# When do we have $B_Y\subset T(B_X)$ if and only if $\overline{B_Y}\subset T(\overline{B_X})$?

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 $$[-1,1]\not\subset T(\overline{B_{l^1}})=(-1,1)$$

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.

• 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. – BigbearZzz Nov 14 '16 at 21:46
• @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 :) – s.harp Jan 4 '17 at 16:01