Separable condition on space for a compact operator

Question

If i have two orthonormal sets $\{e_n\}$ and $\{f_n\}$ in a hilbert space H, we can have the following operator in $B(H)$ such that \begin{equation*} Tx = \sum_{n=1}^{\infty} \alpha_n (x,e_n)f_n. \end{equation*} if $\alpha_n \to 0$ we have that $T$ is compact with dense non-closed image in $H$. So, i would like to know why do we need $H$ to be separable to have the following result

If $H$ is separable then there exists a compact operator on $H$ whose image is dense in $H$ but not equal to $H$.

In general, can we have a compact operator with dense image on a non-separable hilbert space? It seems to me that the operator $T$ defined above is an affirmative answer to ths question or where does the condition of being separable comes to play.

Any help would be appreciated and thank you in advance.

Does this answer your question? Why is the image of a compact operator separable? — Arctic Char, Dec 16 '19 at 02:44
Seems to be a different question, or am i missing something? — Alek Murt, Dec 16 '19 at 02:48
The image of a compact operator is separable. In particular the image cannot be dense if $H$ is non-separable. — Arctic Char, Dec 16 '19 at 02:50
Otherwise in Hilbert space the image of $T$ compact is separable because iterating the map $T\to T-(.,y)Ty$ (where $y,|y|=1$ maximizes $|Ty|$) gives the SVD $Tx= \sum_k (x,y_k)Ty_k$ — reuns, Dec 16 '19 at 03:54

Martin Argerami · Accepted Answer · 2019-12-16T05:29:15.333

The formula \begin{equation*} Tx = \sum_{n=1}^{\infty} \alpha_n (x,e_n)f_n \end{equation*} holds for any compact operator $T:H\to K$. That is, given $T$ compact you can find $\{\alpha_n\}$ with $\alpha_n\searrow0$ and $\{e_n\}$, $\{f_n\}$ orthonormal bases (this is the singular value decomposition). If you consider the subspaces $$ K_m=\operatorname{span}\{f_n:\ n\leq m\} $$ then $K_m\subset K_{m+1}$ and the range of $T$ is contained in $\overline{\bigcup_mK_m}$, which is separable.

Separable condition on space for a compact operator

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