# Boundedness of a continuous linear functional on a topological vector space

Suppose:

$X$ 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

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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