# Embedding a Hilbert space into the bounded operators on a Hilbert space

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.

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

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.