Hilbert spaces - linear and bounded operator $X$ and $Y$ are two Hilbert spaces. $A: X\longrightarrow Y$ is linear and bounded operator.
How to Show that $ImA$ the image of A is not closed in general ( I'd like an example)?

Let $X=Y$ have orthonormal basis $e_1, e_2, \dots$.  Then let $A(e_k) = \frac{1}{k} e_k$.
Every $e_k$ is in the image, so the image is dense.  Also
$$
x = \sum_{k=1}^\infty \frac{1}{k}e_k
$$
is in $X$ because $\sum\frac{1}{k^2}$ converges.  But this $x$ is not in the image of $A$, since the only choice $y$ with $A(y) = x$ would be
$$
y = \sum_{k=1}^\infty e_k
$$
which is not in $X$ because $\sum 1^2$ diverges.
