I am reading a book and I have some questions about the proof. The book wants to show $H^s(\mathbb{R}^d)$ is a Hilbert space iff $s<\frac{d}{2}$. $H^s(\mathbb{R}^d)$ is the homogeneous Sobolev space. For the case $s\geq\frac{d}{2}$, he defines a norm $$N:u\to||\hat{u}||_{L^1(B(0,1))}+||u||_{H^s}$$
He says $H^s(\mathbb{R}^d,N)$ is a Banach space.If we assume that $H^s(\mathbb{R}^d,H^s)$ is also complete then by Banach's theorem, there would be a constant $C$ such that $N(u)\leq C||u||_{H^s}$.
Here is my question. How we know $H^s(\mathbb{R}^d,N)$ is a Banach space? I don't think it is easy to check, maybe there is some way but I don't know. And why the Banach's theorem can be applied here? I don't think it is a finite dimensional space.
Thanks for any help!
Best Answer
First, I am not sure what the author means by Banach's theorem, but it seems like the relevant result from functional analysis which he's using is the bounded inverse theorem: if $T:(X,\left\|\right\|_{X})\rightarrow (Y,\left\|\right\|_{Y})$ is a continuous (i.e. bounded) bijective linear map between two Banach spaces $X$ and $Y$, then $T^{-1}: (Y,\left\|\right\|_{Y})\rightarrow (X,\left\|\right\|_{X})$ is a continuous (i.e. bounded) linear map. Note that the spaces don't need to be finite-dimensional, just Banach; if you don't know this result, you really should consult a text on functional analysis.
This result is used to obtain a contradiction, since if $T=I:(X,\left\|\right\|_{1})\rightarrow (X,\left\|\right\|_{2})$ is the identity map on $X$ and $\left\|\right\|_{1}$ and $\left\|\right\|_{2}$ are two norms on $X$ with respect to which $X$ is complete, then they are equivalent by the preceding result. The author derives a contradiction from this by exhibiting an example $u$ to show that there is no universal constant $C>0$ such that $$\left\|\widehat{u}\right\|_{L^{1}(B(0,1))}\leq C\left\|u\right\|_{\dot{H}^{s}}$$ Here is the example. Let $\mathcal{C}$ be the annulus $\left\{x\in\mathbb{R}^{d} : 1/4\leq\left|x\right|\leq 3/8\right\}$. Note that the annulus $2\mathcal{C}=\left\{1/2\leq\left|x\right|\leq 3/4\right\}$ is disjoitn from $\mathcal{C}$, and $\mathcal{C}\subset B(0,1)$. Define $$\Sigma_{n}:=\left(\sum_{j=1}^{n}\dfrac{2^{n(s+\frac{d}{2})}}{j}\chi_{2^{-j}\mathcal{C}}\right)^{\vee}$$ Since $2^{-j}\mathcal{C}\cap 2^{-i}\mathcal{C}=\emptyset$, $i\neq j$, we have that $$\left\|\widehat{\Sigma}_{n}\right\|_{L^{1}(B(0,1))}=\sum_{j=1}^{n}\dfrac{2^{j(s+\frac{d}{2})}}{j}2^{-jd}C=C\sum_{j=1}^{n}\dfrac{2^{j(s-\frac{d}{2})}}{j},$$ where $C=\left|\mathcal{C}\right|$, and \begin{align*} \left\|\Sigma_{n}\right\|_{\dot{H}^{s}}^{2}&=\int_{\mathbb{R}^{d}}\left|\xi\right|^{2s}\left(\sum_{j=1}^{n}\dfrac{2^{j(s+\frac{d}{2})}}{j}\chi_{2^{-j}\mathcal{C}}(\xi)\right)^{2}d\xi\\ &=\sum_{j=1}^{n}\int_{2^{-j}\mathcal{C}}\left|\xi\right|^{2s}\left(\dfrac{2^{j(s+\frac{d}{2})}}{j}\right)^{2}d\xi\\ &\leq \sum_{j=1}^{n}2^{-2js}2^{-jd}\left(\dfrac{2^{j(s+\frac{d}{2})}}{j}\right)^{2}\int_{\mathcal{C}}d\xi\\ &=C\sum_{j=1}^{n}\dfrac{1}{j^{2}}\leq C_{1}<\infty \end{align*} for all $n$. But since $s\geq d/2$, we see that $\left\|\widehat{\Sigma}_{n}\right\|_{L^{1}(B(0,1))}\uparrow\infty$ as $n\rightarrow\infty$, whence $N(\Sigma_{n})\rightarrow\infty$ as $n\rightarrow\infty$.
