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$X$ and $Y$ are Banach spaces and $T$ is a bounded linear operator from $X$ to $Y$ which sends bounded closed sets to closed sets. Prove that $T(X)$ is closed.

Here I tried to used the fact that a subspace of a complete metric space is closed if and only if it is complete.
My idea was to choose a Cauchy Sequence $\{Tx_n\}$ in $T(X)$ and show that $\{x_n\}$ is bounded.
Proving this will be sufficient, because if this is known then all $\{x_n\}$ are contained a closed ball $B(x,r)$ for some $x \in X$ and some $r > 0$. And this being a bounded closed set, $T(B(x,r))$ is closed in $Y$ and contains all the points $\{Tx_n\}$. Moreover, it is complete, being closed.
Thus $Tx_n$ converges to some point $Tz \in B(x,r)$. Thus $T(X)$ is complete and thus closed in $Y$.

But I haven't been able to get any farther than this. Any help would be appreciated. Thanks!

AlexR
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Prithviraj Chowdhury
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  • The question title is no replacement for the problem statement. I've edited your post to include a more concise title and introduced MathJax to improve readability :) – AlexR Apr 23 '15 at 10:44
  • Thank you,Alex. Won't happen again. :) – Prithviraj Chowdhury Apr 23 '15 at 10:46
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    If $T$ is not injective, then you can always choose $(x_n)$ unbounded so that $(Tx_n)$ is bounded. What you need is that for every bounded $(y_n)$ in $T(X)$ you can choose a bounded sequence $(x_n)$ with $y_n = Tx_n$ for all $n$. But first suppose that $T$ is injective. Then assume you have a bounded sequence $(Tx_n)$ such that $(x_n)$ is unbounded. Can you construct a bounded closed subset $S\subset X$ such that $T(S)$ is not closed from that sequence? – Daniel Fischer Apr 23 '15 at 10:51
  • I think taking S to be the unit sphere works. Because $T(xn/ ||xn||)$ has a subsequence which converges to 0,as $(xn)$ is unbounded. So 0 is a limit point of $T(S)$,but as T is injective $T(s)$ is non zero for every element of S. – Prithviraj Chowdhury Apr 23 '15 at 11:11
  • Right. So you're done with the case that $T$ is injective. Can you reduce the general case to the injective case? (Aside: You can notify users when you respond to their comments using @UserName. As the post author, you are automatically notified of each comment here.) – Daniel Fischer Apr 23 '15 at 11:29
  • @DanielFischer Yes I believe I can! We can quotient X by the kernel of T and get an injective linear operator S from T. And hence,a bounded sequence in $X/ker(T)$ and using this,we can find a bounded sequence in X. Am I right? – Prithviraj Chowdhury Apr 23 '15 at 11:51
  • You need to show that $S$ maps closed and bounded subsets of $X/\ker (T)$ to closed subsets of $Y$, without that you don't know that you get a bounded sequence in $X/\ker (T)$. If you have that, you can either get the bounded sequence in $X$, or note that $T(X) = S(X/\ker T)$ to reach the conclusion. – Daniel Fischer Apr 23 '15 at 12:00

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