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Let $X$ be an infinite dimensional Banach space and $T: X \rightarrow Y$ be compact operator (between Banach spaces).

Question: Does there always exist a sequence $x_n$ of norm $1$ such that $T(x_n)$ converges to zero?

If we assume $X$ to be reflexive, this is the case: Choose a sequence on the unit sphere which converges weakly to zero. The compact operator $T$ then maps this sequence to a sequence converging to zero.

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If your claim were false, $T$ would need to be injective.

Furthermore, it would follow that the constant $$ C := \inf_{\|x\|=1} \|Tx\| $$ is positive. From this, it follows that $T(X)$ is closed and $T : X \to T(X)$ is a bijective linear map between Banach spaces and thus an isomorphism of Banach spaces.

But $T(B_1 (0))$ is precompact, so that $B_1 (0)$ is itself precompact. It follows that $X$ is finite dimensional, a contradiction.

Hence, we can always find a sequence with the desired properties.

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