Abstract
We construct examples of twice differentiable functions in \({\mathbb {R}}^n\) with continuous Laplacian and unbounded Hessian. The same construction is also applicable to higher order differentiability.
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1 Introduction
The standard Schauder theory states that if \(\Delta u=f\) in \(B_1(0) \subset {\mathbb {R}}^n\) and f is Hölder continuous \(( C^{0,\alpha }\), \(0< \alpha <1\)), then u is \(C^{2,\alpha }\). However, it fails when \(\alpha =0\), that is, if \(\Delta u\) is just continuous, then u may not be \(C^{2}\), as shown by a standard example in \({\mathbb {R}}^2\) (see [3]):
This function has continuous Laplacian but is not \(C^2\) because it is not twice differentiable at the origin. (Another such example can be obtained by replacing \(x^2-y^2\) with xy.)
The main goal of this paper is to construct (a family of) functions that are twice differentiable everywhere with continuous Laplacian and unbounded Hessian. These functions have only gained twice differentiablity at the origin over the above example; nevertheless, it appears that some effort is needed to achieve the gain.
Theorem 1.1
Given any \(C^2\) function \(\varphi : (0, \infty ) \rightarrow {\mathbb {R}}\) satisfying
there is a function \(u: {\mathbb {R}}^n \rightarrow {\mathbb {R}}\) depending on \(\varphi\) with compact support, such that it is twice differentiable everywhere in \({\mathbb {R}}^n\), and it has continuous Laplacian and unbounded Hessian. In particular, u is not in \(C^2({\mathbb {R}}^n)\).
Obviously there are many choices of such functions \(\varphi\); for example, \(\varphi (s)= s^{\alpha }\) with \(0<\alpha <1\), \(\varphi (s)=\ln (s)\), or \(\varphi (s)= \ln \ln \cdots \ln s\) if \(s>c\).
As a consequence, the following is a simple application to the Dirichlet problem.
Corollary 1.2
There is a continuous function f such that the unique solution of the Dirichlet problem
is twice differentiable in \(\overline{B_1(0)}\) and has unbounded Hessian.
For any positive integer k, the Schauder theory also asserts that if \(\Delta u\) is \(C^{k, \alpha }\), then u is \(C^{k+2, \alpha }\). Once again, it fails when \(\alpha =0\), that is, if \(\Delta u\) is just \(C^{k}\), then u may not be \(C^{k+2}\). As a result of our construction we have an extension of Theorem 1.1.
Theorem 1.3
Given any \(C^{k+2}\) function \(\varphi : (0, \infty ) \rightarrow {\mathbb {R}}\) satisfying
there is a function \(u:{\mathbb {R}}^n \rightarrow {\mathbb {R}}\) depending on \(\varphi\) with compact support, such that u is \((k+2)\)-times differentiable everywhere in \({\mathbb {R}}^n\), \(\Delta u\) is \(C^k\), but \(D^{k+2}u\) is unbounded. In particular, u is not in \(C^{k+2}({\mathbb {R}}^n)\).
Corollary 1.4
There is a \(C^k\) function f such that the unique solution of the Dirichlet problem
is \((k+2)\)-times differentiable in \(\overline{B_1(0)}\), but \(D^{k+2}u\) is unbounded.
We would like to point out a dichotomy: although the Schauder theory fails when \(\alpha =0\) for each k, it is indeed true if \(k = \infty\), since \(\Delta u \in C^{\infty }\) does imply \(u \in C^{\infty }\) by the elliptic theory.
According to Theorem 1.1, it would be rather natural to ask if unbounded Hessian is the only reason that hinders u from being \(C^2\). Thus we propose the following problem.
Problem: If a function u is twice differentiable everywhere, \(\Delta u\) is continuous, and the Hessian of u is locally bounded, then is u always \(C^2\)?
We remark that the method of construction for Theorem 1.1 will not yield examples of twice differentiable function with continuous Laplacian and bounded Hessian without being \(C^2\). On the other hand, if the function is not required to be twice differentiable everywhere, then there are simple examples of functions with continuous Laplacian and bounded Hessian, such as the following function ( [4]) that is not twice differentiable at the origin.
We also observe that if \(\Delta u\) is continuous, then the modulus of continuity of Du is of \(o(L\log L)\). (If \(\Delta u\) is just bounded, then the modulus of continuity of Du is only of \(O(L\log L)\). [5]) Precisely, the following is true.
Proposition 1.5
Let u be a \(C^1\) solution of \(\Delta u =f\), where f is a continuous function on \(B_1(0)\) in \({\mathbb {R}}^n\). Then for any \(x,y \in B_{\frac{1}{2}}(0)\),
where \(d=|x-y|\), \(\omega (r)=\displaystyle \sup _{|x-y|<r} |f(x)-f(y)|\), and C is a constant depending only on n.
