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).


1 Answer 1


The definition of a compact map has been revised after I posted this answer.

$T(C)$ 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.

  • $\begingroup$ 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? $\endgroup$ Feb 20, 2021 at 23:29
  • $\begingroup$ @ArunBharadwaj I have edited my answer. $\endgroup$ Feb 20, 2021 at 23:37
  • $\begingroup$ 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... $\endgroup$ Feb 21, 2021 at 19:59
  • $\begingroup$ I have also edited the question to provide more clarification about the exact definitions I'm using. $\endgroup$ Feb 21, 2021 at 20:13
  • $\begingroup$ @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. $\endgroup$ Feb 22, 2021 at 5:13

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