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$X$ is topological vector space whose topology is defined by a countable family of separating semi-norms $\|\cdot\|_N$, $N\geq 0$.

Suppose $\Lambda:X\to \mathbb{R}$ is a continuous linear functional.

Question: Does it follow that there exists $N \geq 0$ and a constant $C <\infty$ such that $$|\Lambda \phi| \leq C\|\phi\|_N \text{, for all } \phi \in X \text{ ?}$$

I find it a bit strange if the answer is yes, but if $X=C^\infty(K)$ is the space of smooth functions of compact support $K$, and $\Lambda$ is a distribution, then the answer is yes. I would like to understand why. I'd appreciate any help.

Edit1: The answers suggest the statement is true, but I would like to get some intuition for why it should be so.

Edit2:I am still looking for some more clarification or intuition on this problem. Maybe one of the experts here could have a look at it? The answer below gives the correct statement, but I don't see why it is true. I would really like to know why, and I think it should be an easy statement to prove for folks in functional analysis.

Thank you.

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What does the basis of open neighborhood of $0$ look like? (we can assume that $\rho_n\leqslant \rho_{n+1}$ for all $n$). – Davide Giraudo Dec 11 '12 at 12:35
@DavideGiraudo, sets $V_N=\{\phi : \|\phi\|_N<1/N\}$ form a local basis. – Cantor Dec 11 '12 at 13:01
Intuitively, $\Lambda$ being continuous means if two $\phi$'s are close w.r.t all semi-norms then their image is close. It is strange to me if one can conclude that if two $\phi$'s are close in the $N$-norm their image is close (this is what the inequality implies). I'm looking for some intuition on this. – Cantor Dec 11 '12 at 13:09
up vote 3 down vote accepted

The precise answer is as follows (according to Theorem 3.1 (f) in Conway's A Course in Functional Analysis):

Let $ X $ be a locally convex topological vector space. Suppose that $ \mathcal{P} = \{ \| \cdot \|_{i} \}_{i \in I} $ is a family of seminorms that defines the topology on $ X $. Then $ \Lambda $ is a continuous linear functional on $ X $ if and only if there exist $ i_{1},\ldots,i_{n} \in I $ and positive real numbers $ \lambda_{1},\ldots,\lambda_{n} $ such that $$ \forall x \in X: \quad |\Lambda(x)| \leq \sum_{k=1}^{n} \lambda_{k} \| x \|_{i_{k}}. $$


As it seems that Conway does not provide a proof of the quoted theorem, I shall provide my own proof, as the OP has requested to see one.

There are many definitions of continuity at a point, so I shall pick the one that best suits my needs for the proof.

Definition Let $ X $ and $ Y $ be topological spaces. A function $ f: X \rightarrow Y $ is said to be continuous at $ x $ if and only if for all neighborhoods $ V $ of $ f(x) $, there exists a neighborhood $ U $ of $ x $ such that $ f[U] \subseteq V $.

We shall also use the fact that the continuity of $ \Lambda: X \rightarrow \mathbb{R} $ is equivalent to its continuity at the point $ 0_{X} $.

Let us first establish the ($ \Rightarrow $)-direction of the theorem. As (i) $ \Lambda(0_{X}) = 0 $ and (ii) $ \overline{D}(0;1) $ is a neighborhood of $ 0 $, by the given definition, there exists a neighborhood $ U $ of $ 0_{X} $ such that $ \Lambda[U] \subseteq \overline{D}(0;1) $. Without loss of generality, we may assume that $ U $ is a basic open neighborhood of the form $$ \{ x \in X \,|\, (\forall k \in \{ 1,\ldots,n \})(\| x \|_{i_{k}} < 2 \epsilon) \}, $$ where $ \epsilon > 0 $. Now, for each $ x \in X $, define $ \displaystyle M_{x} \stackrel{\text{def}}{=} \max_{1 \leq k \leq n} \| x \|_{i_{k}} $.

Claim: $ |\Lambda(x)| \leq \dfrac{M_{x}}{\epsilon} $ for all $ x \in X $.

Proof of the claim Let $ x \in X $. We shall consider two cases: (i) $ M_{x} = 0 $ and (ii) $ M_{x} > 0 $.

In Case (i), it is necessarily true that $ \Lambda(x) = 0 $. Suppose otherwise, i.e., $ |\Lambda(x)| = r > 0 $. Then for sufficiently large $ N \in \mathbb{N} $, we have $ |\Lambda(N \cdot x)| = Nr > 1 $. However, $ \| N \cdot x \|_{i_{k}} = N \| x \|_{i_{k}} = 0 $ for all $ k \in \{ 1,\ldots,n \} $, so $ N \cdot x \in U $. We have thus contradicted the earlier statement that $ \Lambda[U] \subseteq \overline{D}(0;1) $. Therefore, we must have $ \Lambda(x) = 0 $, in which case, the inequality $ |\Lambda(x)| \leq \dfrac{M_{x}}{\epsilon} $ automatically holds.

In Case (ii), we have $ \dfrac{\epsilon}{M_{x}} \cdot x \in U $, so $ \left| \Lambda \left( \dfrac{\epsilon}{M_{x}} \cdot x \right) \right| \leq 1 $. It follows immediately that $ |\Lambda(x)| \leq \dfrac{M_{x}}{\epsilon} $.

The claim is now established. //

Using the claim, we obtain $$ |\Lambda(x)| \leq \sum_{k=1}^{n} \frac{1}{\epsilon} \| x \|_{i_{k}}, $$ since $$ \sum_{k=1}^{n} \frac{1}{\epsilon} \| x \|_{i_{k}} = \frac{1}{\epsilon} \sum_{k=1}^{n} \| x \|_{i_{k}} \geq \frac{1}{\epsilon} \cdot M_{x}. $$

For the ($ \Leftarrow $)-direction, the argument is much easier. For any $ \epsilon > 0 $, if you take $ U $ to be the open set of $ X $ defined by $$ \left\{ x \in X \,\Bigg|\, (\forall k \in \{ 1,\ldots,n \}) \left( \| x \|_{i_{k}} < \frac{\epsilon}{n \cdot \max(\lambda_{1},\ldots,\lambda_{n})} \right) \right\}, $$ then you immediately obtain $ \Lambda[U] \subseteq D(0;\epsilon) $. Therefore, as $ \epsilon $ is arbitrary, $ \Lambda $ is continuous at $ 0_{X} $, hence continuous everywhere.

The proof of the theorem is now complete. ////

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Thank you for your answer. Do you have some intuition for why the statement is true? – Cantor Dec 11 '12 at 15:39
Hi Haskell, I finally got hold of the book, but I can't find a proof. Do you know of another source? I like to understand why controlling finitely many seminorms is enough. – Cantor Dec 12 '12 at 10:58
Hi Cantor. I have just provided a proof of the result. – Haskell Curry Dec 19 '12 at 10:19
thank you for your addendum and sorry for the long delay in accepting your answer. I was away for a while. – Cantor Jan 20 '13 at 10:36

From Proposition 7.7 and its corollary in Treves' book, we learn that

A linear form $f$ on a locally convex topological vector space $E$ is continuous if and only if there is a continuous seminorm $p$ on $E$ such that, for all $x \in E$, $$|f(x)| \leq p(x).$$

Your case is a very particular case, in which the continuous seminorms are simply scalar multiples of $\| \cdot \|_N$. By the way, in the same page of the book, your case is discussed.

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