Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $Y$.

share|improve this question
    
What is meant by a bounded subset of $X$? –  Aaron Tikuisis Jun 26 '12 at 20:56
    
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 $X$. –  Allan Yashinski Jun 26 '12 at 22:47
    
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. –  user24527 Jun 27 '12 at 17:28
    
@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? –  Allan Yashinski Jun 27 '12 at 18:10
    
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). –  user24527 Jul 3 '12 at 13:03
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
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. –  Allan Yashinski Jul 4 '12 at 13:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.