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Let $X$ and $Y$ be two Banach spaces. Assume that $f$ is linear bounded mapping from $X$ to $Y$ such that for all $n$, there exists $a_n$ with $||a_n||=1$ and $||f(a_n)||=\frac{1}{n}.$ Prove that $f: X \rightarrow Y$ cannot be both surjective and injective.


I think that when I first assume that $f$ is onto, I will get a contradiction to Open Mapping Theorem. Could you provide me with an initial step to begin? Even though I figured out the scheme. I couldn't find where to begin.

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if $f$ is bijective then $f^{-1}$ is continuous by Open Mapping Theorem. So $\|f^{-1}y|| \leq C \|y\|$ for all $y \in Y$ where $C$ is the norm of $f^{-1}$. Taking $y=fx$ we get $\|x\|\leq C\|f(x)\|$ for all $x \in X$. What happens to this when you take $x=a_n$ and let $n \to \infty$?

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  • $\begingroup$ We get a contradiction as we get $1 < 0$. Thank you a lot!. $\endgroup$
    – user722271
    May 14 at 23:28

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