In $C(\mathbb{R})$ space we define two families of seminorms:

$p_x(f)=|f(x)|$ where $ x\in\mathbb{Q}$ and $q_x(f)=|f(e^x)|$ where $ x \in \mathbb{R}$

I have to check if above families induce locally convex topological vector space in $C(\mathbb{R})$. I have also to check the continuity of functional $C(\mathbb{R}) \ni f \rightarrow f(\sqrt{2}) \in \mathbb{R}$ in any of those locally convex topologies.

I have no idea how to do it.


1 Answer 1


In view of the other question, we know that any family of seminorms induces a locally convex topology, but not necessarily a Hausdorff topology. So to check whether the induced topologies are Hausdorff, we need to check whether the families are separating.

A family $\mathscr{S}$ of seminorms is separating if and only if

$$\bigl(\forall p\in \mathscr{S}\bigr)\bigl(p(f) = 0\bigr) \implies f = 0.$$

That may be used as the definition of "separating", or be proved as a characterisation starting from a different(ly worded) definition.

Looking at $\mathscr{P} = \{ p_x : x \in \mathbb{Q}\}$, if $p_x(f) = 0$ for all $x\in\mathbb{Q}$, then $f$ vanishes at all rational points. By continuity, $f \equiv 0$, since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus $\mathscr{P}$ is separating and induces a Hausdorff locally convex topology on $C(\mathbb{R})$.

For $\mathscr{Q} = \{ q_x : x \in \mathbb{R}\}$, it may be useful to describe the seminorms a little differently, $q_x = r_{e^x}$, where $r_y(f) = \lvert f(y)\rvert$, and then $\mathscr{Q} = \{ r_y : y \in \mathbb{R}, y > 0\}$. So the family $\mathscr{Q}$ doesn't monitor what happens on the negative ray at all, and is not a separating family of seminorms. The topology induced by $\mathscr{Q}$ on $C(\mathbb{R})$ is not Hausdorff.

A linear functional $\lambda$ on a topological vector space $E$ is continuous if and only if

$$\{ x \in E : \lvert \lambda(x)\rvert \leqslant 1\}$$

is a neighbourhood of $0$ in $E$. If the topology of $E$ is induced by a family $\mathscr{S}$ of seminorms, that means that there are finitely many $p_1,\dotsc,p_k \in \mathscr{S}$ and an $\varepsilon > 0$ such that

$$\bigcap_{\kappa = 1}^k p_\kappa^{-1}([0,\varepsilon]) \subset \{ x \in E : \lvert \lambda(x)\rvert \leqslant 1\}$$

(because the sets of the form on the left form a neighbourhood basis of $0$ in $E$), or equivalently,

$$\lvert\lambda(x)\rvert \leqslant \frac{1}{\varepsilon} \max \{ p_\kappa(x) : 1 \leqslant \kappa \leqslant k\}$$

for all $x\in E$.

In our particular case, for the linear functional $\eta_{\sqrt{2}} \colon f \mapsto f(\sqrt{2})$, we trivially have

$$\lvert \eta_{\sqrt{2}}(f)\rvert = \lvert f(\sqrt{2})\rvert = r_{\sqrt{2}}(f) = q_{\frac{1}{2}\log 2}(f),$$

so for the family $\mathscr{Q}$, we can take $k = 1$ and $\varepsilon = 1$. Thus $\eta_{\sqrt{2}}$ is continuous with respect to the topology induced by $\mathscr{Q}$ (which, let's repeat it, is not a Hausdorff topology).

For the topology induced by $\mathscr{P}$, the fact that $\sqrt{2}$ is irrational is important. Evaluation in a rational $x$ would be easily seen to be continuous in the same way as above. One may now (correctly) suspect that evaluation in an irrational $x$ is not continuous with respect to the topology induced by $\mathscr{P}$, and then needs to show it. Thus one picks an arbitrary finite subset $I$ of $\mathbb{Q}$, and has to show that "being small on $I$ does not imply being small at $x$". Typically, one considers functions vanishing at all points of $I$, and needs to show that such a function need not vanish at $x$.

  • $\begingroup$ Hi Daniel,I don't know how to build the equivalence between $\bigcap_{\kappa = 1}^k p_\kappa^{-1}([0,\varepsilon]) \subset \{ x \in E : \lvert \lambda(x)\rvert \leqslant 1\}$ and $\lvert\lambda(x)\rvert \leqslant \frac{1}{\varepsilon} \max \{ p_\kappa(x) : 1 \leqslant \kappa \leqslant k\}$ Can you provide a bit hint $\endgroup$
    – yi li
    Nov 5, 2020 at 8:12
  • 1
    $\begingroup$ @yi_li Pick $x \in E$, let $m = \max \{ p_{\kappa}(x) : 1 \leqslant \kappa \leqslant k\}$. If $m = 0$, then $t \cdot x$ belongs to the intersection for every $t \in \mathbb{R}$, hence $\lvert \lambda(tx)\rvert \leqslant 1$ for all $t$, and that implies $\lambda(x) = 0$, hence $\lvert \lambda(x)\rvert \leqslant \frac{1}{\varepsilon} m$. And if $m > 0$, then $\frac{\varepsilon}{m}\cdot x$ lies in the intersection, hence $\lvert\lambda(x)\rvert = \frac{m}{\varepsilon}\bigl\lvert\lambda\bigl(\frac{\varepsilon}{m}x\bigr)\bigr\rvert \leqslant \frac{1}{\varepsilon}m$. The converse is direct. $\endgroup$ Nov 5, 2020 at 11:56
  • $\begingroup$ nice proof,thanks $\endgroup$
    – yi li
    Nov 5, 2020 at 12:09

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