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.
Best Answer
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 $$ \forall x \in X: \quad |\Lambda(x)| \leq \sum_{k=1}^{n} \lambda_{k} \| x \|_{i_{k}}. $$
Addendum
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. ////