Here we notice that
which can be easily proved by considering two cases: \(\displaystyle \lim _{d \rightarrow 0} \int _{d}^1 \frac{\omega (r)}{r} dr < \infty\) or \(\displaystyle \lim _{d \rightarrow 0} \int _{d}^1 \frac{\omega (r)}{r} dr = \infty .\) It is this \(o(L\log L)\) observation that motivated us to Theorems 1.1 and 1.3.
Theorems 1.1 and 1.3 will be proved in Sects. 3 and 4, respectively, after a thorough study of a building block function in Sect. 2. One of the ideas in the construction has its origin in [2] (and also [1]), where the inhomogeneous Cauchy-Riemann equation in the complex plane was considered. Since the proof of Proposition 1.5 is almost identical to that for Corollary 1 in [5], we include a detailed proof in the Appendix for the convenience of the reader.
2 A building block function
In this section we will look at a function that will become a building block and provide some crucial insights for the construction of examples for Theorem 1.1.
Recall that \(\varphi\) is a function satisfying
Thus for \(x=(x_1,\ldots ,x_n) \in {\mathbb {R}}^n\), \(\displaystyle \lim _{|x| \rightarrow 0} \varphi (-\ln |x|^2) = \infty\). Nevertheless, the product of \(\varphi (-\ln |x|^2)\) with positive powers of |x| is well controlled, as shown by the following two simple lemmas that will be repeatedly used later in the construction.
Lemma 2.1
For any \(\beta >0\) and \(\varphi\) satisfying (1),
Proof
Letting \(s=\frac{1}{|x|}\), \(\displaystyle \lim _{|x| \rightarrow 0} |x|^\beta \varphi \left( -\ln |x|^2 \right) = \lim _{s \rightarrow \infty } \frac{\varphi (\ln s^2)}{s^{\beta }}\) is of \(\frac{\infty }{\infty }\) type. By the L’Hopital’s Rule,
as long as \(\beta >0\). \(\square\)
Lemma 2.2
For any \(0 < \beta \le 1\) and \(\varphi\) satisfying (1), there is a constant \(C_{\varphi }\) depending only on \(\varphi\), such that
Proof
By Lemma 2.1, we know
When \(|x|=\frac{2}{3}\),
which is a constant independent of t.
It remains to show that the local maximum of \(\beta |x|^{\beta } \left| \varphi \left( -\ln |x|^2 \right) \right|\) is also bounded by a constant depending only on \(\varphi\). Denote \(s=\frac{1}{|x|}\), then \(s > 1\), and
is equivalent to
We will find the local extremum of \(\lambda (s)\). Because
a critical point \(s_0\) must satisfy
At this point the local extremum of \(\lambda\) is
and so
From (1) we know that \(\displaystyle |\varphi '(\ln s_0^2)|\) is bounded by a constant depending only on \(\varphi\). Thus the local maximum of \(|\lambda (s)|= \beta |x|^{\beta } \left| \varphi \left( -\ln |x|^2 \right) \right|\) is also bounded by a constant depending only on \(\varphi\). This completes the proof.\(\square\)
It is worth noting that Lemma 2.2 would not be true without the coefficient of \(\beta\) in the function. For example, let \(\displaystyle \{ x_k \}\) be a sequence of points in \({\mathbb {R}}^n\) with \(|x_k|=e^{-2^k}\), and let \(\beta _k=\frac{1}{2^k}\), then \(|x_k| \rightarrow 0\) and \(\beta _k \rightarrow 0\) as \(k \rightarrow \infty\), but
This difference will be crucial to our construction of functions with continuous Laplacian and unbounded Hessian.
For any \(|x| \le \frac{1}{2}\), define a function v(x) by
This function will be a building block for our construction, it generalizes the function w at the beginning of Sect. 1 from \({\mathbb {R}}^2\) to \({\mathbb {R}}^n\). It satisfies almost all the conditions in Theorem 1.1, except one that it is not twice differentiable at the origin.
Lemma 2.3
The function v defined by (2) has continuous Laplacian and unbounded Hessian, but it is not twice differentiable at 0.
Proof
By definition, v(x) is \(C^2\) for all \(x \ne 0\), and its derivatives are the following. (In the case \(n \ge 3\), we use i and j to denote indices that are greater than or equal to 3.)
We observe that each term in these derivatives is of the form \(p(x)\varphi \left( -\ln |x|^2 \right)\), \(p(x)\varphi ' \left( -\ln |x|^2 \right)\), or \(p(x)\varphi '' \left( -\ln |x|^2 \right)\), where p(x) is homogeneous in x. For first derivatives, the degree of homogeneity is 1, and for second derivatives, the degree of homogeneity is 0. Because of this, by Lemma 2.1 and the choice of \(\varphi\),
Thus all the first and second derivatives of v approach 0 as \(|x| \rightarrow 0\), except for
As \(|x| \rightarrow 0\), its first term goes to \(\infty\) and all the other terms go to 0, thus \(\frac{\partial ^2 v}{\partial x_1 \partial x_2} (x)\) is unbounded near the origin, which causes the Hessian of v to be unbounded.
