In class the other day my professor made the following claim about the Schwartz class $\mathcal S$:

Let $u: \mathcal S \to \mathbb C$ be linear. $u$ is continuous iff there exist $C,N>0$ such that for all $\varphi \in \mathcal S$, $$|u(\varphi)| \leq C \sup_{|\alpha|,|\beta| \leq N} ||x^\alpha D^\beta \varphi||_{L^\infty}$$ where $\alpha,\beta$ denote multiindices.

(so $||x^\alpha D^\beta \varphi||_{L^\infty}$ is just the $(\alpha,\beta)$th seminorm of $\mathcal S$).

He claims this follows immediately from uniform boundedness for Frechet spaces. I'm largely unfamiliar with this area of functional analysis, so it isn't at all clear to me. In particular, I don't see why we'd only need to use finitely many seminorms.

So, is the claim

Let $F$ be a Frechet space with seminorms $||\cdot||_j$. A linear map $u: F \to \mathbb C$ is continuous iff there exist $C,N>0$ such that for all $f \in F$, $$|u(f)| \leq C\sup_{j \leq N} ||f||_j$$

valid, or is this just something true for the Schwartz class? Does it follow from Banach-Steinhaus?


1 Answer 1


The result is true in any locally convex space $F$. It is a consequence of the fact that the topology is given by a family of seminorms.

Since $u$ is continuous at $0$, the set $u^{-1}(B_1(0))$ is open (preimage of the unit disk). So there exists a neighbourhood $V$ of zero, of the form $V=\{f:\ \|f(x)\|_j<\varepsilon,\ j=1,\ldots,N\}$, with $V\subset u^{-1}(B_1(0))$. Given any $f\in F$, we have $$ \frac{\varepsilon f}{2\sum_{j=1}^N\|f\|_j}\in V, $$ so $$ \left|u\left(\frac{\varepsilon f}{2\sum_{j=1}^N\|f\|_j}\right)\right|\leq1, $$ which we may write as $$ |u(f)|\leq \sum_{j=1}^N\frac2\varepsilon\,\|f\|_j\leq\frac{2N}\varepsilon\,\sup_{j\leq N}\|f\|_j. $$

There is a small exception to the above, which is the case in which $\|f\|_j=0$ for $j=1,\ldots,N$. But in that case $mf\in V$ for all $m\in\mathbb N$, so $|u(mf)|\leq1$ and so $|u(f)|\leq1/m$, forcing $u(f)=0$.

  • $\begingroup$ Thank you, this is exactly what I needed! $\endgroup$ Jan 24, 2019 at 0:36

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