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Assume that we have a vector space $X$ over reals with a countable sequence of seminorms $p_n$ on $X$ such that: $$ p_n(x)\leq p_{n+1}(x) \textrm{ for } n\in \mathbb N, x\in X, $$ $$ \textrm{ for } x\in X\setminus \{0\} \textrm{ there is } n\in \mathbb N \textrm{ such that } p_n(x)\neq 0. $$ Then $X$ is a metric space with the metric $$ d(x,y)=\sum_{n=1}^\infty \frac{1}{2^n} \frac{p_n(x-y)}{1+p_n(x-y)}, \ x,y\in X. $$ Such a space $X$ is called the countably seminormed space. Let's consider in a complete countably seminormed space $X$ a sequence $T_k:X\rightarrow \mathbb R$ of linear continuous mappings such that for each $x\in X$ the sequence of numbers $(T_k(x))_{k \in \mathbb N}$ is bounded. Then by the Banach-Steinhaus theorem for the Frechet spaces the family $(T_k)_{k\in \mathbb N}$ is equi-continuous. Moreover, by properties of linear functionals on countably seminormed spaces, for each $k\in \mathbb N$ there exists $N_k$ such that $T_k$ is continuous with respect to the seminorm $p_{N_k}$. I.e. $|T_k(x)|\leq M p_{N_k}(x)$ for all $x\in X$, where $M>0$ is some constant depending on $k$.

Does there then exist an $N\in \mathbb N$ and $M>0$, not depending on $k\in \mathbb N$, such that $$ |T_k(x)|\leq M p_N(x) \textrm{ for } x\in X, k \in \mathbb N ? $$

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A.B
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2 Answers2

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Yes, that is what equicontinuity means. That the family is equicontinuous means there is one neighbourhood $V$ of $0$ such that

$$T_k(V) \subset \mathbb{D}$$

for all $k$ (where $\mathbb{D}$ is the open unit disk/interval). Since the family of seminorms is increasing, there is one $N$ with $V \supset \{ x : p_N(x) < r\}$ for some $r > 0$. But then

$$\lvert T_k(x)\rvert \leqslant \frac1r p_N(x)$$

for all $x$ and $k$.

Daniel Fischer
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Yes, this is equicontinuous. Since the topology is defined by a sequence of seminorms, so you can define the equicontinuity by the $p_k$.

This is the problem about Banach-Steinhaus theorem on Frechet space (or metric linear spaces.) Maybe you can see Rolewicz'book "Metric Linear Spaces" or goolge for more about this problem.

Lei Li
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