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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 $\delta>0$.

0xbadf00d
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2 Answers2

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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 $$\|\varphi\|_{E'} < \delta \implies \varphi \in \bigcap_{k=1}^n \{\phi \in E' : p_{K_k}(\phi) < \delta_k\} \subseteq U \implies \sup_{f \in \mathcal{C}} |f(\varphi)-f(0)|<\varepsilon.$$

mechanodroid
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  • Why are the $K_k$ norm-bounded (unless $E$ is finite-dimensional)? Aren't they only totally bounded (which should be enough for your concern)? – 0xbadf00d Dec 28 '20 at 08:06
  • @0xbadf00d A compact set $K$ in a normed space is always norm-bounded: you can cover $K$ with finitely many balls of radius $1$ $$K \subseteq \bigcup_{i=1}^n B(x_i,1)$$ so for all $x \in K$ we have $|x| \le \max{x_1, \ldots, x_n} + 1$. (btw. totally bounded always implies bounded, the converse holds only in finite-dimensional spaces). – mechanodroid Dec 28 '20 at 14:21
  • Please take note of the edit made to my answer below. I've fixed the issues we've discussed in the other post. BTW, the result should follow immediately by noting that the topology generated by $P$ is coarser than the topology generated by $\overline p$, as we've discussed there, right? – 0xbadf00d Jan 05 '21 at 05:56
  • And I'm curious about one last other thing: As noted before, $X:={\varphi\in E':\left|\varphi\right|{E'}\le\delta}$ is $\sigma_c(E',E)$. Now, if $F:X\to\mathbb R$ is $\sigma_c(E',E)$-continuous, it is $\left|;\cdot;\right|{E'}$-continuous by your result. But does the other implication hold as well? I know that on $X$ the topology of compact convergence (which is $\left.\sigma_c(E',E)\right|X$) and the topology of pointwise convergence coincide. Maybe on $X$ these topoolgies coincide with the topology generated by the norm $\left|;\cdot;\right|{E'}$ as well? – 0xbadf00d Jan 05 '21 at 08:07
  • @0xbadf00d This is a very interesting question, I'm not too sure but I believe the topology $\sigma_c(E',E)$ is strictly weaker than the dual norm topology on both $E'$ and ${f\in E' : |f|{E'} \le \delta}$. Namely, for every $K \subseteq E$ compact we have $$p_K(f) = \sup{x\in K}|f(x)| \le |f|{E'} \underbrace{\sup{x \in K}|x|}_{<+\infty}$$ since compact sets are bounded. This implies that $\sigma_c(E',E)$ is weaker than the dual norm topology. – mechanodroid Jan 08 '21 at 21:01
  • If the two topologies were equal, then there would be $M>0$ and compact sets $K_1, \ldots, K_n \subseteq E$ such that for all $f \in E'$ we have $$|f|{E'} \le M \max{p{K_1}(f),\ldots, p_{K_n}(f)} = Mp_K(f)$$ where $K = K_1 \cup \cdots \cup K_n$ is also compact. Hence $$\sup_{x \in \text{unit sphere}} |f(x)| \le M\sup_{x \in K}|f(x)|.$$ – mechanodroid Jan 08 '21 at 21:02
  • Now this should immediately look suspicious since the unit sphere is not compact. Indeed, if we take $E$ to be nonseparable, then $\overline{\operatorname{span}K} \ne E$ so this proper closed subspace is contained in a closed maximal proper subspace $H$ of $E$. Now we can find a bounded linear functional $f \in E'$ such that $\ker F = H$. Then LHS above is nonzero but RHS is zero which is a contradiction. – mechanodroid Jan 08 '21 at 21:02
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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.

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.

0xbadf00d
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