On the other hand, because each diagonal entry of the Hessian has a removable discontinuity at the origin,
Lastly, we check the differentiablity of v at 0. It is differentiable because by Lemma 2.1,
Therefore, all first derivatives of v equal 0 at the origin, and v is \(C^1\) throughout \({\mathbb {R}}^n\). Computing the partial derivative \(\frac{\partial ^2 v}{\partial x_1 ^2} (0)\) by definition, we have
Similarly, we also have
Therefore, \(\Delta v (0)=0\). Consequently, \(\Delta v\) is continuous at 0. However, v is not twice differentiable at 0. To see that, we check the differentiability of \(\frac{\partial v}{\partial x_1 }\) at 0:
Let \(x_1=x_3=\cdots =x_n=0\), then \(|x_2|=|x|\) and
hence \(\frac{\partial v}{\partial x_1 }\) is not differentiable at 0. Similarly, \(\frac{\partial v}{\partial x_2 }\) is not differentiable at 0 either.
Interestingly,
thus \(\frac{\partial v}{\partial x_i }\) for all \(i \ge 3\) are differentiable at 0.
Therefore, v fails to be twice differentiable at 0 because \(\frac{\partial v}{\partial x_1 }\) and \(\frac{\partial v}{\partial x_2 }\) are not differentiable at 0. This completes the proof of Lemma 2.3.\(\square\)
3 Construction for Theorem 1.1
In this section, we will first “smooth out" v into a function that is \(C^2\) at the origin, then we will combine a sequence of such functions through scaling and translation to create a desired function that is twice differentiable everywhere with continuous Laplacian and unbounded Hessian, thus proving Theorem 1.1.
Definition 3.1
Let \(\eta : [0, \infty ) \rightarrow [0,1]\) be a fixed, non-increasing \(C^{\infty }\) function such that
For any \(0<t\le \frac{1}{2}\), define a function \(u_t: {\mathbb {R}}^n \rightarrow {\mathbb {R}} \,\, (n \ge 2)\) by
It follows immediately from Lemma 2.1 and (3) that \(u_t\) is continuous everywhere. Actually, it can be shown that \(u_t \in C^2 \left( {\mathbb {R}}^n \right)\), but we will not verify it here because it is not to be used in our construction.
What will be essential to our construction is the fact that all the first derivatives of \(u_t\) and second derivatives of the form \(\frac{\partial ^2 u_t}{\partial x_j ^2}\) are uniformly bounded by constants independent of t. However, that is not the case for \(\frac{\partial ^2 u_t}{\partial x_1 \partial x_2}\), as will be shown later in this section.
Lemma 3.2
There is a constant \(C_{\eta , \varphi }\) depending only on \(\eta\) and \(\varphi\), such that
Proof
By the definition of \(u_t\),
and
so we assume \(|x| \le \frac{2}{3}\). Furthermore, since \(0<t \le \frac{1}{2}\), when \(\frac{1}{2} \le |x| \le \frac{2}{3}\), we know \(\left| u_t (x) \right|\), \(\left| \frac{\partial u_t}{\partial x_j} (x) \right|\), and \(\left| \frac{\partial ^2 u_t}{ \partial x_j^2 } (x) \right|\) are all bounded by a constant depending on \(\eta\) and \(\varphi\) and independent of t. Therefore it remains to show that when \(|x| < \frac{1}{2}\), \(\left| u_t (x) \right|\), \(\left| \frac{\partial u_t}{\partial x_j} (x) \right|\), and \(\left| \frac{\partial ^2 u_t}{ \partial x_j^2 } (x) \right|\) are all bounded by a constant independent of t.
First, when \(|x| < \frac{1}{2}\), we have \(|u_t(x)| \le |x|^2 \varphi (-\ln |x|^2 )\). Then since \(\displaystyle \lim _{|x| \rightarrow 0} |x|^2 \varphi \left( -\ln |x|^2 \right) =0\) by Lemma 2.1, we know \(\displaystyle \sup _{|x| < \frac{1}{2} } \left| u_t (x) \right|\) is bounded by a constant independent of t. Thus (5) is true.
Since \(\eta (|x|) \equiv 1\) for \(|x| \le \frac{1}{2}\), the derivatives of \(u_t\) when \(0< |x| < \frac{1}{2}\) are the following.
In the case \(n \ge 3\), for any \(i \ge 3\),
The first and second terms in (8) are bounded by
Because \(|x| < \frac{1}{2}\), we have \(|x|^{2t+1} \le |x|,\) so
By Lemma 2.1, \(|x| \varphi (-\ln |x|^2 )\) has a removable discontinuity at 0, therefore on the set \(|x| < \frac{1}{2}\) it is bounded by a constant depending only on \(\varphi\). Thus the first and second terms in (8) are bounded by a constant depending only on \(\varphi\). The last term in (8),
is bounded by
It is further bounded by
since \(|x| < \frac{1}{2}\). Because of (1), we know \(|x| \varphi ' (-\ln |x|^2 )\) has a removable discontinuity at 0, therefore on the set \(|x| < \frac{1}{2}\) it is bounded by a constant depending only on \(\varphi\). Thus the last term in (8) is also bounded by a constant depending only on \(\varphi\). Therefore, \(\frac{\partial u_t}{\partial x_1}\) is bounded by a constant depending on \(\varphi\) only. In the same way, we can prove that \(\frac{\partial u_t}{\partial x_2}\) and \(\frac{\partial u_t}{\partial x_i} (i \ge 3)\) are also bounded by a constant depending on \(\varphi\) only. This proves (6).
