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Let $T$ be a continuous linear functional from Banach space $X$ to Banach space $Y$. Is it true that $\overline{T(\overline{B_X})} = \overline{T(B_X)}$? Here $B_X$ denotes the open unit ball of $X$, and the overline denote the closure operation (in respective strong topologies).

It seems obviously true by a diagonal argument, but I need a rigorous proof.

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  • $\begingroup$ Where did you find this question? Perhaps you could try rigorising the argument you already had, or at least state your diagonal argument? Typically, questions require more context or effort shown. $\endgroup$ – mdave16 Aug 18 '17 at 23:12
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Yes. If $x_n \in \overline{B_X}$, $y_n = (1 - 1/n) x_n \in B_X$, and $\|T(x_n) - T(y_n)\| \to 0$, so if $z = \lim_{n \to \infty} T(x_n)$ then $z = \lim_{n \to \infty} T(y_n)$.

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  • $\begingroup$ The proof is shorter than the one I had. Since the map is from $X$ to $Y$, it's probably better to denote $y_n$ by $x_n'$, and denote $z$ by $y$. Then it would be perfect since $y\in Y$, and $x_n' \in X$. Thank you very much! $\endgroup$ – minqiang Aug 19 '17 at 1:29

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