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Let $X$ and $Y$ be locally convex, Hausdorff topological vector spaces and let $[a,b] \subset \mathbb{R}$. Let $f: [a,b] \to \hom(X,Y)$ be continuous, where $\hom(X,Y)$ is the space of continuous linear maps from $X$ to $Y$ with the topology of uniform convergence on bounded subsets of $X$. Let $H \subset \hom(X,Y)$ be the image of $f$.

  1. Is $H$ always an equicontinuous subset of $\hom(X,Y)?$ This is true in the case where $X$ is barelled by the Banach-Steinhaus theorem, but I do not want to make any additional assumptions on $X$ or $Y$.

  2. If $H$ need not be equicontinuous in general, can we make any modifications to $f$ to ensure it is? e.g. assuming $f$ is infinitely differentiable, or continuous with respect to a different topology on $\hom(X,Y)$. Again, I do not want to make any additional assumptions about the spaces $X$ and $Y$.

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  • $\begingroup$ What is meant by a bounded subset of $X$? $\endgroup$
    – Aaron Tikuisis
    Commented Jun 26, 2012 at 20:56
  • $\begingroup$ A subset $B \subset X$ is bounded if $\sup_{x \in B} p(x) < \infty$ for every seminorm $p$ in the family that generates the locally convex topology on $X$. $\endgroup$
    – Allan Yashinski
    Commented Jun 26, 2012 at 22:47
  • $\begingroup$ I have not checked, what follows might be completely wrong. H is compact in a space of continuous functions, subspace of the continuous functions on X (with the weak topology) with the compact-open topology. Use Ascoli - Arzela. $\endgroup$
    – user24527
    Commented Jun 27, 2012 at 17:28
  • $\begingroup$ @NN - Thanks for the comment. Which version of Arzela-Ascoli are you referring to? Usually there is a strong restriction on the type of space $X$ is, e.g. $X$ is a compact Hausdorff space. Did you consider the weak topology on $X$ so that it would satisfy some weaker hypothesis? $\endgroup$
    – Allan Yashinski
    Commented Jun 27, 2012 at 18:10
  • $\begingroup$ Ascoli theorem in Engelking, general topology, where domain is a k-space. Note that X and Y can be supposed to be complete (linear functions extend to the completion). However, I forgot that the k-modification of the weak topology is something well known to be useful only for Banach spaces (Eberlein-Smullyan). In any case, the obtained equicontinuity is for a different topology on X (I realize now that you can change the topology on the space of functions, not in X). Sorry (also for the delay; I had lost the question until a posted answer re-put today the question in the first page). $\endgroup$
    – user24527
    Commented Jul 3, 2012 at 13:03

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Assume that there is a sequence $T_n\in hom(X,Y)$ which converges to $0$ uniformly on all bounded subsets of $X$ but is not equicontinuous (such a sequence should exist if $X$ fails to be $c_0$-quasibarrelled, see chapter 8.2 of the book Barrelled Locally Convex Spaces of J. Bonet and P. Perez Carreras). I believe that the function $f:[0,1] \to hom(X,Y)$ defined by $f(0)=0$, $f(1/n)=T_n$ and affine-linear interpolation (e.g. $f(t/n + (1-t)/(n+1)) = tT_n + (1-t) T_{n+1}$ for $t\in [0,1]$ and $n\in \mathbb N$) should be continuous but the range is not equicontinuous. It should also be possible to make a smooth variant out of this example.

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  • $\begingroup$ Thanks. I think what I was really after (though I didn't ask it) was if a compact subset of $\hom(X,Y)$ is automatically equicontinuous. Your convergent sequence alone answers that question. $\endgroup$
    – Allan Yashinski
    Commented Jul 4, 2012 at 13:50

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