Approximation by Hölder Functions in Besov and Triebel–Lizorkin Spaces

Abstract

In this paper, we show that Besov and Triebel–Lizorkin functions can be approximated by a Hölder continuous function both in the Lusin sense and in norm. The results are proved in metric measure spaces for Hajłasz–Besov and Hajłasz–Triebel–Lizorkin functions defined by a pointwise inequality. We also prove new inequalities for medians, including a Poincaré type inequality, which we use in the proof of the main result.

Appendix

The fact that Hajłasz–Besov and Hajłasz–Triebel–Lizorkin spaces are complete (Banach spaces when $p,q\ge 1$ and quasi-Banach spaces otherwise) has not been proved in earlier papers.

Theorem 5.1

The spaces $N^s_{p,q}(X)$ and $M^s_{p,q}(X)$ are complete for all $0<s<\infty$, $0<p,q\le \infty$.

Proof

We prove the Besov case; the proof for Triebel–Lizorkin spaces is similar. Let $(u_i)_i$ be a Cauchy sequence in $N^s_{p,q}(X)$. Since $L^p(X)$ is complete, there exists a function $u\in L^p(X)$ such that $u_i\rightarrow u$ in $L^p(X)$ as $i\rightarrow \infty$. We will show that $(u_i)_i$ converges to u in $N^s_{p,q}(X)$.

We may assume (by taking a subsequence) that

$$\begin{aligned} \Vert u_{i+1}-u_i\Vert _{N^s_{p,q}(X)}\le 2^{-i} \end{aligned}$$

for all $i\in \mathbb {N}$ and that $u_i(x)\rightarrow u(x)$ as $i\rightarrow \infty$ for almost all $x\in X$. Hence, for each $i\in \mathbb {N}$, there exists $(g_{i,k})_k\in l^q(L^p(X))$ and a set $E_i$ of zero measure such that

$$\begin{aligned} |(u_{i+1}-u_i)(x)-(u_{i+1}-u_i)(y)| \le d(x,y)^s(g_{i,k}(x)+g_{i,k}(y)) \end{aligned}$$

for all $x,y\in X{\setminus } E_i$ satisfying $2^{-k-1}\le d(x,y)<2^{-k}$, and that $\Vert (g_{i,k})\Vert _{l^q(L^p(X))}\le 2^{-i}$. This implies that

$$\begin{aligned} |(u_{i+k}-u_i)(x)-(u_{i+k}-u_i)(y)| \le d(x,y)^s\left( \sum _{j=i}^\infty g_{j,k}(x)+ \sum _{j=i}^\infty g_{j,k}(y)\right) \end{aligned}$$

for all $i,k\ge 1$ for almost all $x,y\in X$. This together with the pointwise convergence shows that, letting $k\rightarrow \infty$, we have

$$\begin{aligned} |(u-u_i)(x)-(u-u_i)(y)| \le d(x,y)^s\left( \sum _{j=i}^\infty g_{j,k}(x)+ \sum _{j=i}^\infty g_{j,k}(y)\right) . \end{aligned}$$

Hence $u-u_i$ has a fractional s-gradient $(\sum _{j=i}^\infty g_{j,k})_k$.

When $p,q\ge 1$, we have $\Vert (\sum _{j=i}^\infty g_{j,k})_k\Vert _{l^q(L^p(X))}\le 2^{-i+1}$. If $0<\min \{p,q\}<1$, then, by (2.1), there is $0<r<1$ such that

$$\begin{aligned} \left\| \left( \sum _{j=i}^\infty g_{j,k}\right) _k\right\| _{l^q(L^p(X))}^r \le C\sum _{j=i}^\infty \Vert (g_{j,k})_k\Vert _{l^q(L^p(X))}^r \le C2^{-ri}. \end{aligned}$$

Hence, in both cases, $u-u_i\in N^s_{p,q}(X)$ and $u_i\rightarrow u$ in $N^s_{p,q}(X)$. Thus $u\in N^s_{p,q}(X)$, and the claim follows. $\square$