Sobolev function has continuous representative

Question

I am working on the following exercise from Stein's functional analysis: The space $H^m(\mathbb{R}^d)$ consists of the functions $f \in L^2(\mathbb{R}^d)$ whose derivatives $\partial^\alpha_x f$ taken in the sense of distributions satisfy $\partial^\alpha_x f\in L^2(\mathbb{R}^d)$ for $|\alpha|\leq m$. On $H^m(\mathbb{R}^d)$ we define the following inner product: $$(f,g)_{H^m(\mathbb{R}^d)} = \sum_{|\alpha|\leq m}(\partial^\alpha_x f ,\partial^\alpha_x g)_{L^2(\mathbb{R}^d)}$$ where $$(f,g)_{L^2(\mathbb{R}^d)} = \int_{\mathbb{R}^d}f(x) \overline{g(x)}dx$$ Prove that $H^m(\mathbb{R}^d)$ with the following norm $$||f||_{H^m(\mathbb{R}^d)} = \sqrt{(f,f)_{H^m(\mathbb{R}^d)}}$$ is a Hilbert space.

(a). Prove further that $f \in H^m(\mathbb{R}^d)$ if and only if $\hat{f}(\xi) (1+|\xi|)^m \in L^2(\mathbb{R}^d)$ where $\hat{f}$ is the Fourier transform. And prove that the two norms $||f||_{H^m(\mathbb{R}^d)}$ and $||\hat{f}(\xi)(1+|\xi|)^m||_{L^2(\mathbb{R}^d_\xi)}$ are equivalent.

(b). If $m>d/2$, then additionally prove that $f$ can be corrected on a set of measure zero so that $f$ becomes continuous and is in fact in $C^k$ for $k<m-d/2$. Hint: Use Fourier inversion and notice that
$$\hat{f}(\xi)(1+|\xi|)^{|\alpha|} \in L^1(\mathbb{R}^d)$$ if $|\alpha|<m-d/2$. You then can use this to show that $f$ is bounded in the standard $C^k$ norm.

I have proven that $H^m(\mathbb{R}^d)$ is a Hilbert space as well as part (a). However, I am struggling with part (b). I know that this is the Sobolev embedding theorem, but all of the proofs I have seen for this elsewhere online do not make much sense to me. Any help with this would be greatly appreciated.

Answer 1

Recall from your definition of the Fourier transform, $$ |g(x)|\leq c_d \| \hat{g}\|_{L^1}, $$ for any $g$ such that $\hat{g}\in L^1(\mathbb{R}^d)$.

Now we compute \begin{equation} \begin{split} |f(x)| & \leq c_d \int_{\mathbb{R}^d} |\hat{f}(\xi)|\, d\xi = c_d\int_{\mathbb{R}^d} |\hat{f}(\xi)|(1+|\xi|^2)^{m/2}(1+|\xi|^2)^{-m/2}\, d\xi \\ &\leq C_d \| f\|_{H^m} \int_{\mathbb{R}^d} (1+|\xi|^2)^{-m}\, d\xi. \end{split} \end{equation} Now use that $m>d/2$. To get the continuity, argue as here.

This settles the case $k=0$. For $k>0$ use that $\widehat{\partial^\beta f}(\xi)= c_{d,\beta} \xi^\beta \hat{f}(\xi)$ for any multi-index $\beta$, and if $|\beta|\leq k<m-d/2$ then $$ \int_{\mathbb{R}^d} \dfrac{|\xi^\beta|^2}{(1+|\xi|^2)^m} d\xi \leq \int_{\mathbb{R}^d} \dfrac{|\xi|^{2|\beta|}}{(1+|\xi|^2)^m} d\xi<\infty, $$ so you can argue as before, with $g= \partial^\beta f$.