Lastly, we prove (7). When \(0< |x| < \frac{1}{2}\),
In the case \(n \ge 3\), for any \(i \ge 3\),
To prove (7) we need to estimate each term of (11), (12), and (13).
We start with (11). Note that since \(|x| < \frac{1}{2}\), the 3rd through 6th terms are bounded by
which is further bounded by
By (1), \(\varphi ' (-\ln |x|^2 )\) has a removable discontinuity at 0, therefore on the set \(|x| < \frac{1}{2}\) it is bounded by a constant depending only on \(\varphi\). Similarly the 7th term is also bounded by a constant depending only on \(\varphi\).
The first term,
and the second term,
are bounded by
By Lemma 2.2,
where \(C_{\varphi }\) depends only on \(\varphi\). Hence the first and second terms are bounded by a constant depending on \(\varphi\). Therefore, we have proved that all the terms in (11) are uniformly bounded by a constant independent of t.
All the terms in (12) and (13) can be estimated in the same way, so this completes the proof of (7).\(\square\)
Now we are ready to construct the main function, u, by “piecing together" a sequence of functions \(u_{t_k}\) as follows.
Choose two decreasing sequences of numbers \(R_k \rightarrow 0\) and \(r_k \rightarrow 0\), such that
and for geometric reasons that will be explained later we also require
for example, we may choose \(R_k=10^{-k}\) and \(r_k=10^{-(k+1)}\).
We use \(\zeta _0\) to denote the point \(\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},\ldots , \frac{1}{\sqrt{2}} \right)\) in \({\mathbb {R}}^n\) and choose a sequence \(\{t_k\}\) such that \(0< t_k < \frac{1}{4}\) and \(\displaystyle \lim _{k \rightarrow \infty } t_k =0\). Define the function u(x) by
where the only conditions on \(\epsilon _k\) for now are \(\epsilon _k>0\) and \(\displaystyle \lim _{k \rightarrow \infty } \epsilon _k =0\), we do not need to assign specific values to \(\epsilon _k\) until near the end of this section.
Condition (14) ensures that the balls centered at the points \(R_k\zeta _0\) with radii \(r_k\) are mutually disjoint.
For each \(k \in {\mathbb {N}}\), let \(B_k\) be the ball centered at the point \(R_k \zeta _0\) with radius \(\frac{2}{3}r_k\), then these \(B_k\) are also mutually disjoint. By (3) and (4), the support of each function \(u_{t_k} \left( \frac{x-R_k \zeta _0}{r_k} \right)\) is the ball \(\{x \in {\mathbb {R}}^n: |x-R_k \zeta _0| \le \frac{2}{3}r_k \}\), which is \(B_k\). Therefore, although the definition of u(x) appears to be an infinite sum, it actually is only a single term. For any given \(x \in {\mathbb {R}}^n\), if x is not in any of the \(B_k\), then
otherwise
As \(k \rightarrow \infty\), the radius of \(B_k\) goes down to 0 and its center moves toward the origin, but none of the balls \(B_k\) contains the origin. In fact, for any \(j=1,\ldots , n\), the \(x_j\)-th coordinate hyperplane does not intersect any of the ball \(B_k\). To see this, let
be an arbitrary point on the \(x_j\)-th coordinate hyperplane. The distance from this point to the center of the ball, \(R_k\zeta _0 = \left( \frac{R_k}{\sqrt{2}},\ldots , \frac{R_k}{\sqrt{2}} \right)\), is
since \(\frac{1}{\sqrt{2}} \approx 0.71\) and \(\frac{2}{3} \approx 0.67\). Thus \(u = 0\) on all of the n coordinate hyperplanes, and consequently \(u(0)=0.\) By (5) \(u_{t_k}\) is uniformly bounded by a constant independent of \(t_k\), hence by definition \(\displaystyle \lim _{|x| \rightarrow 0} u(x)=0\). Therefore u is continuous at the origin, and thus continuous everywhere in \({\mathbb {R}}^n\).
Lemma 3.3
The function u(x) as defined in (15) is twice differentiable everywhere in \({\mathbb {R}}^n\), and all its first and second order partial derivatives at the origin are equal to 0.
