operators around Sobolev spaces

Sobolev spaces arise naturally in mathematical physics, as most of the textbooks on this subject say. Indeed, one of the most important tools of mathematics for physicists is perhaps the differential equations. So an ideal topological space, or even topological vector space, or even sometimes Banach space, must take into account the differentials of its elements. There are some similar spaces, like the Schwartz spaces(the space of functions whose derivative of any order multiplied by any power of the variables descent rapidly). The topology on these spaces are, in some sense, too restrictive. So we expect to replace the point-wise convergence by some global convergence. This is what a Sobolev space does.

There are various ways of defining a Sobolev space. The one that is often used is using the Fourier transform to define a Sobolev space $H^s(\mathbb{R}^n)$ over the Euclidean space $\mathbb{R}^n$. There is an obvious shortcoming of this kind of definition, that is, it can not be generalized to other spaces except that we can do the Fourier transform on the space.

The name of this post is ‘operators around Sobolev spaces’. We will first introduce the basic concepts in Sobolev spaces, and then say something about the inclusion operator, the injection operator, and perhaps the interpolation operators.

We denote $S(\mathbb{R}^n)$ the Schwartz space, in which every function $f:\mathbb{R}^n\rightarrow\mathbb{C}$ satisfies that $\forall k,m\in\mathbb{N}, \sup_{x\in\mathbb{R}^n}|x^k\partial^{\alpha}f(x)|<\infty(|\alpha|=m)$. We denote the Fourier transform on $S(\mathbb{R}^n)$ as $\mathfrak{F}:S(\mathbb{R}^n)\rightarrow S(\mathbb{R}^n), f(x)\mapsto \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}f(x)e^{-i}\text{d}y$. We know that in fact $\mathfrak{F}$ is an automorphism on $S(\mathbb{R}^n)$. Now for any real number $s\in \mathbb{R}$, we define $H^s(\mathbb{R})$ to be the closure of $S(\mathbb{R}^n)$ under the inner product, $f,g\in S(\mathbb{R}^n), (f,g)=\int_{\mathbb{R}^n}(1+|y|^2)^s \mathfrak{F}(f)(y)\overline{\mathfrak{F}(g)}(y)\text{d}y$(it is indeed an inner product). For simplicity, from now on, we will denote the Fourier transform of $f$ as $\hat{f}$. That is to say, $H^s(\mathbb{R}^n)$ is a Hilbert space under this scalar product, and they are the so-called Sobolev spaces.

Another equivalent definition for these spaces is to use the tempered distribution, $S(\mathbb{R}^n)'$. We can define the Fourier transform on this dual space, and we say that a tempered distribution $f\in S(\mathbb{R}^n)$ lies in $H^s(\mathbb{R}^n)$ if it satisfies that $||f||_{H^s}^2\int_{\mathbb{R}^n}(1+|y|^2)^s|\hat{f}(y)|^2\text{d}y$. We will not prove the equivalence in this post.

First we can say something about the relations between these Sobolev spaces. Note that, the function $g:\mathbb{R}\rightarrow\mathbb{R}, s\mapsto a^s(a>1)$ is an increasing function. So if two real numbers $s, then if $f\in H^t$(we omit the underlying space $\mathbb{R}^n$ when there is no confusion), then $||f||^2_{H^s}=\int (1+|y|^2)^t(1+|y|^2)^{s-t}|\hat{f}(y)|^2\leq||f||^2_{H^t}\infty$, so we obtain that

$H^t$ injects continuously in $H^s$ when $s.

So we see that these Sobolev spaces behave better than those $L^p$ spaces on $\mathbb{R}^n$.

