Let $H_{1}$ and $H_{2}$ be two Hilbert spaces. My question is, when is the set of linear and bounded operators $\mathcal{L}(H_{1}, H_{2})$ with the usual norm a Hilbert space? I think I've proved that whenever $H_{1}$ or $H_{2}$ are $1$-dimensional the space of linear and bounded operators is a Hilbert space. But, what happens in arbitrary dimensions?

Thanks in advance.

  • 2
    $\begingroup$ Since you excluded $1$-dimensional spaces, you should try $H_1=H_2=\mathbb R^2$. Note that Hilbert spaces are characterized by the parallelogram law. $\endgroup$
    – Jochen
    May 16 at 16:11

1 Answer 1


In higher dimension, i.e. $\dim H_1\ge 2$ and $\dim H_2\ge 2,$ the space $\mathcal{L}(H_1,H_2)$ is not a Hilbert space. For example, let $H_1=H_2=\mathbb{C}^2$ and $$A=\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \qquad B=\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$$ Then $\|A+\lambda B\|=1$ for $0<\lambda \le 1.$ Assume for a contradiction that the operator norm is associated with an inner product. Then $$\displaylines{1=\|A+\lambda B\|^2=\langle A+\lambda B,A+\lambda B\rangle\\ =\|A\|^2+2\lambda\, \Re \langle B,A\rangle +\lambda^2\|B\|^2}$$ The expression on the right hand side is constant for $0<\lambda\le 1.$ Hence $\|B\|=0.$


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