Proof
By definition u(x) is \(C^2\) for all \(x \ne 0\), so we only need to show it is twice differentiable at the origin. Because \(u=0\) on all the coordinate hyperplanes, for any \(i,j=1,\ldots ,n\),
Thus
Recall that the balls \(B_k\) are mutually disjoint, so for any given \(x \in {\mathbb {R}}^n\), either
or
By (5) we have \(|u_{t_k}| \le C_{\eta , \varphi }\) which only depends on \(\eta\) and \(\varphi\). Thus
Hence we know that
which means u is twice differentiable at the origin. This completes the proof.\(\square\)
Lastly, we will show that u has continuous Laplacian but unbounded Hessian.
Lemma 3.4
The function u(x) as defined in (15) has continuous Laplacian everywhere in \({\mathbb {R}}^n\), and the partial derivative \(\frac{\partial ^2 u}{\partial x_1 \partial x_2}\) is unbounded near the origin.
Proof
Because \(u=0\) on all the coordinate hyperplanes,
For any given \(x \in {\mathbb {R}}^n\), either
or
Thus \(\displaystyle \lim _{|x| \rightarrow 0} \frac{\partial ^2 u}{\partial x_j^2}(x) = 0\), which implies that \(\displaystyle \frac{\partial ^2 u}{\partial x_j^2}(x)\) is continuous at 0. Since it is also continuous for all \(x \ne 0\), it is continuous everywhere. This proves that \(\displaystyle \Delta u = \sum \nolimits _{j=1}^n \frac{\partial ^2 u}{\partial x_j^2}\) is continuous everywhere in \({\mathbb {R}}^n\).
Next we will show \(\displaystyle \frac{\partial ^2 u}{\partial x_1 \partial x_2}\) is unbounded near the origin. For general t,
For each k, choose \(x^{(k)} \in {\mathbb {R}}^n\) such that
Then \(x^{(k)}\) is in the ball \(B_k\) and
Because \(t_k < \frac{1}{4}\), we know \(e^{-\frac{1}{4t_k}}< e^{-1} < \frac{1}{2}\), so in a neighborhood of \(e^{-\frac{1}{4t_k}}\), \(\eta (|x|) \equiv 1\) and \(\eta '(|x|) = \eta '' (|x|) =0.\) Also note that \(x_2=0\) at the point \(\left( e^{-\frac{1}{4t_k}}, 0,\ldots 0 \right)\), then by (16) we have
where we purposefully did not simplify the second term. Thus (17) becomes
The second term in (18) goes to 0 because
by Lemma 2.2. The third term in (18) goes to 0 because \(\frac{1}{2t_k} \rightarrow \infty\) and \(\displaystyle \lim _{s \rightarrow \infty } \varphi '(s)=0\).
Now choose
The first term in (18) becomes
Therefore,
This shows that \(\frac{\partial ^2 u}{\partial x_1 \partial x_2}\) is not bounded near the origin, and the lemma is proved.\(\square\)
4 Construction for Theorem 1.3
The idea for constructing higher order examples for Theorem 1.3 is the same as that for Theorem 1.1, and we only need to replace \(x_1x_2\) by the real or imaginary part of \((x_1+ix_2)^{k+2}\), where \(k \in {\mathbb {N}}\). For example, if \(k=1\), then
so we may use either \(x_1^3 - 3x_1x_2^2\) or \(3x_1^2x_2-x_2^3\) in the construction. For general k,
Evidently, the expressions for its real and imaginary parts are inconvenient to compute. Thus to simplify the calculations we use complex variable for the first two components of x: for any \(x=(x_1, x_2, x_3,\ldots ,x_n) \in {\mathbb {R}}^n\), denote
Then
and consequently
Our strategy is to create a complex-valued function such that it is \((k+2)\)-times differentiable in \({\mathbb {R}}^n\), its Laplacian is \(C^k\), but \(D^{k+2}u\) is unbounded. The real and imaginary parts of u are two real-valued functions, and at least one of them would be a desired function that satisfies all the conditions in Theorem 1.3. The proof is similar to that for Theorem 1.1, so we will only present the key calculations and noticeable differences without repeating the entire proof.
Recall that in the higher order case \(\varphi (s)\) needs to be \((k+2)\)-times differentiable and satisfy
The building block function in this case needs to be modified into
The Laplacian of v is
It can be verified that \(\Delta v\) is \(C^k\), the partial derivative \(\dfrac{\partial ^{k+2} v}{ \partial z^{k+2}}\) is unbounded, and v is not \((k+2)\)-times differentiable. Because this fact is not to be used in our constructions, we will not verify it here.
As in Sect. 3, the next step is to smooth out the function v. Define \(u_t: {\mathbb {R}}^n \rightarrow {\mathbb {R}} \,\, (n \ge 2)\) by
where \(\eta\) is the same as in (3) and \(\varphi\) satisfies (19). This \(u_t\) is a complex-valued \(C^{k+2}\) function. The following is a key fact that will be used later.
Lemma 4.1
For \(u_t\) defined by (20), the k-th partial derivatives of \(\Delta u_t\) are all bounded by a constant independent of t.