To see that $(\dot{H}^{s}(\mathbb{R}^{d}),N)$ is a Banach space, let $(u_{n})_{n}$ be a sequence in $\dot{H}^{s}(\mathbb{R}^{d})$ which is Cauchy with respect to $N$. The definition of $N$ implies that $\widehat{u}_{n}$ is a Cauchy sequence in $L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$, which is complete, whence there exists an element $f\in L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$ such that $(\widehat{u}_{n})$ converges to $f\in L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$. Similarly, since $L^{1}(B(0,1))$ is complete, there exists an element $g\in L^{1}(B(0,1))$ such that $\widehat{u}_{n}$ converges to $g$ in $L^{1}(B(0,1))$.
I claim that $f=g$ a.e. on $B(0,1)$. To show this, we need the following simple lemma.
Proof. Since $\left|\xi\right|^{2s}$ is nonnegative, it is evident that $\left|\xi\right|^{2s}d\xi$ is absolutely continuous with respect to Lebesgue measure. Conversely, suppose $\int_{A}\left|\xi\right|^{2s}d\xi=0$. Then for any $\epsilon>0$, $$\int_{A}d\xi=\int_{A\cap B(0,\epsilon)}d\xi+\int_{A\setminus B(0,\epsilon)}d\xi\leq\int_{B(0,\epsilon)}d\xi+\epsilon^{-2s}\underbrace{\int_{A}\left|\xi\right|^{2s}d\xi}_{=0}\leq \left|B(0,1)\right|\epsilon^{d}$$ Letting $\epsilon\downarrow 0$ completes the proof. $\Box$
Since $\widehat{u}_{n}$ convereges to $f$ in $L^{2}(\mathbb{R}^{d};\left|\xi\right|^{2s}d\xi)$, passing to a subsequence if necessary, we may assume that $\widehat{u}_{n}\rightarrow f$ a.e. with respect to the measure $\left|\xi\right|^{2s}d\xi$. Similarly, since $\widehat{u}_{n}\rightarrow g$ in $L^{1}(B(0,1))$, passing to a further subsequence if necessary, we may assume that $\widehat{u}_{n}\rightarrow g$ (Lebesgue) a.e. on $B(0,1)$. The desired conclusion follows immediately from the above lemma.
We now use this result to show that the $L_{loc}^{1}(\mathbb{R}^{d})$ function $f$ is a tempered distribution. Then we can define $u$ to be the inverse Fourier transform of $f$, and we will have shown that $u\in\dot{H}^{s}$ and $u_{n}\rightarrow u$ with respect to the norm $N$.
For any Schwartz function $\varphi\in\mathcal{S}(\mathbb{R}^{d})$, we have \begin{align*} \int_{\mathbb{R}^{d}}\left|f(\xi)\varphi(\xi)\right|d\xi&\leq\int_{B(0,1)}\left|g(\xi)\varphi(\xi)\right|d\xi+\int_{\mathbb{R}^{d}}\left|\xi\right|^{s}\left|f(\xi)\right|\left|\varphi(\xi)\right|d\xi\\ &\leq\left\|g\right\|_{L^{1}(B(0,1))}\left\|\varphi\right\|_{L^{\infty}}+\left\|f\right\|_{L^{2}(\left|\xi\right|^{2s}d\xi)}\left\|\varphi\right\|_{L^{2}}, \end{align*} where we use Holder's inequality in the last step. Both $\left\|\varphi\right\|_{L^{\infty}}$ and $\left\|\varphi\right\|_{L^{2}}$ are controlled by a finite sum of Schwartz seminorms, from which the conclusion follows.