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Theorem 1.3.2

A linear form $u$ on $C_c^\infty(X)$ is a distribution if and only if $\lim_{j \to \infty} \langle u, \phi_j \rangle = 0$ for every sequence $\phi_j$ which converges to zero in $C_c^\infty(X)$ as $j \to \infty$.

Proof:

  1. Necessity: If $u$ is a distribution, then for any sequence $\{\phi_j\}$ converging to zero in $C_c^\infty(X)$, the pairing $\langle u, \phi_j \rangle$ converges to zero. Hence, $\lim_{j \to \infty} \langle u, \phi_j \rangle = 0$.

  2. Sufficiency: Assume $u$ is sequentially continuous. Suppose $u$ is not a distribution. Then there exists a compact set $K \subset X$ such that: $$ |\langle u, \phi \rangle| > N \sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi| $$ for $\phi$ \in $C_c^{\infty}(K)$. For each $N$, choose $\phi_N$ such that: $$ |\langle u, \phi_N \rangle| > N \sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N| $$ Define $\psi_N(x) = \frac{\phi_N(x)}{\sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N|}$. Notice $\psi_N$ is supported in $K$, and: $$ |\partial^\beta \psi_N| \leq \frac{\sup |\partial^\beta \phi_N|}{\sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N|} \leq \frac{1}{N} $$ for $|\beta| \leq N$. Thus, $\psi_N \to 0$ in $C_c^\infty(X)$ as $N \to \infty$.

    However, from: $$ |\langle u, \psi_N \rangle| \geq \frac{|\langle u, \phi_N \rangle|}{\sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N|} > \frac{N \sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N|}{\sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N|} = N $$ we get a contradiction. Hence, $u$ must be a distribution if it is sequentially continuous.

My Question:

In the proof of Theorem 1.3.2 from "Introduction to the Theory of Distributions" by F. G. Friedlander and M. Joshi, how is the bound: $$ |\partial^\beta \psi_N| \leq \frac{1}{N} $$ derived? I understand that $\psi_N$ is defined as: $$ \psi_N(x) = \frac{\phi_N(x)}{\sum_{|\alpha| \leq N} \sup |\partial^\alpha \phi_N|} $$ but I'm not sure how this leads to the desired bound.

Any insights or explanations would be greatly appreciated! If, for example, I could manage to replace the denominator with $N \cdot \sup \vert \partial ^{\beta} \phi_N \vert $ I would have obtained the estimate I need

Matteo Aldovardi
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2 Answers2

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The space $C_c^\infty(X)$ of test functions is the colimit in the category of locally convex spaces of the metrizable spaces $\mathscr D(K)$ of smooth functions with support in a finxed compact subset $K\subseteq X$. For metrizable spaces as a source, continuity is the same as sequential continuity (this is shown in the proof for the particular situation of $\mathscr D(K)$). Also, a linear map $T:E\to F$ on a colimit $E=$colim $E_\alpha$ is continuous if and only all restrictions $T|_{E_\alpha}$ are continuous. Combining these two facts, you get without any calculations that a linear map $T:C_c^\infty(X)\to Y$ (with values in any locally convex space) is continuous if and only if it is sequentially continuous.

Jochen
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  • thanks for the answer, I’m not familiar with the concept of a colimit, but I can look up its definition. I was hoping to complete the calculation proof, as it would make it easier for me to adapt it to prove another theorem: that a sequentially continuous form on $C_c^m$ always restricts to a distribution of order $m$, and vice versa. – Matteo Aldovardi Oct 23 '24 at 04:34
  • @Jochen: Since you know TVS theory very well, I meant to ask you what is the status of being sequential or not for $\mathscr{D}$. For other spaces of distributions, I have seen contradictory accounts in the literature, see my comments in https://math.stackexchange.com/questions/4961702/exact-meaning-of-a-bounded-set-in-the-space-of-tempered-distributions-mathcal The related space $\oplus_{\mathbb{N}}\mathbb{R}$ is sequential, and $\mathscr{D}$ is isomorphic to $\oplus_{\mathbb{N}}s$ where $s$ is the space of rapidly decaying sequences. – Abdelmalek Abdesselam Oct 26 '24 at 18:06
  • I doubt that this is the right place to discuss this question but I strongly believe that there are sequentially closed linear subspaces of $\mathscr D$ which are not closed. – Jochen Oct 26 '24 at 21:44
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Suppose that there exists a compact set $ K $ such that the ratio

$$ \frac{|\langle u, \varphi \rangle|}{\sum_{|\alpha| \leq m} \sup_{x \in \Omega} |\partial^\alpha \varphi(x)|} $$

is unbounded as $ \varphi $ ranges over $ \mathcal{D}_K(\Omega) $. Use this to select $ \varphi_m $ such that $ \langle u, \varphi_m \rangle = 1 $. For the definition of $ K $, for every $ m \in \mathbb{N} $, we have

$$ \frac{\langle u, \varphi_m \rangle}{\sum_{|\alpha| \leq m} \sup_{x \in \Omega} |\partial^\alpha \varphi(x)|} > m, $$

or equivalently

$$ \sum_{|\alpha| \leq m} \sup_{x \in \Omega} |\partial^\alpha \varphi(x)| < \frac{1}{m}. $$

Now, we have a sequence of test functions $ (\varphi_m) $ that converges to $ 0 $ in the topology of $ \mathcal{D}(\Omega) $, but the values $ \langle u, \varphi_m \rangle $ do not tend to $ 0 $.

Credits:

I reposted an answer from 'user357151' to another question titled 'A characterization of being sequentially continuous' by 'DonQuixote'. I did this so that anyone reading knows that I found an answer. Additionally, I didn't want to remove my post because I believe it is a valuable contribution. The title is explanatory, and the post contains two beautiful answers: one very elegant and abstract, and the other simpler and more direct.

Matteo Aldovardi
  • 328