Proof
As in the proof of Lemma 3.2, we only need to prove that when \(|x| < \frac{1}{2}\), all k-th partial derivatives of \(\Delta u_t\) are bounded by a constant independent of t. Since \(\eta (|x|) \equiv 1\) when \(|x| \le \frac{1}{2}\), the Laplacian of \(u_t\) is
where
and for \(j=3,\ldots ,n,\)
We will show that all the k-th partial derivatives of each term in (21) and (22) are bounded by a constant independent of t. In the subsequent discussions in this section, we will use C to denote a constant that depends on \(\varphi\), n, k and is independent of t.
We start with (21). The first term of (21) is bounded by
Its first derivatives may be taken with respect to z, \({\bar{z}}\), or \(x_j \,\, (j \ge 3)\).
-
If we take its derivative with respect to z, then
$$\begin{aligned} & \frac{\partial }{\partial z} \left( (k+3)t z^{k+2} |x|^{2t-2} \varphi (-\ln |x|^2 ) \right) \\ & \quad = (k+3)(k+2)t z^{k+1} |x|^{2t-2} \varphi (-\ln |x|^2 ) + (k+3)tz^{k+2}(2t-2)|x|^{2t-3}\left( \frac{{\bar{z}}}{2|x|} \right ) \varphi (-\ln |x|^2 ) + (k+3)t z^{k+2} |x|^{2t-2}\left( \varphi ' (-\ln |x|^2 ) \frac{-{\bar{z}}}{|x|^2} \right) , \end{aligned}$$where the first term and the second term are bounded by
$$\begin{aligned} Ct|x|^{k-1+2t} \left| \varphi (-\ln |x|^2 ) \right| , \end{aligned}$$and the third term is bounded by
$$\begin{aligned} Ct|x|^{k-1+2t} \left| \varphi ' (-\ln |x|^2 ) \right| . \end{aligned}$$Therefore, this derivative is bounded by
$$\begin{aligned} Ct|x|^{k-1+2t} \left| \varphi (-\ln |x|^2 ) \right | + Ct|x|^{k-1+2t} \left | \varphi ' (-\ln |x|^2 ) \right| . \end{aligned}$$ -
If we take its derivative with respect to \({\bar{z}}\), then
$$\begin{aligned} & \frac{\partial }{\partial {\bar{z}}} \left( (k+3)t z^{k+2} |x|^{2t-2} \varphi (-\ln |x|^2 ) \right) \\ & \quad = (k+3)t z^{k+2} \left( (2t-2)|x|^{2t-3}\frac{z}{2|x|} \right) \varphi (-\ln |x|^2 ) \\ & \qquad + (k+3)t z^{k+2} |x|^{2t-2} \left( \varphi ' (-\ln |x|^2 ) \frac{-z}{|x|^2} \right) . \end{aligned}$$By similar argument we know that this derivative is also bounded by
$$\begin{aligned} Ct|x|^{k-1+2t} \left| \varphi (-\ln |x|^2 ) \right | + Ct|x|^{k-1+2t} \left | \varphi ' (-\ln |x|^2 ) \right| . \end{aligned}$$ -
If we take its derivative with respect to \(x_j\), then
$$\begin{aligned} & \frac{\partial }{\partial x_j} \left( (k+3)t z^{k+2} |x|^{2t-2} \varphi (-\ln |x|^2 ) \right) \nonumber \\ & \quad = (k+3)t z^{k+2} \left( (2t-2)|x|^{2t-3}\frac{x_j}{|x|} \right) \varphi (-\ln |x|^2 ) \\ & \qquad + (k+3)t z^{k+2} |x|^{2t-2} \left( \varphi ' (-\ln |x|^2 ) \frac{-2x_j}{|x|^2} \right) . \end{aligned}$$Again, this derivative is bounded by
$$\begin{aligned} Ct|x|^{k-1+2t} \left| \varphi (-\ln |x|^2 ) \right | + Ct|x|^{k-1+2t} \left | \varphi ' (-\ln |x|^2 ) \right| . \end{aligned}$$
In conclusion, regardless of which variable we differentiate with, the first partial derivative of \((k+3)t z^{k+2} |x|^{2t-2} \varphi (-\ln |x|^2 )\) is bounded by
Comparing the first term of this bound, \(Ct|x|^{k-1+2t} \left| \varphi (-\ln |x|^2 ) \right|\), to the right hand side of (23), we see that the power of |x| decreased from \(k+2t\) to \(k-1+2t\). By the same type of calculations, the k-th partial derivatives of \((k+3)t z^{k+2} |x|^{2t-2} \varphi (-\ln |x|^2 )\) will be bounded by
Because
\(\sum \nolimits_{l=1}^k Ct|x|^{k-l+2t} \left| \varphi ^{(l)} (-\ln |x|^2 ) \right|\) is bounded by a constant independent of t. By Lemma 2.2, \(\displaystyle Ct|x|^{2t} \left| \varphi (-\ln |x|^2 ) \right|\) is also bounded by a constant independent of t. Therefore, the k-th derivatives of the first term of (21) are bounded by a constant independent of t.
