 2^{e(n)-1},2^{e(n)} [ \end{aligned}

is equal to $$c_{e}\eta _{e}$$ or equivalently $$\displaystyle h=\sum \nolimits _{e=1}^{\infty }c_{e}\eta _{e}$$ and we have proved that B is closed. $$\square$$

The next result is obtained by combining the results of Theorems 3.4 and 4.2.

### Theorem 4.3

For every $$1<p<+\infty$$ the set $$W^{m,p}\left( ]0,+\infty [\right) \backslash \bigcup _{q\ne p}W^{m,q}\left( ]0,+\infty [\right)$$ is latticeable.

### Remark 4.4

Arguing as was pointed out in Remark 3.5 the same result is also true for the set

\begin{aligned} W_0^{m,p}\left( ]0,+\infty [\right) \backslash \bigcup _{q\ne p}W_0^{m,q}\left( ]0,+\infty [\right) . \end{aligned}

### Proof

Let us denote by $$\lambda _{s}$$ with $$s\in [0,1]$$ to a family of $$\mathfrak {c}$$ linearly independent functions in the set $$W^{m,p}\left( ]0,1[\right) \backslash \bigcup _{q>p}W^{m,q}\left( ]0,1[\right)$$ whose existence is guaranteed by Theorem 3.4. Moreover, we may assume that $$\lambda _{s}\in C([0,1])$$ and $$\lambda _{s}(0)=\lambda _{s}(1)=0$$. Similarly we denote by $$\upsilon _{s}$$ with $$s\in [0,1]$$ to a family of $$\mathfrak {c}$$ linearly independent functions in the set $$W^{m,p}\left( ]1,+\infty [\right) \backslash \bigcup _{q<p}W^{m,q}\left( ]1,+\infty [\right)$$ whose existence ensure Theorem 4.2. In this case we may assume that $$\upsilon _{s}\in C([1,+\infty [)$$ and $$\upsilon _{s}(1)=0$$.

Next we define the functions

\begin{aligned} \Gamma _{s}(x):= {\left\{ \begin{array}{ll} \lambda _{s}(x), &{} \text {if }x\in ]0,1[, \\ \upsilon _{s}(x), &{} \text {if }x\in ]1,+\infty [, \\ 0, &{} \text {if }x=1, \end{array}\right. } \end{aligned}

which generate a vector space contained in $$W^{m,p}\left( ]0,+\infty [\right) \backslash \bigcup _{q\ne p}W^{m,q}\left( ]0,+\infty [\right)$$. In particular, this set is maximal latticeable. $$\square$$

As a corollary we deduce the same result for Sobolev Spaces in the real line.

### Corollary 4.5

For every $$1<p<+\infty$$ the set $$W^{m,p}\left( \mathbb {R}\right) \backslash \bigcup _{q\ne p}W^{m,q}\left( \mathbb {R}\right)$$ is latticeable.

### Corollary 4.6

Assume that I is an unbounded open interval and $$1\le p<+\infty$$. Then, the set

\begin{aligned} W^{m,p}\left( I\right) \backslash \bigcup _{q\ne p}W^{m,q}\left( I\right) \end{aligned}

is maximal dense-lineable.

### Proof

Assume without loss of generality that $$I=\mathbb {R}$$. We use again Theorem 3.7 with $$X=W^{m,p}\left( \mathbb {R} \right)$$, $$A=W^{m,p}\left( \mathbb {R} \right) \backslash \bigcup _{q \ne p}W^{m,q}\left( \mathbb {R} \right)$$ which, according to Corollary 4.5, is in particular lineable, and $$B=C_c^\infty (\mathbb {R})$$ which, by Lemma 3.2, is dense in $$W^{m,p}\left( \mathbb {R} \right)$$. We now show that A is stronger than B. i.e., $$A+B\subseteq A$$. First we observe that $$A+B\subseteq W^{m,p}\left( \mathbb {R} \right)$$ and, given $$f\in A$$ and $$g\in B$$ we have that $$(f+g)\not \in \bigcup _{q\ne p}W^{m,q} \left( \mathbb {R} \right)$$. Otherwise, for some $$q_0\ne p$$, $$(f+g) \in W^{m,q_0} \left( \mathbb {R}\right)$$ and, since $$g\in W^{m,q_0} \left( \mathbb {R}\right)$$, it follows that $$f=(f+g)-g \in W^{m,q_0} \left( \mathbb {R}\right)$$ which contradicts that $$f\in A$$. $$\square$$