Now that $H^s$ is a Hilbert space, its topological dual is itself. But we can have another description of its dual. We can define a $\mathbb{R}$-linear application $E: H^s\times H^{-s}\rightarrow \mathbb{C}, (f,g)\mapsto \int_{\mathbb{R}^n}\hat{f}\overline{\hat{g}}$. Note that this functional is $\mathbb{C}$-linear in the first variable, while anti $\mathbb{C}$-linear in the second variable. What is more, using the Cauchy-Schwartz inequality, we can see that $|E(f,g)|\leq ||f||_{H^s}||g||_{H^{-s}}$, and there is also the identity $||f||_{H^s}=\sup_{g\in H^{-s},||g||_{H^{-s}}\leq1}|E(f,g)|$. So in this sense, we have that $(H^s)'=H^{-s}$.

Recall what is our intention of creating these spaces: to better deal with the convergence and the derivations. In deed, when $s$ is a positive integer, for example, $s=1$, then for $f\in H^s$, $||f||_{H^1}^2=\int (1+|y|^2)|\hat{f}(y)|^2=\int |f(x)|^2+|\nabla f(x)|^2$, the first term is due to the Parseval identity, and the second term is due to the fact that Fourier transform exchange the derivation and the multiplication by the free variable. So, for those positive integer value $s$, $H^s$ indeed talks about the derivation on the global scale. And the greater $s$ is, the higher order derivation is involved.

This task having been accomplished, the next is to see if we can go back from these spaces to the routine spaces, like those $latexL^p$ spaces. Surprisingly, Sobolev spaces also behave very well in this aspect. That is the Sobolev embedding theorem.

Before doing anything more interesting, we want to say something on the $H^s$-norm. We see that these norms are not very flexible, in the sense that for any function $f\in H^s$, if we define $f_a(x)=f(ax)(0, then comparing these two norms $||f||_{H^s},||f_a||_{H^s}$, it seems that there is no simple relations in between. Whereas in the case of $L^p$ spaces($p\geq1$), we see easily that $||f_a||_{L^p}=a^{-n/p}||f||_{L^p}$, a simple relation valid for any element in the space. The reason for this shortcoming of $H^s$ is that there is a term non-homogeneous in the integral. This inspires us to define a corresponding homogeneous $H^s$-norm, that is, for $f\in H^s$, $||f||_{h^s}^2=\int |y|^{2s}|\hat(f)(y)|^2$. This time we have that $||f_a||_{h^s}^2=\int |y|^{2s}|\hat{f_a}(y)|^2=\int |y|^{2s}|a^{-n}\hat{f}(y/a)|^2=a^{-n+2s}\int |y|^2|\hat{f}(y)|^2=a^{-n+2s}||f||_{h^s}^2$, that is $||f_a||_{h^s}=a^{s-n/2}||f||_{h^s}$.

Compare the two scalar factors, $a^{-n/p}, a^{s-n/2}$, if $s+\frac{n}{p}=\frac{n}{2}$, we see that these two norms $L^p$ and $H^s$ are, in some sense, comparable. In fact, it is in deed the case, this is the following result:

If $s+\frac{n}{p}=\frac{n}{2}$(with $0), then $\forall f\in S(\mathbb{R}^n)$, we have that $||f||_{L^p}\leq c_{s,d}||f||_{h^s}$.

Note that although this result concerns only with the functions in the Schwartz space, but using some density, we can extend this inequality to greater spaces.

The proof of this result depends heavily on the frequency analysis on the frequency counterpart of the Schwartz space. That is, for this $f\in S$, we will decompose $\hat{f}$ in several parts to simplify analysis. But, first of all, let’s see how can we deal with the $L^p$ norms. In general, these norms are difficult to deal with since they are not at all linear. But there is some special cases, for example, $p=1,2,\infty$. So we should convert the $L^p$ integral into something more linear, or easier to deal with. Perhaps for this we should return to the definition of Lebesgue integrals. Recall that for a positive function, we have that $\int f(x)^p\text{d}x=\sum (\lambda_{n+1}^p-\lambda_n^p)m(\{x|\lambda_n $\sum m(\{x|\lambda_n\lambda)$. So that the evaluation of $||f||_{L^p}$ reduces to the evaluation of $m(f>\lambda)$. This, in some sense, is more intuitive.