All the other terms in (21) and (22) can be estimated in the same way. This completes the proof of the lemma.\(\square\)
Then we define u by
where \(R_l\), \(r_l\), \(\zeta _0\), \(t_l\), and \(\epsilon _l\) are the same as in Sect. 3; namely, \(R_l\) and \(r_l\) are decreasing sequences and
Thus by the same argument as in Sect. 3 we know that this infinite sum actually only has a single term for any given x value.
We first show that \(\Delta u\) is \(C^k\). Note that here the power of \(r_l\) is \(k+2\) as opposed to 2 in Sect. 3, so
By Lemma 4.1, the k-th derivatives of \(\Delta u_{t_l}\) are uniformly bounded by a constant independent of \(t_l\). Then since \(\displaystyle \lim _{l \rightarrow \infty } \epsilon _l = 0\), we conclude that the k-th derivatives of \(\Delta u\) all approach 0 as \(|x| \rightarrow 0\). Recall that by construction \(u = 0\) on all of the coordinate hyperplanes, so all partial derivatives of u of any order is 0 at the origin. In particular, all the k-th derivatives of \(\Delta u\) at the origin is 0. Therefore, all the k-th derivatives of \(\Delta u\) are continuous at the origin, and consequently \(\Delta u\) is \(C^k\) throughout \({\mathbb {R}}^n\).
Next, we show that some of the \((k+2)\)-th derivatives of u is unbounded. Precisely, we will show that \(\dfrac{\partial ^{k+2} u}{\partial z^{k+2}}\) is unbounded. We start with a close look at the first and second partial derivatives of \(u_t\) with respect to z.
Note that the power of z in
is \(k+2\). After one differentiation with respect to z, one of the terms in its derivative is
which is the first term in the following formula for \(\frac{\partial u_t}{ \partial z}\):
After another differentiation with respect to z, there will be one term,
where the power of z is k. That is the first term in the following formula for \(\frac{\partial ^2 u_t}{ \partial z^2}\):
After \((k+2)\)-times of differentiation with respect to z, one of the terms in \(\dfrac{\partial ^{k+2} u_t}{ \partial z^{k+2}}\) is
As will be shown later, this term is crucial to proving that \(\dfrac{\partial ^{k+2} u}{\partial z^{k+2}}\) is unbounded.
For each l, as we did in Sect. 3, choose \(x^{(l)} \in {\mathbb {R}}^n\) such that
As discussed in Sect. 3, in a neighborhood of \(\left( e^{-\frac{1}{4t_l}}, 0,\ldots ,0 \right) ,\) \(\eta (|x|) \equiv 1\) and \(\eta '(|x|) = \eta ''(|x|) = 0\). Therefore, when we evaluate \(\frac{\partial u_t}{ \partial z}\) and \(\frac{\partial ^2 u_t}{ \partial z^2}\) in a neighborhood of \(\left( e^{-\frac{1}{4t_l}}, 0,\ldots , 0 \right) ,\) all the terms in (24) and (25) that have an \(\eta '\) or \(\eta ''\) factor will disappear. For that reason in the discussion that follows, we will only consider the terms that have an \(\eta\) factor.
Then (24) becomes
Except the first term, the other terms in (24) are bounded by
And (25) becomes
Except the first term, all the other terms in (25) are bounded by
By the same process, in a neighborhood of \(\left( e^{-\frac{1}{4t_l}}, 0,\ldots 0 \right) ,\) \(\dfrac{\partial ^{k+2} u_{t_l}}{\partial z^{k+2}}\) is equal to
plus some other terms that are bounded by
By Lemma 2.2 and the fact that
we know (26) is bounded by a constant independent of \(t_l\).
Now we look at
If we evaluate (26) at the point \(\left( e^{-\frac{1}{4t_l}}, 0,\ldots , 0 \right)\) and multiply the value with \(\epsilon _l\), the result will go to 0 as \(\epsilon _l \rightarrow 0\).
The first term of \(\dfrac{\partial ^{k+2} u}{ \partial z^{k+2}} (x^{(l)})\) is equal to
Recall that
hence \(\epsilon _l \varphi \left( \frac{1}{2t_l}\right) \rightarrow \infty\) as \(l \rightarrow \infty\). Consequently, \(\dfrac{\partial ^{k+2} u}{ \partial z^{k+2}} (x^{(l)}) \rightarrow \infty\) as \(l \rightarrow \infty\), which implies that \(\dfrac{\partial ^{k+2} u}{ \partial z^{k+2}}\) is unbounded near the origin. Since \(\dfrac{\partial }{\partial z} = \dfrac{1}{2} \left( \dfrac{\partial }{\partial x_1} - i \dfrac{\partial }{\partial x_2} \right) ,\) as a result we know that some of the partial derivatives of u with respect to the \(x_1\) and \(x_2\) variables are unbounded near the origin.
