Proving this set is compact in $L^2(\Omega)$

functional-analysissobolev-spacessolution-verificationweak-derivatives

Let $\Omega = [0,1]^d \subset \mathbb{R}^d, d \ge 3$, and for any $x
\in \Omega$ define $$Tf(x) \equiv \int_{\Omega} e^{-
\frac{(y-x)^2}{2}}f(y) \lambda (dy) $$ for any $f : \Omega \to
\mathbb{R}$ for which the RHS is well defined. Using Sobolev
inequality or otherwise, show that the following set $$ \{Tf :
\|f\|_{L^2} \le 2 \}$$ is compact in $L^2(\Omega)$.

My questions:

  1. Is my attempt below correct?
  2. How would we even use the Sobolev inequality here? I thought that the only thing we need is the Rellich-Kondrachov theorem.
  3. Is there an easy way to see the above set is closed?

My attempt:

I proved that $Tf \in L^2(\Omega)$ and also $\|Tf\|_{L^2(\Omega)} \le C$ for every $f$ in the set above, but now I need to find the weak derivative $\partial_{x_i} Tf$ and show it is in $L^2(\Omega)$ as well. But I am not sure how to find $\partial_{x_i} Tf$. My attempt with this:

Since our function is smooth then the weak derivative is equal to the ordinary partial derivative. So we calculate:

$$\partial_{x_i} Tf = \int_{\Omega} \partial_{x_i}e^{- \frac{(y-x)^2}{2}}f(y) \lambda (dy) = \int_{\Omega} (y-x)e^{- \frac{(y-x)^2}{2}}f(y) \lambda (dy)$$
And now:
$$|\partial_{x_i} Tf(x)| \le \int_{\Omega} |2| \cdot 1 \cdot |f(y)| \lambda (dy) \le 2 \cdot \|f\|_2 $$ so
$$\|\partial_{x_i} Tf\|_{L^2(\Omega)} \le 2\lambda(\Omega)^{\frac{1}{2}} \|f\|_2 \le 4\lambda(\Omega)^{\frac{1}{2}}$$ Hence any weak derivative of first order is in $L^2(\Omega)$ and in fact since the above bound works for any $f$ with $\|f\|_2 \le 2$ then we have that our set is bounded in $H^{1}(\Omega)$. By Rellich-Kondrachov theorem this set is precompact in $L^2(\Omega)$ but since it is closed (I am not sure how to prove this actually) then it is compact in $L^2(\Omega)$.

Best Answer

You do not need Rellich-Kondrachov's theorem. You only need Ascoli-Arzelà theorem. As you write in your attempt: $$\partial_{x_i} Tf (x) = \int_{\Omega} \partial_{x_i}e^{- \frac{(y-x)^2}{2}}f(y) \lambda (dy) = \int_{\Omega} (y-x)e^{- \frac{(y-x)^2}{2}}f(y) \lambda (dy)$$ and thus for all $i \in \{1,\dots,d\}$ and for all $x \in \Omega$ we have that: $$|\partial_{x_i} Tf(x)| \le \int_{\Omega} |2| \cdot 1 \cdot |f(y)| \lambda (dy) \le 2 \cdot \|f\|_2$$ end then $\|\nabla Tf\|_{L^{\infty}(\Omega)} \leq 2q \|f\|_2 \leq 4d$ for all $f$ such that $\|f\|_2 \leq 2$. The family $\{Tf : \|f\|_{L^2} \le 2 \}$ is uniformly $4d$-Lipschitz continuous. What follows from Ascoli-Arzelà's theorem is that $\{Tf : \|f\|_{L^2} \le 2 \}$ is relatively compact. Hope this will help you.

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