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$T : X \to Y$ 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.

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    I don't think this holds. Consider the Banach space $c_0$ of all real (or complex) sequences that tend to $0$ equipped with the supremum norm $\|\cdot\|_{\infty}$. Consider the linear functional $T:c_0\rightarrow \mathbb R, (a_n)\mapsto \sum_{n\geq 0} \frac{a_n}{2^n}$, then we obvioulsy have $\|T\|=2$, and yet $T(\overline{B(0,1)})=(-2,+2)$ which is not closed.2012-07-20

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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.

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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.