Now return to the proof for the above result. For any positive real number $R\in\mathbb{R}$, we decompose $f=\mathfrak{F}^{-1}(\hat{f}\chi_{B(0,R)}+\hat{f}(1-\chi_{B(0,R)}))$. That is, we decompose the Fourier transform of $f=f_{1,R}+f_{2,R}$ into a high frequency part and low frequency part. So we have an inclusion $\{|f|>\lambda\}\subset\{|f_{1,R}|>\lambda/2\}\bigcup\{|f_{2,R}|>\lambda/2\}$. So the problem is converted to estimate the $L^{\infty}$ norms of these two partial functions. For the first, $||f_{1,R}||_{L^{\infty}}\leq c_d(1)||\hat{f_{1,R}}||_{L^1}=c_d(1)\int_{B(0,R)}|\hat{f}|$. Now recall the definition of the $H^s$ norms, we can utilize the Cauchy Schwartz inequality to get $||f_{1,R}||_{L^{\infty}}\leq c_d(1)|||y|^{-s}||_{L^2(B(0,R))}|||y|^s\hat{f}(y)||_{L^2(B(0,R))}$. Note that the first norm is a finite number due to the fact that $2s, a crucial fact, and the second integral is smaller than $||f||_{h^s}$. So, all in all, we have that $||f_{1,R}||_{L^{\infty}}\leq c_d(2)R^{n/2-s}||f||_{h^s}$. So, if we choose $R$ such that the second term of the last inequality is smaller than $\lambda/4$(equivalently, that is to say, if $R(\lambda)=(\frac{\lambda}{4c_d(2)||f||_{h^s}})^{\frac{1}{n/2-s}}$), then we would get that $m(f_{1,R}>\lambda/2)=0$. So now things become $\int |f|^p\leq p\int_0^{\infty}\lambda^{p-1}m(f_{2,R(\lambda)}\geq\lambda/2)$. But note that(the inequality of Chebyshev) $m(f_{2,R}\geq\lambda/2)\leq\frac{||f_{2,R}||^2_{L^2}}{(\lambda/2)^2}$. So we have that $\int |f|^p\leq 4p\int\lambda^{p-3}\text{d}\lambda\int |\hat{f}(1-\chi_{B(0,R(\lambda))})|^2$,using the definition of $R(\lambda)$, the second term changes to $4p\int |\hat{f}(x)|^2\text{d}x\int_{0\leq\lambda\leq 4c_d(2)|x|^{n/2-s}||f||_{h^s}}\lambda^{p-3}$, that is, $\frac{4p}{p-2}(4c_d(2))^{p-2}||f||_{h^s}\int |\hat{f}(x)^2||x|^{2s}=c_d(3)||f||^p_{h^s}$. So combining these things together, we get that $||f||_{L^p}\leq c_d(4)||f||_{h^s}$, thus proving the above result. In the reasoning of proofs involving the Sobolev spaces, the exponent sometimes give surprising combinatoric properties. For example, in the process above, we used the identity that $(p-2)(n/2-s)=n(p-2)/p=n-2n/p=2(n/2-n/p)=2s$. Often the last term is very simple, not at all like what we would have thought. This seems like speed superposition formula in the special relativity, $u=\frac{v-w}{1-\frac{vw}{c^2}}$. The calculations in an introductory special relativity course are often magical. But we know that, when put in the four-vectors formulation, things become much more transparent. The essential things behind these exponent $n,s,p$ for the present is perhaps still the dimensional analysis.

The above result is called the homogeneous Sobolev embedding. Before going to state the general Sobolev embedding theorem, we need to prove another result, the interpolation theorem:

For $s, and $f\in S$, we have that $||f||_{H^{as+(1-a)t}}\leq ||f||_{H^s}^a||f||_{H^t}^{1-a}$

