Is it true that $ T(\overline{B(0,1)}) = \overline{T(B(0,1))} $ for continuous operator $T$? $T : X \to Y$ be a surjective linear continuous operator($X, Y$ : Banach Space). Then does $$ T(\overline{B(0,1)}) = \overline{T(B(0,1))} $$ hold? Here $B(0,1)$ means an open ball in $X$, and Bar means its closure.
 A: This is a comment made into an answer.
Consider the Banach space $c_0$ of all real-valued sequences that tend to $0$ equipped with the supremum norm $\|\cdot\|_{\infty}$. Consider the linear functional $$T:c_0\rightarrow \mathbb R, (a_n)\mapsto \sum_{n\geq 0} \frac{a_n}{2^n}.$$ We obvioulsy have $\|T\|=2$, and yet $$T(B(0,1))=T(\overline{B(0,1)})=(-2,+2)$$ which is not closed.
A: In general, this is not true. Consider arbitrary normed space $X$ and non-zero functional $\varphi$ of norm $1$, such that norm is not attained. Then $\varphi$ gives the desired counterexample.
Indeed, since $\varphi$ is non-zero, its image is the whole scalar field $\mathbb{C}$. Since its norm one, but norm of $\varphi$ is not attained, then $\varphi(B_X(0,1))=\varphi(\overline{B_X(0,1)})=B_\mathbb{C}(0,1)$. And we see that
$$
\varphi(\overline{B_X(0,1)})\neq\overline{\varphi(B_X(0,1))}
$$
For more concrete counterexample, take $X=C([0,1])$ and consider
$$
\varphi:X\to\mathbb{C}:f\mapsto \int\limits_{0}^{1/2}f(t)dt-\int\limits_{1/2}^1f(t)dt
$$
Here you can find a proof that $\varphi$ doesn't attain its norm.
