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Let $\{T_n\in\mathcal{B}(X)\,|\,\text{rank}(T_n)=R\,\}^{\infty}_{n=1}$ is a sequence of linear bounded finite-rank operators on a Banach space with the same rank $R$. Let it converge uniformly to an operator $T\in\mathcal{B}(X)$, that is $\Vert T_n - T\Vert_{\mathcal{B}(X)}\longrightarrow 0\;(n \rightarrow\infty)$, that is $\forall\epsilon>0\;\exists N\in\mathbb N\;\forall n\ge N\;\forall x\in X\;|\;\Vert x\Vert_{X}\le 1$ holds $\Vert T_n x - T x\Vert_X\le\epsilon$.

Prove that $\text{rank}(T)\le R$.

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    Michael, you will usually get more effective answers here if you explain what you tried. As is, my suggestion is: try arguing by contradiction.2012-12-26
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    @PavelM Thanks! I'll try.2012-12-26

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