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I believe that the following is true: Let $X$ and $Y$ be normed spaces, both of dimension at least $2$. Then, the space of bounded linear operators $L(X,Y)$ is not a Hilbert space. Is there a nice reference for this result?

Sketch of a proof: Suppose that $L(X,Y)$ is a Hilbert space. By considering $y \otimes x^* \in L(X,Y)$, $(y \otimes x^*)(x) := \langle x^*, x\rangle_X y$ for fixed $y \in Y$ or fixed $x^* \in X^*$, $X^*$ and $Y$ are subspaces of $L(X,Y)$ and, thus, pre-Hilbert spaces. Thus, also $X$ is pre-Hilbert. By considering orthonormal sets $\{e_1, e_2\} \subset X$ and $\{f_1, f_2\} \subset Y$ it is easy to check that the parallelogram identity in $L(X,Y)$ fails and this yields a contradiction.

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  • $\begingroup$ Perhaps this helps math.stackexchange.com/questions/4451818/… $\endgroup$ Commented May 19, 2022 at 14:17
  • $\begingroup$ Your proof is correct. In the linked question it is assumed that $X,Y$ are Hilbert, which is not the case here. $\endgroup$
    – daw
    Commented May 20, 2022 at 5:52

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