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When having a Fréchet space $X$ whose topology is generated by the semi norms $p_1,\ldots,p_n,...$ we can define the metric $$d(x,y) = \sum_{i\geq 1}\frac{1}{2^i}\frac{p_i(x-y)}{1+p_i(x-y)}.$$

It's well know from the theory of locally convex topological vector spaces that a linear functional $u:X\to \mathbb R$ is continuous if and only if there exist $C>0$ and $N \in \mathbb N$ such that $$|u(x)|\leq C(p_1(x)+\cdots+p_N(x)),\forall x \in X.$$

My question is: Is every continuous linear functional uniformly continuous with respect to the metric $d$?

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  • $\begingroup$ Why wouldn't it be,since $d$ induces the same topology as that induced by the seminorms, (through the usual subbase construction). or maybe I am misunderstanding your question. $\endgroup$ – Matematleta Aug 27 '16 at 19:54
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You can deduce this immediately from the translation invariance of $d$, i.e. the fact that $d(a,b)=d(a+c,b+c)$ for any $a,b,c\in X$: let $u$ be a continuous linear functional and fix any $\epsilon>0$. Since $u$ is continuous at $0$, there exists some $\delta>0$ such that $d(x,0)<\delta$ implies $|u(x)|<\epsilon$. Thus, whenever $d(y,z)<\delta$, you have (applying the above with $x:=y-z$ and observing that $d(y-z,0)=d(y,z)<\delta$) $$ |u(y)-u(z)|=|u(y-z)|<\epsilon. $$

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  • $\begingroup$ Oh, that was much simpler than i though. Thank you very much. $\endgroup$ – Hugocito Aug 27 '16 at 20:10

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