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.
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.