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Let $C:=C([0,1],\mathbb{R})$ be the space of real-valued continuous functions defined on $[0,1]$. Could we find a topological vector space topology $\pi$ on $C$ such that the following two conditions are satisfied:

  1. The unit ball $\{f\in C: ||f||:=\sup_{t\in [0,1]}|f(t)|\le 1\}$ is compact.

  2. For every sequence $(f_n)\subset C$ converging to $f\in C$ under $\pi$, then one has $$f_n(0),~ f_n(1) \mbox{ and } \int_0^1f_n(t)dt \mbox{ converge respectively to } f(0),~ f(1) \mbox{ and } \int_0^1f(t)dt.$$

I was thinking about the weak topology and the weak star topology.

First, notice that $[0,1]$ is compact, then the dual space of $C$ is $rca([0,1])$, i.e. the space of (regular) Borel measures on [0,1] of bounded total variation, see e.g. page 255 Dunfond and Schwratz. Then clearly, as the dual space of $C$, $rca([0,1])$ is endowed with a topology induced by the following norm:

$$||\pi||=\sup_{||f||\le 1}\int_{[0,1]}f(t)\pi(dt). (\ast)$$

Moreover, it is easy to show that if $rca([0,1])$ is endowed with the weak topology, see e.g. page 175 Bogachev, then the corresponding dual space is $C$. But the weak topology is not equivalent to the topology $||\cdot||$.

Second, I may identify $C$ as a subspace of $rca([0,1])$ equipped with the topology induced by $(\ast)$. Denote by $rca([0,1])^{\ast}\supset C$ its dual space, and endowing the dual space with the weak star topology, it is easy to show the unit ball of $C$ is weakly compact, i.e. any sequence has a convergent subsequence under weak star topology, but I can not show the limit belongs to $C$.

If some one has an idea, please let me know. Thanks a lot!

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  • $\begingroup$ I'd improved the formatting, and added the condition that the topology is a TVS topology, please fix if I'm mistaken (but otherwise I can't figure out how "weakly compact" makes sense). $\endgroup$ – YCor Jul 16 '15 at 14:52
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    $\begingroup$ Sure. Take the weak topology generated by the three functionals ptwise evaluation at 0, ptwise evaluation at 1, and integration. $\endgroup$ – Bill Johnson Jul 16 '15 at 14:55
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    $\begingroup$ If you want a Hausdorff locally convex topology, the answer is no, because it would make $C[0,1]$ into a dual Banach space, and $C[0,1]$ is not even isomorphic to a dual Banach space. $\endgroup$ – Bill Johnson Jul 16 '15 at 14:57
  • $\begingroup$ @YCor Thanks for pointing out that. I mean compact and I have modified. $\endgroup$ – CodeGolf Jul 16 '15 at 15:25
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    $\begingroup$ Assume we have such a topology with (2) replaced with the slightly stronger: the function $f\mapsto (f(0),f(1),\int_{[0,1]}f)$ is continuous. Since addition by 1 is a homeomorphism, the set of continuous functions valued in $[0,2]$ should be closed. Then let $(f_n)$ be zero on $[1/n,1]$, and affine on $[0,1/n]$ with $f_n(0)=1$. By compactness, let $f$ be an accumulation point of $(f_n)$. By the conditions, we have $f(0)=1$ and $\int_{[0,1]}f=0$, and by what's just written, $f\ge 0$. Since $f$ is continuous, we get a contradiction. So there is no such topology. $\endgroup$ – YCor Jul 16 '15 at 17:31

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