This result in fact says that the function $T_f:\mathbb{R}\rightarrow\mathbb{R}, s\mapsto \log(||f||_{H^s})$ is convex. The proof for this result is not difficult.The cases $a=0,1$ are trivial, so we suppose that $a\in (0,1)$. Note that(we set $r=as+(1-a)t$) $\int (1+|y|^2)^r|\hat{f}(y)|^2=\int (1+|y|^2)^{as}|\hat{f}(y)|^{2a}(1+|y|^2)^{(1-a)t}|\hat{f}(y)|^{2(1-a)}$. Now use the inequality of Holder for the exponent pair $1=\frac{1}{1/a}+\frac{1}{1/(1-a)}$, we get that $||f||_{H^r}^2\leq||(1+|y|^2)^{as}|\hat{f}(y)|^{2a}||_{L^{1/a}}||(1+|y|^2)^{(1-a)t}|\hat{f}(y)|^{2(1-a)}||_{L^{1/(1-a)}}=||f||_{H^s}^{2a}||f||_{H^t}^{2(1-a)}$, therefore proving the theorem.

The philosophy behind this theorem is very important, in that it controls the $H^r$ norms with just two extreme norms.

Now return to the Sobolev embedding theorem. We have shown that for $0, we have that for any $f\in S$, there is $||f||_{L^{p_0}}\leq c||f||_{h^s}$. Note that for $s>0$, there is $||f||_{h^s}\leq ||f||_{H^s}$. So we get that

If $0, then $H^s$ injects continuously in $L^{p_0}$.

Note that again for the case, $s>0$, if $f\in H^s$, then using the identity of Plancherel, we have that $f\in L^2$ with that $||f||_{L^2}\leq||f||_{H^s}$. So,

If $s>0$, then $H^s$ injects continuously in $L^2$.

Now using the inequality of Holder, and note that for $s>0$, there is $p_0>2$, so for any $p\in [2,p_0]$, we have that, $H^s$ injects continuously in $L^p$.

In the above, we are talking in the range that $0. What about the case $s>d/2$? In fact, this case is much easier than the first case, the case we considered above. We can show that the functions in $H^s$ are continuous functions. How to prove that a function is continuous? Here we can know some norms on the Fourier transform of a function, then to show that this function is continuous, one idea is to show that the $L^1$ norm of its Fourier transform is finite(and then utilize the continuity of Lebesgue integration).

Indeed, if $f\in H^s$, then $||\hat{f}||_{L^1}=\int (1+|y|^2)^{s/2}|\hat{f}(y)|(1+|y|^2)^{-s/2}$.Then use the Cauchy-Schwartz inequality, we have that $||\hat{f}||^2_{L^1}\leq \int |\hat{f}(y)|^2(1+|y|^2)^{2s} \int (1+|y|^2)^{-s}=||f||_{H^s}^2\int (1+|y|^2)^{-s}$. The second integral is finite because $s>d/2$, that is the key point in all this proof. Thus we showed that $||\hat{f}||_{L^1}$ is finite, therefore $f(x)$ is continuous, and tend to $0$ when $|x|\rightarrow \infty$.

Perhaps an example helps illustrate the power of these continuous embeddings. For example, we are asked to show that for $u\in H^1(\mathbb{R}^2)$, we have that $||u||_{L^4}^2\leq c||u||_{L^2}||\nabla u||_{L^2}$.

Without the knowledge of the Sobolev embeddings, this inequality is really hard to prove. But let’s see what we can do with the help of Sobolev embeddings. Note that the right hand side is in fact $c||u||_{h^0}||u||_{h^1}$. An application of the interpolation theorem for the triple $s=0,t=1,r=s/2+t/2$ gives that $||u||_{h^{1/2}}^2\leq||u||_{h^0}||u||_{h^1}$. So we have to show that $||u||_{L^4}\leq c||u||_{h^{1/2}}$. Note that $n=2,s=1/2$, so $p_0=4$, so the embedding theorem reads $||u||_{L^4}\leq c||u||_{h^{1/2}}$, and we are done.

There are others questions to be addressed, for example, are these embeddings compact?In general, the answer is negative. We will see that in fact, the topology defined by $L^p$ norms are, in some sense, too fine. We need another topology more coarse than this one, but which is still useful. And under this topology, we will have the compacity of these embeddings.