Finally, we need to show that u is \((k+2)\)-times differentiable at the origin. Recall that because \(u=0\) on all the coordinate hyperplanes, all partial derivatives of any order of u at the origin is 0. Then
Note that \(|x| \ge R_l - \frac{2}{3}r_l\), and similar to (5) in Sect. 3 we can prove \(u_{t_l}\) is uniformly bounded by a constant \(C_{\eta , \varphi }\) depending only on \(\eta\) and \(\varphi\), therefore
It follows that \(\displaystyle \lim _{|x| \rightarrow 0}\frac{ u (x) }{|x|^{k+2}} =0\), which implies u is \((k+2)\)-times differentiable at the origin.
Thus we can conclude that as a complex-valued function, u is \((k+2)\)-times differentiable at 0, \(\Delta u\) is \(C^k\) throughout \({\mathbb {R}}^n\), but \(D^{k+2}u\) is unbounded near 0. The real and imaginary parts of u are two real-valued functions that are \((k+2)\)-times differentiable at 0, their Laplacian are \(C^k\) throughout \({\mathbb {R}}^n\), and at least one of them has some unbounded \((k+2)\)-th partial derivatives. Therefore, we have found a function that satisfies all the conditions in Theorem 1.3.
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Appendix: Proof of Proposition 1.5
Appendix: Proof of Proposition 1.5
The proof of Proposition 1.5 is based on a method that was introduced in [5] and elaborated in detail in [3]. Note that after a translation we can assume x or y is at the origin, so we only need to prove that for \(|z| < \frac{1}{16}\), (here z is a point in \({\mathbb {R}}^n\), not a complex variable as was used in the previous section), we have
For \(|z| \ge \frac{1}{16}\) the estimate is also true by a covering argument (see [3]).
First, we recall three elementary estimates (see [3]) that will be used frequently in this proof.
If a function v satisfies \(\Delta v=0\) in \(B_r\), then for any positive integer k,
where C only depends on n and k.
If a function v satisfies \(\Delta v=\lambda\) in \(B_r\), where \(\lambda\) is a constant and \(r <1\), then
If a function v satisfies \(\Delta v=f\) in \(B_r\), where f is a given bounded function, then the scaled maximum principle states that
Now we are ready to prove (27). For \(k=0,1,2,\ldots\), let \(u_k\) be the solution to
Then \(\displaystyle \Delta (u_k-u) = f(0)-f\) in \(B_{2^{-k}}\) and \(u_k-u=0\) on \(\partial B_{2^{-k}}\). By the scaled maximum principle it follows that
and therefore
Then since \(u_{k+1} - u_k\) is harmonic, by (28) we have
For any \(|z| \le \frac{1}{16}\), choose \(k \in {\mathbb {N}}\) such that
We will estimate \(\left| Du(z) - Du(0) \right|\) by
We are going to estimate these three terms separately. First, we claim that
To see this, let \({\tilde{u}}(x)= u(0)+x\cdot Du(0)\) be the linear approximation of u at 0. Then \(Du(0)=D{\tilde{u}}(0)\) and \(|{\tilde{u}}(x)-u(x)|=o(|x|)\). Thus
Then we can write \(\displaystyle Du_k(0) -Du(0)=\sum _{j=k}^{\infty } \left( Du_j(0)-Du_{j+1}(0) \right) ,\) and consequently
Next, we estimate the term \(\left| Du(z) - Du_k(z) \right|\). Let \(v_j\) be the solution of
By the same argument as before we can show
Because \(\Delta (u_k - v_k) = f(0)-f(z)\) in \(B_{2^{-k}}(0) \cap B_{2^{-k}}(z)\) and \(B_{2^{-k-1}}(z) \subset B_{2^{-k}}(0) \cap B_{2^{-k}}(z)\),
Then
By (31) we know
and similarly we can prove
so
Using this in (36) we have
Now we only need to estimate \(|Du_k(z) - Du_k(0)|\). Let
\(h_j\) is harmonic, so by (28)
Thus,
Consequently,
Now we need to estimate \(\Vert D^2u_1 \Vert _{L^{\infty }(B_{\frac{1}{4}})}\).
Define a function
Then \(\zeta\) is harmonic because \(\Delta \zeta = \Delta u_1 - f(0) =0\), and \(\zeta = u_1=u\) on \(\partial B_{\frac{1}{2}}(0)\).
Furthermore, \(D_{ij} \zeta = D_{ij} u_1\) when \(i \ne j\), and \(D_{ii} \zeta = D_{ii} u_1 - \frac{f(0)}{n}\).
Therefore,
It follows that
Combining (34), (35), (37), and (38), we have
Finally, note that since \(\omega (r)\) is increasing with r increasing,
Thus
Therefore, we have proved that
and this implies (27).
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Pan, Y., Yan, Y. Examples of twice differentiable functions with continuous Laplacian and unbounded Hessian. Collect. Math. 75, 453–479 (2024). https://doi.org/10.1007/s13348-023-00395-8
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DOI: https://doi.org/10.1007/s13348-023-00395-8