Show that this family is equicontinuous at $0$

Question

Let $E$ be a normed vector space, $$p_K(\varphi):=\sup_{x\in E}|\varphi(x)|\;\;\;\text{for }\varphi\in E'$$ for compact $K\subseteq E$ and $\sigma_c(E',E)$ denote the initial topology with respect to $(p_K,K\subseteq E\text{ is compact})$, i.e. the subspace topology on $E'$ inherited from the topology of compact convergence on $C(K)$.

Let $\mathcal C\subseteq C(E')$ be uniformly $\sigma_c(E',E)$-equicontinuous.

Why can we conclude that $$\forall\varepsilon>0:\exists\delta>0:\forall\varphi\in E':\left\|\varphi\right\|_{E'}<\delta\Rightarrow\sup_{f\in\mathcal C}\left|f(0)-f(\varphi)\right|<\varepsilon?\tag1$$

Most probably the desired claim is trivial to obtain, but I'm not able to see how due to the rather complicated setting.

$(1)$ is obviously some kind of equicontinuity at $0$. I'm not sure if it's relevant, but by the Banach-Alaoglu theorem $\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$ is $\sigma_c(E',E)$-compact for all $\delta>0$.

Answer 1

Recall the definition of uniform equicontinuity of $\mathcal{C}$ as a set of maps $(E',\sigma_c(E',E)) \to \Bbb{R}$:

For every neighbourhood $V \subseteq \Bbb{R}$ of $O$ there is a neighbourhood $U$ of $0$ in $(E',\sigma_c(E',E))$ such that $$\varphi,\psi \in V \implies f(\varphi)-f(\psi) \in V, \, \text{for all }f \in \mathcal{C}.$$

Now for $\psi = 0$ and $V = \left\langle-\frac\varepsilon2, \frac\varepsilon2\right\rangle$, we get a neighbourhood $U$ of $0$ such that $$\varphi \in U \implies |f(\varphi)-f(0)|<\frac\varepsilon2, \, \text{for all }f \in \mathcal{C} \implies \sup_{f \in \mathcal{C}} |f(\varphi)-f(0)|\le\frac\varepsilon2<\varepsilon$$ $U$ being a neighbourhood of $0$ contains an intersection of finitely many open balls around the origin of radii $\delta_1, \ldots, \delta_k$ with respect to the seminorms of compact sets $K_1, \ldots, K_n \subseteq E$: $$\bigcap_{k=1}^n \{\phi \in E' : p_{K_k}(\phi) < \delta_k\} \subseteq U.$$ Sets $K_k$ are bounded in norm by some $M_k > 0$ so if we set $$\delta := \min_{1 \le k \le n}\frac{\delta_k}{M_k}$$ then for any $\varphi \in E'$ we have $$\|\varphi\|_{E'} < \delta \implies p_{K_k}(\varphi) = \sup_{x \in K_k}\|\varphi(x)\| \le \|\varphi\|_E'\sup_{x \in K_k}\|x\| < \delta M_k \le \delta_k$$ for all $k=1, \ldots, n$ so $$\|\varphi\|_{E'} < \delta \implies \varphi \in \bigcap_{k=1}^n \{\phi \in E' : p_{K_k}(\phi) < \delta_k\} \subseteq U \implies \sup_{f \in \mathcal{C}} |f(\varphi)-f(0)|<\varepsilon.$$

Answer 2

If I'm not mistaken, this should be an instance of a more general result: Let

• $(X,\tau)$ be a topological space;
• $Y$ be a normed $\mathbb R$-vector space;
• $$\overline p(f):=1\wedge\sup_{x\in X}\left\|f(x)\right\|\;\;\;\text{for }f\in C(X,\tau;Y);$$
• $$p_K(f):=\sup_{x\in K}\left\|f(x)\right\|_Y\;\;\;\text{for }f\in C(X,\tau;Y)$$ for $\tau$-compact $K\subseteq X$ and $$P:=\{p_K:K\subseteq X\text{ is }\tau\text{-compact}\}.$$
• $(Z,d)$ be a metric space;
• $F:C(X,\tau;Y)\to Z$ be continuous with respect to the locally convex topology on $C(X,\tau;Y)$ generated by $P$ and the metric $d$ on $Z$.

Then we easily see that $f$ is continuous with respect to the norm $\overline p$ on $C(X,\tau;Y)$ generated by $P$ and the metric $d$ on $Z$: Let $f\in C(X,\tau;Y)$ and $\varepsilon>0$. By the continuity assumption on $F$, there is a $P$-neighborhood $N$ of $f$ with $$d(F(f),F(g))<\varepsilon\;\;\;\text{for all }g\in N\tag1.$$ Let $U_p$ denote the open unit ball in $$C(X,\tau;Y)$$ with respect to $p\in P$. We can write $N=f+N_0$ for some $P$-neighborhood $N_0$ of $0$. Moreover, there are $k\in\mathbb N_0$, $\tau$-compact $K_1,\ldots,K_k\subseteq X$ and $\delta_0>0$ with $$B_0:=\delta_0\bigcap_{i=1}^kU_{p_{K_i}}\subseteq N_0\tag2.$$ Now let $\delta\in(0,1)$ with $\delta\le\delta_0$. Then, $$\delta U_{\overline p}\subseteq B_0\tag3$$ and hence $$d(F(f),F(g))<\varepsilon\;\;\;\text{for all }g\in f+\delta U_{\overline p}\tag4;$$ i.e. $f$ is continuous at $f$ with respect to the locally convex topology on $C(X,\tau;Y)$ generated by $P$ and the metric $d$ on $Z$.

Alternatively, the result would have been followed immediately by noting that the topology generated by $P$ is coarser than the topology generated by $\overline p$, as discussed here.

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Now, if $X$ is a normed $\mathbb R$-vector space and $\tau$ is topology generated by $\left\|\;\dot\;\right\|_X$, then $$\left\|A\right\|=1\wedge\sup_{x\in X}\left\|Ax\right\|_Y\le\left\|A\right\|_{\mathfrak L(X,Y)}\tag5\;\;\;\text{for all }A\in\mathfrak L(X,Y)$$ and hence the topology generated by $\left\|\;\cdot\;\right\|$ is coarser than the uniform operator topology (i.e. the topology generated by $\left\|\;\cdot\;\right\|_{\mathfrak L(X,Y)}$). So, we immediately obtain that $F$ is continuous with respect to the topology generated by $\left\|\;\cdot\;\right\|_{\mathfrak L(X,Y)}$ and the metric $d$ on $Z$ as well.