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 $B\_Y,\overline{B\_Y}$.


It is not hard to show that $\overline{B\_Y}\subset T(\overline{B\_X}) \implies B\_Y\subset T(B\_X)$. However the converse is false. Indeed, consider the counter example $X=l^1(\Bbb N), Y=\Bbb R$ and 
$$
T(x\_1,x\_2,x\_3,\dots):=\sum\_{n=1}^{\infty}\frac{nx\_n}{n+1}.
$$ 



> 
> How do we characterize the spaces $X,Y$ such that $B\_Y\subset T(B\_X) \implies \overline{B\_Y}\subset T(\overline{B\_X})$?
> 
> 
> 


I ask the same question in MSE a few days ago but got no satisfactory answer. The only answer showed that reflexivity is sufficient but what I want is an "iff" statement. Here is the [link](https://math.stackexchange.com/questions/2010382/when-do-we-have-b-y-subset-tb-x-if-and-only-if-overlineb-y-subset-t-ove) to my question.



When do we have $B_Y\subset T(B_X)$ if and only if $\overline{B_Y}\subset T(\overline{B_X})$?
$X$ reflexive; $Y$ at least one dimensional. This follows from James' characterization of reflexivity. If $X$ is not reflexive there is a norm one linear functional on $X$ that does not achieve its norm. 


