The closure of the image of the unit sphere The following seems very credible to me but is it correct? If $E$ and $F$ are Banach spaces, $T:E\to F$ linear and continuous and $\epsilon>0$, 
$$\overline{TB_E}\subseteq(1+\epsilon)TB_E\,,$$
where $B_E$ is the unit ball of $E$. 
 A: Here is a quick proof if $F = E^*$: for any $\epsilon > 0$ you have
$$ \overline{TB_E} = [-\|T\|,\|T\|] \subset (1+\epsilon) (-\|T\|,\|T\|).$$

What about general $F$? The answer is generally no for nonreflexive spaces $E$. Suppose that $E = C([0,1])$ with $\|u\|_E = \|u\|_\infty$, $F = L^1([0,1])$ with $\|u\|_F = \|u\|_1$, and $T : E \to F$ is given by $$ (Tu)(x) = \int_0^x u(t) \, dt.$$
Define
$$v(x) = \left\{ \begin{array}{ccl} 0 & \text{if} & 0 \le x \le \frac 12 \\ x - \frac 12 & \text{if} & \frac 12 \le x \le 1 \end{array} \right.$$ 
Then $v \in F$, and moreover $v \in \overline{TB_E}$ since $v$ can be approximated uniformly by smooth increasing functions in $E$ with supremum norm not exceeding $1$. (Details omitted: the idea is to round the corner in the graph of $v$ smoothly in such a way that the slope never exceeds $1$.)
Unfortunately $v$ is not differentiable and thus not in the range of $T$, so $v \notin (1+\epsilon)TB_E$ for any $\epsilon > 0$.
The situation is different if $E$ is reflexive, since in that case the closed unit ball $\overline B_E$ is weakly compact. 
For any $\epsilon > 0$ you have $\overline{B_E} \subset (1+\epsilon)B_E$ which implies that $T \overline{B_E} \subset (1+\epsilon) TB_E$.
A continuous map is weakly continuous so that $T \overline{B_E}$ is weakly closed in $F$, hence closed in the original topology. Thus $$ \overline{TB_E} \subset \overline{T \overline{B_E}} = T \overline{B_E} \subset (1+\epsilon) TB_E.$$
