How Does the Compact Support Theorem Relate to Distribution Inequality?

[evinda]

Hello! (Wave)

Theorem: Let $u \in D'(\Omega)$ and $K \subset \Omega$, $K$ compact

$$\exists \lambda \in \mathbb{N} \text{ and } c \geq 0 \text{ such that } \\ |\langle u, \phi \rangle| \leq c \sum_{|a| \leq \lambda} ||\partial^{a} \phi||_{L^{\infty}}, \forall \phi \in C_C^{\infty}(K)$$

Proof

Suppose that it doesn't hold then for all $\lambda, c$ there will be a test function $\phi_{\lambda}$ such that

$1=|\langle u, \phi_{\lambda} \rangle|> \lambda \sum_{|a| \leq \lambda} ||\partial^{a} \phi_{\lambda}||_{L^{\infty}(\Omega)}$

We divide by $\lambda$, so:

$$||\phi_{\lambda}||_{\infty} \leq \sum_{|a| \leq \lambda} ||\partial^{a} \phi_{\lambda}||_{L^{\infty}(\Omega)} < \frac{1}{\lambda}$$

We send $\lambda \to +\infty$. So $\partial^{a} \phi_{\lambda} \to 0 \Rightarrow \langle u, \phi_{\lambda} \rangle \to 0$ from $(\star)$. And so we have a contradiction.

$u$: distribution
$(\star)$: $u(\phi_j) \to u(\phi)$ for each $\phi \in C_{C}^{\infty}(\Omega)$ for each sequence $\{ \phi_j\}$ with $\partial^a{\phi_j} \to \partial^a \phi$.First of all, why can we consider that there is a test function $\phi_{\lambda}$ such that $|\langle u, \phi_{\lambda} \rangle|=1$ ?

Then, how do we get that $||\phi_{\lambda}||_{\infty} \leq \sum_{|a| \leq \lambda} ||\partial^a \phi_{\lambda}||_{L^{\infty}(\Omega)}$ ?

Also how do we deduce from $(\star)$ that $\partial^a{\phi_{\lambda}} \to 0$ imples that $\langle u, \phi_{\lambda} \rangle \to 0$ ?

 

[jvicens]

Hello! (Wave)
Thank you for your post. I would like to address your questions and clarify the proof for you.

Firstly, in order to prove the theorem, we need to assume that there is no test function $\phi_{\lambda}$ such that $|\langle u, \phi_{\lambda} \rangle|=1$. This is because if such a function exists, then the inequality in the theorem would not hold for that particular $\lambda$ and $c$. Therefore, we assume that there is no such function, and proceed with the proof.

Secondly, we get the inequality $||\phi_{\lambda}||_{\infty} \leq \sum_{|a| \leq \lambda} ||\partial^a \phi_{\lambda}||_{L^{\infty}(\Omega)}$ by dividing both sides of the given inequality by $\lambda$, and then taking the supremum over all possible test functions $\phi_{\lambda}$. This is because the supremum of a sum is always less than or equal to the sum of the suprema.

Lastly, $(\star)$ is the definition of a distribution. It states that for any sequence of test functions $\phi_j$ that converges to a test function $\phi$, the corresponding sequence of distributions $u(\phi_j)$ converges to $u(\phi)$. In this case, we have a sequence of test functions $\phi_{\lambda}$ with the property that $\partial^a \phi_{\lambda} \to 0$. Therefore, by $(\star)$, we have that $u(\phi_{\lambda}) \to u(0) = 0$, which implies that $\langle u, \phi_{\lambda} \rangle \to 0$. This contradicts our assumption that $|\langle u, \phi_{\lambda} \rangle| = 1$, and thus proves the theorem.

I hope this helps clarify the proof for you. Let me know if you have any further questions or concerns.