Consider a norm-decreasing linear map from a Hilbert space $H$ to the bounded linear operators $B(H')$ on another Hilbert space $H'$. Can it happen that the image of $H$ in $B(H')$ is not closed - can somebody please give an example.

  • $\begingroup$ Do you have an example of a norm-decreasing linear map with non-closed image (between, say, two Banach spaces of your choosing)? $\endgroup$
    – s.harp
    Jul 22, 2019 at 22:47
  • $\begingroup$ no, this would be good to see. $\endgroup$ Jul 22, 2019 at 23:44

1 Answer 1


Let $H=\ell^2(\mathbb N)$. Let $T:H\to B(H)$ be the linear map induced by $$ Te_n=\tfrac1{n}\,E_{nn}, $$ where $\{e_n\}$ is the canonical basis of $H$ and $\{E_{kj}\}$ are the canonical matrix units in $B(H)$. Any $x=\{x_n\}\in H$ is mapped by $T$ to $\sum_n\tfrac{x_n}{n}E_{nn}\in D_0$, the subalgebra of compact diagonal operators. The image is dense because it contains the dense subset $\operatorname{span}\{E_{nn}:\ n\}$. But it's not everything: the element $$ G=\sum_n\tfrac1n\,E_{nn} $$ is not in the image of $T$. As $T$ is injective, if $Tx=G$ we need to have $x_n/n=1/n$. That requires $x_n=1$ for all $n$, and no such $x\in H$ exists.


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