Sum of the sets in a Banach space I have the following question. Let $M$ be the closed unit ball with center in $0$ and radius $1$ in a Banach space $X$, so $M=B_1(0)$. Let $f,g$ be linear und bounded $X\to X$. Is it true, that $\overline{ (f+g)(M) }\subset \overline{f(M)}+\overline{g(M)}$ (here $\overline{A}$ means closure of $A$).
Thanks in advance.
 A: Let $B = l_1(\mathbb N)$ and
$$f(e_i) = \begin{cases}
0, i = 0 \\
\frac{e_0}{2} + e_{3i - 2} + \frac{e_{3i - 1}}{i} - \frac{e_{3i}}{i}
\end{cases}$$
and
$$g(e_i) = \begin{cases}
0, i = 0 \\
\frac{e_0}{2} - e_{3i - 2} + \frac{e_{3i - 1}}{i} + \frac{e_{3i}}{i}
\end{cases}$$
We have $(f+g)(e_i) = e_0 + \frac{2 e_{3i - 1}}{i} \to e_0$, thus $e_0 \in \overline{(f + g)(M)}$.
If $f(x_n) \to y$ then $x_n$ converges coordinate-wise (except may be in the zeroth coordinate), and if coordinate-wise limit of $x_n$ has non-zero $i$-th coordinate, then so is $3i-2$-nd (and next two) coordinates of $y$. And if all but may be zeroth coordinates of $x_n$ converges to $0$ then $y = 0$.
So, if $y \in \overline{f(M)}$ then either $y = 0$ or for some $i$ we have $|y_{3i - 1}| > 0$, $|y_{3i}| > 0$ and $y_{3i} = - y_{3i - 1}$.
Finally, if $y \in \overline{f(M)}$, $z \in \overline{g(M)}$ then either one of them is zero and $y + z \neq e_0$, or either $y_{3i - 1} \neq -z_{3i - 1}$ and $(y + z)_{3i - 1} \neq 0$ or $y_{3i} \neq z_{3i}$ and $(y + z)_{3i} \neq 0$, in both cases $y + z \neq e_0$.
So, $e_0 \notin \overline{f(M)} + \overline{g(M)}$.
Such situation is impossible if $M$ is compact (for example if our space is finite dimensional), because if $(f + g)(x_n) \to y$ then there is converging subsequence $x_{n_k}$ and $f(x_{n_k}) \to a$, $g(x_{n_k}) \to b$ s.t. $a + b = y$. In our case, however, $f(x_{n_k})$ diverges for any subsequence.
