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Let $X$ be a Banach Space and $Y$ be a normed linear space. Let $(A_n)$ and $A$ be bounded linear operators from $X\rightarrow Y$ such that $\Vert (A_n-A)x \Vert\rightarrow0$ for every $x\in X$. If $K:X\rightarrow X$ is compact, then show $\Vert (A_n-A)K \Vert \rightarrow 0$.

I'm not sure how to proceed in this question. Any help will be appreciated.

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    Since $K$ is compact, $A_nK$ and $AK$ are also compact. – 2017-02-19
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    @Aweygan How can I use that here? – 2017-02-19

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Assume the claim is not true. Then there is $\delta>0$ and $x_n$ with $\|x_n\|_X=1$ such that $$ \|(A_n-A)Kx_n\| \ge\delta $$ for infinitely many $n$. Now $(x_n)$ is bounded, so $(Kx_n)$ contains a converging subsequence $Kx_n \to y$. So $$ \|(A-A_n)Kx_n \|\le \|(A-A_n)Ky\| +\|(A-A_n)(Kx_n-y) \|\\ \le \|(A-A_n)Ky\| +\|A-A_n\|\cdot\|Kx_n-y \|. $$ Now the first claim goes to zero by assumption, the second goes to zero as $(A_n)$ is bounded (uniform boundedness principle).

[The answer is a little bit sloppy - one has choose subsequencess all the time. I hope you get the idea anyway.]

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    In the first term, shouldn't it be just $y$ instead of $Kx_n$? – 2017-02-19
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    @SahibaArora No, but there is an error. The first line should read $$\|(A-A_n)Kx_n \|\le \|(A-A_n)y\| +\|(A-A_n)(Kx_n-y) \|.$$ – 2017-02-19
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    Yes, that's what I meant. – 2017-02-19
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    @SahibaArora corrected. thanks for pointing out – 2017-02-19