We conclude this section by extending the obtained results to $$\mathbb {R}^N$$. In fact, the results of the previous section can be easily extended to the case of Sobolev spaces $$W^{m,p}(I)$$ where $$I=I_1\times \cdots \times I_N$$ is a N-dimensional bounded cube of $$\mathbb {R}^N$$. Indeed, since I is bounded we have $$W^{m,p}(I_i)\subset W^{m,p}(I)$$ by considering extensions $$\tilde{w}(x_1,\ldots , x_N)= w(x_i)$$ for each $$w\in W^{m,p}(I_i)$$. In addition $$\Vert {\tilde{w}}\Vert _{m,p} \le C \Vert w\Vert _{m,p}$$. Thus, since $$\displaystyle W^{m,p}(I_i){{\setminus }} \bigcup _{q>p}W^{m,q}(I_i)$$ is spaceable latticeable and

\begin{aligned} W^{m,p}(I_i){\setminus } \bigcup _{q>p}W^{m,q}(I_i) \subset W^{m,p}(I){\setminus } \bigcup _{q>p}W^{m,q}(I), \end{aligned}

we have that $$W^{m,p}(I){{\setminus }} \bigcup _{q>p}W^{m,q}(I)$$ is spaceable latticeable.

Moreover, for any N-dimensional cube I, non necessarily bounded, we have that $$W_0^{m,p}(I_1)\times \cdots \times W_0^{m,p}(I_N)\subset W_0^{m,p}(I)$$ by considering $$\tilde{w}(x_1,\ldots , x_N)= \prod _{i=1}^N w_i(x_i)$$ for every $$(w_1, \ldots , w_N)\in W_0^{m,p}(I_1)\times \cdots \times W_0^{m,p}(I_N)$$. In addition, $$\Vert {\tilde{w}}\Vert _{m,p} \le C \prod _{i=1}^N \Vert w_i\Vert _{m,p}$$. Thus, if $$I_1$$ is bounded, using that

\begin{aligned} \left( W_0^{m,p}(I_1){{\setminus }} \bigcup _{q>p}W_0^{m,q}(I_1)\right) \times W_0^{m,p}(I_2)\times \ldots \times W_0^{m,p}(I_N) \subset W_0^{m,p}(I){{\setminus }} \bigcup _{q>p}W_0^{m,q}(I), \end{aligned}

it follows that $$W_0^{m,p}(I){{\setminus }} \bigcup _{q>p}W_0^{m,q}(I)$$ is spaceable latticeable.

Analogously, if $$I_1$$ is unbounded we have that $$W_0^{m,p}(I){\setminus } \bigcup _{q\ne p}W_0^{m,q}(I)$$ is spaceable latticeable for every $$1<p<\infty$$. These results are also true for a general open subset $$\Omega \subset \mathbb {R}^N$$ since for every N-dimensional cube $$I\subset \Omega$$, $$W_0^{m,p}(I)\subset W_0^{m,p}(\Omega )$$ by means of extending functions by zero.

Collecting everything we have proved the following result.

### Theorem 4.7

Assume $$I=I_1\times \cdots \times I_N$$ that for some real intervals, $$I_1, \ldots , I_N$$. Then

1. 1.

If I is bounded, $$W^{m,p}(I){{\setminus }} \bigcup _{q>p}W^{m,q}(I)$$ is spaceable latticeable.

2. 2.

If I is bounded in one direction, $$W_0^{m,p}(I){{\setminus }} \bigcup _{q>p}W_0^{m,q}(I)$$ is spaceable latticeable.

3. 3.

If I is unbounded in one direction, $$W_0^{m,p}(I){{\setminus }} \bigcup _{q\ne p}W_0^{m,q}(I)$$ is spaceable latticeable.

4. 4.

If $$\Omega \subset \mathbb {R}^N$$ is open then $$W_0^{m,p}(\Omega ){{\setminus }} \bigcup _{q>p}W_0^{m,q}(\Omega )$$ is spaceable latticeable.

5. 5.

If $$\Omega \subset \mathbb {R}^N$$ is open and it contains an unbounded cube I then

\begin{aligned} W_0^{m,p}(\Omega ){{\setminus }} \bigcup _{q\ne p}W_0^{m,q}(\Omega ) \end{aligned}

is spaceable latticeable.