Let $X$,$Y$ be normed spaces, $T:X\to Y$ be a bounded linear operator. Denote the open and closed unit balls by $$ B_X:=\{ x\in X\ |\ \|x\|<1\} \\ \overline{B_X}:=\{ x\in X\ |\ \|x\|\le1\} $$ and similarly for $B_Y,\overline{B_Y}$.

I thought that $B_Y\subset T(B_X)$ if and only if $\overline{B_Y}\subset T(\overline{B_X})$ is a reasonable statement and tried to prove it. I got that $\overline{B_Y}\subset T(\overline{B_X}) \implies B_Y\subset T(B_X)$ but not the other way around. After a while I found a counterexample:

Let $X=l^1(\Bbb N), Y=\Bbb R$ and $$ T(x_1,x_2,x_3,\dots):=\sum_{n=1}^{\infty}\frac{nx_n}{n+1}. $$ It is not hard to see that $\|T\|=1$ but $|T(x)|<\|x\|_1$ for any $x\in l^1(\Bbb N)$. In this case we have $$ (-1,1)\subset T(B_{l^1})=(-1,1) $$ but $$ [-1,1]\not\subset T(\overline{B_{l^1}})=(-1,1) $$

This leads to my question:

What conditions do we need in order to conclude $B_Y\subset T(B_X) \implies \overline{B_Y}\subset T(\overline{B_X})$?

Completeness is out of the question since both $l^1(\Bbb N) $ and $\Bbb R$ are complete. I believe it might have something to do with uniform (or strictly) convexity of the spaces since $l^1(\Bbb N)$ lacks these properties.


If $X$ is reflexive for example it holds. To see it, consider:

  1. $T(\overline{B_X})\supset \overline{B_Y}$ iff for any $y\in Y$ you have an $x\in X$ with $T(x)=y$ and $\|y\|≥\|x\|$
  2. $T(B_X)\supset B_Y$ iff for every $y$ there exists a sequence $x_n\in X$ with $T(x_n)=y$ and $\lim_n\|x_n\|≤\|y\|$.

If $X$ is a reflexive Banach space then every bounded sequence has a weakly converging subsequence by the Banach Alaoglu theorem. For that reason if the condition of 2. holds then for every $y$ we have a weakly convergent sequence $x_n$ to some $x$ with $T(y)=T(x_n)$. Since $T$ is norm-norm continuous it is also weak-weak continuous and $T(x_n)\to T(x)$ in the weak topology on $Y$. But $T(x_n)=y$ is constant so $T(x)=y$ must also hold.

Now we have a point $x$ so that $T(x)=y$ and $x$ is the weak limit of a sequence $x_n$ so that $\|x_n\|$ converges to something smaller than $\|y\|$. Its again an application of the Banach Alaoglu theorem that $\|x\|≤\lim_n\|x_n\|≤\|y\|$, because the ball of radius $\lim_n\|x_n\|+\epsilon$ is compact in the weak topology, so the sequence $x_n$ must converge in it for all $\epsilon$.

So the condition of 1. is verified.

  • $\begingroup$ Very nice indeed! I wonder if reflexivity is a necessary condition for the inclusion to hold. What I actually wanted is a characterization of spaces such that the property holds. $\endgroup$ – BigbearZzz Nov 14 '16 at 21:46
  • $\begingroup$ @BigbearZzz A bit late, sorry: I think this is a very interesting question but I have no idea whether or not it is an iff $X$ reflexive thing. I can't think of an example where $X$ is not reflexive and it holds (either for all $Y$, or just for all maps to a specified $Y$). If you can think of such an example I'm sure it will be light the way in some way :) $\endgroup$ – s.harp Jan 4 '17 at 16:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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