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Let $X$ be an F-space (space with topology given by complete, translation-invariant metric). Let $E\subset X^*$, such that $$\rho(f):=\sup_{\varphi\in E} |\varphi(f)|<\infty.$$

Show that $\rho$ is a continuous seminorm.

The fact that $\rho$ is a seminorm is trtivial, but I don't have any idea how to prove the continuity. We have a metric $d$ from $X$. If there would be formula:$$d(x,y)=\rho(x-y)$$

we'd get contiunuity form: $$|\rho(f)-\rho(g)|=||\rho(f)|-|\rho(g)||\le |\rho(f-g)|<\delta,$$ so for $\varepsilon=\delta$ we'd have continuity. But we don't use anywhere the fact how the seminorm is defined (by supremum), so I don't think that this lead to solution (I've just written my observations).

Any hint to this problem?

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    What versions of the Banach-Steinhaus theorem do you know? – 2017-02-17
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    $X$ - F-space, $Y$-TVS, $\Pi=\left\{ T_{\omega}\right\}_{\omega\in\Omega}$, where $T_\omega :X\rightarrow Y$ - continuous and linear for all $\omega\in\Omega$. If $\Pi$ is pointwise bounded, then $\Pi$ is equicontinuous. – 2017-02-17
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    Good. Now let $Y = \mathbb{C}$ (or $\mathbb{R}$, whatever is your scalar field) and $\Pi = E$. Can you go on from there? – 2017-02-17
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    okay, so we want to show that $\Pi=E$ is pointwise bounded. Take $f\in X$. Then $\rho(f)=\sup_{\varphi} |\varphi(f)|$ and it is finite. Since supremum over functionals is bounded, then each functional is. So we have equcontinuity of $E$ and every $\varphi\in E$ is continuous. Finally since supremum preserves continuity, we get the statment. Is this okay? – 2017-02-17
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    The supremum doesn't (necessarily) preserve continuity. The supremum of a family of continuous functions is lower semicontinuous, but not always continuous. You need to use the equicontinuity. – 2017-02-17
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    From the equicontinuity we know that $\forall\varepsilon>0 \ \exists\delta>0 \ \forall \ f,g\in X \forall\varphi\in E$ if $(\Vert f-g\Vert<\delta)\implies (\Vert\varphi(f)-\varphi(g)\Vert<\varepsilon).$ So we get that $\varepsilon>\Vert \sup_{\varphi}|\varphi(f)-\varphi(g) |\Vert \ge \Vert \sup_{\varphi}|\varphi(f)| -\sup_{\varphi}|\varphi(g)|\Vert$ and this gives us required continuity. Ok? – 2017-02-17
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    More or less. If $E$ isn't finite, you can only assert that $\lvert \rho(f) - \rho(g)\rvert \leqslant \varepsilon$, not $< \varepsilon$. A more idiomatic argument is that a seminorm is continuous if and only if its unit ball is a neighbourhood of $0$. Which in this case is equivalent to the equicontinuity of $E$. – 2017-02-17

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