Picard Condition (searching for an idea) The so-called Picard-condition is:
Let X,Y be Hilbertspaces and $T\colon X\to Y$ is a compact operator with singular value decomposition system $\left\{(\sigma_j,u_j,v_j)\right\}$. An element $x\in\overline{\operatorname{range}(T)}$ is an element in $\operatorname{range}(T)$ exactly then when 
$$
\sum\limits_j^{\infty}\frac{\lvert\langle x,v_j\rangle\rvert^2}{\sigma_j^2}<\infty.
$$
Task: Show that from the Picard-condition it follows that 
$$
\operatorname{range}(T)\neq\overline{\operatorname{range}(T)}.
$$

Let $x\in \overline{\operatorname{range}(T)}$. When $\sum\limits_j^{\infty}\frac{\lvert\langle x,v_j\rangle\rvert^2}{\sigma_j^2}$ is not smaller than $\infty$, then it follows that $x\notin\operatorname{range}(T)$ and so automatically $\operatorname{range}(T)\neq\overline{\operatorname{range}(T)}.$.
So assume that $\sum\limits_j^{\infty}\frac{\lvert\langle x,v_j\rangle\rvert^2}{\sigma_j^2}<\infty$. Let again $x\in \overline{\operatorname{range}(T)}$, then it follows from the Picard-condition, that $x\in\operatorname{range}(T)$, which means $\overline{\operatorname{range}(T)}\subseteq\operatorname{range}(T)$
Now I have to show that it is not the other way round, which means I have to show that $x\in \operatorname{range}(T)\nrightarrow x\in\overline{\operatorname{range}(T)}$? If yes, how can I show that?
Or what Di I have to show?
 A: If only finitely many $\sigma_j$ are nonzero, the result is not true: in that case, the range of $T$ is finite-dimensional, and so $\mbox{range}(T)=\overline{\mbox{range}(T)}$.
When infinitely many $\sigma_j$ are nonzero (i.e. when $T$ is not of finite rank), we always have the proper inclusion 
$$
\mbox{range}(T)\subsetneq\overline{\mbox{range}(T)}.
$$
To see this, note that since $\sigma_j\to0$, we can choose a subsequence $\sigma_{j_k}$ such that $\sigma_{j_k}<1/k^2$. Now, for each $k$,
$$
v_{j_k}=T\left(\frac1{\sigma_{j_k}}\,u_{j_k}\right)\in\mbox{range}(T).
$$
Consider 
$$
x=\sum_k\sigma_{j_k}v_{j_k}\in Y
$$
(this is well-defined since $\sum_k\sigma_{j_k}^2<\sum_k\frac1{k^2}<\infty$).
Then $x\in\overline{\mbox{range}(T)}$, as the partial sums in the series belong to $\mbox{range}(T)$ (since it is a subspace that contains the $v_{j_k}$). On the other hand,
$$
\sum_j\frac{|\langle x,v_j\rangle|^2}{\sigma_j^2}\geq\sum_k\frac{|\langle x,v_{j_k}\rangle|^2}{\sigma_{j_k}^2}=\sum_k\frac{\sigma_{j_k}^2}{\sigma_{j_k}^2}=\infty,
$$
so $x\not\in\mbox{range}(T)$
