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.
 A: 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. ////
A: Cantor, I agree that it's weird to be able to bound the functional using only 1 (or finitely many) seminorm(s). This is how I reconciled the issue. For reference, I'm looking at Rudin's Functional Analysis Theorem 1.37 and exercise 1.8(b).
The key point is that although these are seminorms, and thus may "fail to see" some non-zero vectors as non-zero, they still see "most vectors" according to the topology. That is, in the topology generated by a collection of seminorms, $\mathcal{P}$, a basic open set is a finite intersection of sets of the form 
$$
V(p,n) = \{ x : p(x) < 1/n\}, \quad \text{ where $p\in \mathcal{P}$}
$$
This topology is counter intuitive because a basic open set may look unbounded if you're universe is a set that you have other ideas about. Here's a quick example to demonstrate the idea.
Consider the seminorms $p_i : \mathbb{R}^2 \to \mathbb{R}$ given by $p_i(x_1,x_2) = |x_i|$. 
The set $V(p_1,1)$ is the infinite vertical strip and yet it's a basic open set. This looks nothing like the typical open ball in $\mathbb{R}^2$ with the usual topology. In this example, we are lucky because the seminorm topology coincides with the euclidean metric topology: 
$$
V(p_1, 1/2) \cap V(p_2,1/2) \subset \{x : ||x||_2 <1\}
$$
But what about extending this example to $\mathbb{R}^\infty$? Suddenly basic open sets will always have infinite rays shooting out. But that's okay because if someone says, "$\Lambda$ is contiuous w.r.t. this topology" it's already taken into account.
So, in the first answer to your question, when Haskell Curry says, $\Lambda[U] \subseteq \overline{D(0,1)}$ and later that
$$
\bigcap_{i=1}^k V(p_i, n_i) \subseteq U
$$
What's being implicitly stated is that we don't need to worry about the possibly huge sets $p_i^{-1}(\{0\})$ because the topology doesn't worry about it and because the continuity doesn't worry about it.
A: 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.
