I've recently encountered Fréchet spaces for the first time, and I was wondering if it remains true that a functional $T \rightarrow \mathbb{R}$ with $S$ a Fréchet space, is continuous if for all $\phi \in S$ we have

$$|T\phi| \leq c\|\phi\|_S$$

where $\|\cdot\|_S$ denotes a semi-norm on the $S$ space. The reason I'm asking is that I have only ever seen the above result formulated for Banach spaces, and I was curious if it extends naturally?

  • $\begingroup$ @kimchilover Let's say that $S$ is the Schwarz space, and the norm is one of the semi-norms on that space. $\endgroup$ Dec 9, 2017 at 17:06
  • $\begingroup$ The way you had phrased it made it seem like you didn't know the difference between a B space and an F space. I'll remove my earlier comments, and then this one later. $\endgroup$ Dec 9, 2017 at 17:10

1 Answer 1


This is true, with the caveat that the semi-norm need not be one from whatever collection of seminorms was originally used to define the topology on $S$. Also, this is true for any locally convex space; it need not be a Fréchet space (which requires in addition a compatible complete translation-invariant metric).

Claim: In a locally convex space $X$ whose topology is defined by seminorms $\{p_\alpha\}$, a linear functional $f:X\to\mathbb{R}$ is continuous if and only if it is majorized by a finite combination of seminorms: $$|f(x)|\le M\max (p_1(x),\dots, p_n(x))$$ Proof: The set $U=\{x : |f(x)|<1\}$ is an open set containing $0$. By the definition of LCS topology, $U$ it contains a neighborhood of $0$ the form $\{x \colon \max_{k=1, \dots, n} p_k(x)<\epsilon\}$. The conclusion with $M=1/\epsilon$. $\quad \Box$

It remains to left $p(x) = \max (p_1(x),\dots, p_n(x))$ and you get the desired seminorm.


This is an example to show that one can't always pick just one of "original" seminorms. Let $S$ be the space $C(\mathbb{R})$ of all continuous functions on $\mathbb{R}$ equipped with seminorms $$ p_n(f) = \sup_{[n, n+1]}|f|,\qquad n\in\mathbb{Z} $$ These seminorms induce the topology of uniform convergence on compact sets. The linear functional $T(f) = \int_{0}^3 f(x)\,dx$ is continuous on $S$, but there is no single seminorm $p_n$ that dominates $|T(f)|$; one has to use $\max(p_0, p_1, p_2)$.


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