# Compact map of a closed convex set

Question: $$X$$ is a $$B^*$$ space and $$C$$ is a closed convex set of $$X$$. Let $$T: C\to C$$ be a compact map (defined below). Prove that $$T(C)$$ is sequentially pre-compact (That is, any sequence in $$T(C)$$ has a Cauchy subsequence).

(The real goal is to show that $$T$$ has a fixed point in $$C$$, but showing $$T(C)$$ is sequentially pre-compact is good enough because of Schauder's fixed point theorem)

Compact map: Given $$X$$ a $$B^*$$ space and $$E \subset X$$, Let $$T:E \to X$$. We call $$T$$ a compact map if $$T$$ is continuous and for all $$K\subset E$$ with $$K$$ bounded, we have that $$T(K)$$ is sequentially compact. (That is, every sequence in $$T(K)$$ has a convergent subsequence with limits in $$T(K)$$.).

My attempt: If $$C$$ has only one element there is nothing to prove. So assume there are at least two element in $$C$$. I am able to show that the following is true, $$\Lambda = \bigcup_{\substack{K \subseteq C\\ K \text{ bounded}}}K = C$$ (The forward inclusion is trivial. For the reverse inclusion, if we assume $$x \in C\setminus\Lambda$$, since there is some other $$y$$ also in $$C$$; we can look at the convex hull of $$\{x,y\} \subseteq C$$ which is totally bounded and therefore bounded and thus is $$\subseteq \Lambda$$, which puts $$x\in \Lambda$$ which is a contradiction!)

This also gives me $$T(C) = \bigcup_{\substack{K \subseteq C\\ K \text{ bounded}}} T(K)$$

But I am sort of lost here trying to show that any sequence in $$T(C)$$ has a Cauchy subsequence. Intuitively, since the union is arbitrary, it seems to me that it should be possible to create a sequence such that for any $$N \in \mathbb{N}$$ and for any $$K \subseteq C$$ bounded, there is some $$k > N$$ such that $$x_k \not\in T(K)$$.

Correction/Edit: The question (which originally appeared in my homework set) has been updated to include the additional hypothesis that $$C$$ is bounded.

My solution: Now the entire question is trivial since we have the bounded subset $$C\subseteq C$$ such that $$T(C)$$ is sequentially compact. Finally, in order to use Schauder's fixed point theorem, we need the set to be complete; but this is achieved easily. First we notice that the sequential compactness of $$T(C)$$ ensures the compact-ness of $$T(C)$$. So if we consider a Cauchy sequence $$\{x_k\} \subset C$$ that doesn't coverge in $$C$$, by the continuity of $$T$$, we obtain a Cauchy sequence $$\{T(x_k)\} \subset T(C)$$ that does not converge in $$T(C)$$ - but this contradicts the compactness of $$T(C)$$! So, $$C$$ is complete and we can use Schauder; and so we win!

Further question: I am still curious to know if the statement still holds if we remove the bounded assumption on $$C$$. According to the answer by @Kavi Rama Murthy, it does look like we don't always have sequential pre-compactness of $$T(C)$$ in this case; but the question about the fixed point itself remains elusive (as noted in my reply comment).

$$T(C)$$ need not be sequentially compact: Let $$C=X$$ and $$Tx=f(x)x_0$$ where $$x_0$$ is a fixed nonzero elmeent of $$X$$ and $$f$$ is a fixed non-zero continuous linear functional on $$X$$. Then the hypothesis is satisfed but $$T(C)$$ is the span of $$x_0$$ which is not pre-compact.
Take $$X=C=\mathbb R$$ and $$Tx=x+1$$ to see that $$T$$ may not have any fixed point.
• Ah this makes sense! This was kind of what I was intuitively thinking, but I wasn't able to write it out properly. Thank you so much! Going back to square 1, can I perhaps get a direction/hint on how to show that $T$ might have a fixed point in $C$, since that was the ultimate goal anyway? Feb 20, 2021 at 23:29
• Apologies for the delayed reply, I just had time to think about this clearly. Thank you for your answer; but maybe I am not seeing this fully: What if I consider $K = (2,3)$ a bounded subset of $C$, then $T(K) = (3,4)$; which is sequentially precompact, but not sequentially compact (since I can have a sequence $x_k = 3+\frac{1}{2k} \to 3$ but $3$ is not in $T(K)$, so there is a sequence in $T(K)$ with none of its subsequences convergent with limits in $T(K)$.) So the hypothesis is not fully satisfied for this example... Feb 21, 2021 at 19:59
• @ArunBharadwaj Your definition says every sequence in $T(K)$ has a subsequence convergin g to a point of $E$, not in $T(K)$. In my example $E=C=\mathbb R$ and this condition is satisfied. Feb 22, 2021 at 5:13