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Let $X$ be a Fréchet space whose topology is given by the separating family of seminorms $p_1 \leq p_2 \leq p_3 \leq \cdots.$ I want to prove for a linear functional $f$ on $X$ the following characterization of continuity (see the Wikipedia article on locally convex topological vector spaces, or the related problem in the first chapter of Rudin's Functional Analysis):

$\quad \qquad \qquad$ $f$ is continuous $\iff$ $\exists M > 0 \quad \exists n \in \mathbb{N} \quad \forall x \in X \quad |f(x)| \leq Mp_n(x).$

I can see why the backward implication is true. For the converse, one naturally mimicks the proof that continuity implies boundedness in normed spaces. Namely, by hypothesis, there exists $n \in \mathbb{N}$ and $\varepsilon > 0$ such that $|f(x)| \leq 1$ whenever $p_n(x) \leq \varepsilon$. Hence, for any $x \in X$,

$$|f(x)| = \left|\frac{p_n(x)}{\varepsilon} f\bigg(\frac{\varepsilon}{p_n(x)} x\bigg)\right| = \frac{1}{\varepsilon}p_n(x) \left|f\bigg(\frac{\varepsilon}{p_n(x)} x\bigg)\right| \leq \frac{1}{\varepsilon}p_n(x)$$

provided that $p_n(x) \neq 0$. But what about those $x$ with $p_n(x) = 0$? If $p_n$ were a norm, then the only such $x$ would be $0$, for which the desired inequality trivially holds since $f(0) = 0.$ However, in general, $p_k$ need not be a norm for any $k$ even though the collection $\{p_k\}$ is separating. How does one complete the proof then?

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up vote 1 down vote accepted

If $p_n(x)=0$ then $p_n(mx)=0$ for all $m\in\mathbb N$. Hence $|f(mx)|\le 1$ which implies $|f(x)|\le 1/m$ for all $m$ and thus $f(x)=0$.

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Thanks to Prof. Jochen Wengenroth's answer, I could prove the following general characterization: A linear function $f : X \to Y$ between two locally convex Hausdorff topological vector spaces is continuous if and only if for each continuous seminorm $q$ on $Y$ there exists a continuous seminorm $p$ on $X$ such that $q \circ f \lesssim p$. –  Murat Güngör Sep 1 '13 at 8:22
    
Of course one can replace $\lesssim$ by $\leq$ above. –  Murat Güngör Sep 2 '13 at 6:21
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