Let $H$ be a hilbert space with the sequence of Orthonormal sets, $\{e_n\}$.
Define $D_A=\{x\in H: \sum\limits_{n=1}^\infty |n \langle x, e_n \rangle|^2 < \infty\}$.
Operator $A:D_A\to H$ such that: $A(e_n)=ne_n$.
Prove that the Graph of $A$ (that is the set $\{x\oplus A(x):x\in H\}$) is closed.
I'm trying to use the elementary definitions of a set being closed. That is if $\{y_j\}$ is a sequence in Graph($A$) that converges to $y$ then $y\in$Graph($A$).
Also I know that each element $x\in H$ can be written as, $x = \sum\limits_{n=1} \langle x, e_n \rangle e_n$. Thus $A(x) = \sum\limits_{n=1}n \langle x, e_n \rangle e_n$.
so we can write $y_j = \sum \langle x_j, e_n \rangle e_n \oplus \sum n \langle x_j, e_n \rangle e_n$.
Now for this product of Banach spaces I use the norm $\|a\oplus b\|_2 = (\|a\|^2+\|b\|^2)^{\frac{1}{2}}$ [I think, that will simplify the things than the other equivalent norms]
Hence $y_j\rightarrow y$ means $\exists j_0$ such that $\forall j\geq j_0,$ $|y_i-y|<\epsilon$.
And one more thing that I can write is that $\|y_j\|^2 = \sum |\langle x_j, e_n \rangle|^2 + \sum |n \langle x_j, e_n \rangle|^2$.
But after that I'm completely lost.
Appreciate your help.
Even a completely different approach is also fine
One other thing That I tried is to see the Graph(A) as $D_A\oplus A(D_A)$.
And to prove that both $D_A$ and $A(D_A)$ are colsed. But it was also not succesful
Let $(x_k)$ and $(Ax_k)$ be sequences in $D(A)$ and $H$, respectively, with $x_k \to x$ and $Ax_k \to y$ in $H$. It remains to show $Ax=y$.
Take $e_n$. Then $\langle Ax_k,e_n\rangle = n \langle x_k,e_n\rangle = \langle y_k,e_n\rangle$. For $k\to \infty$, we obtain $n \langle x,e_n\rangle = \langle y,e_n\rangle$ for all $n$. Hence, $$ \sum_n n^2 |\langle x_n,e_n\rangle |^2 \le \sum_n |\langle y_n,e_n\rangle |^2 < \infty, $$ and $x\in D(A)$. Then $Ax = \sum_n n\langle x_n,e_n\rangle e_n= \sum_n \langle y,e_n\rangle e_n = y$.