I need to prove the following characterization for bounded diagonal linear operator:
For $\mathbb K \in \{\mathbb R, \mathbb C\}$, any $\mathbb K$-Hilbert Space $(H,\langle\cdot,\cdot\rangle,||\cdot||_{H})$, orthonormal basis $\mathbb H \subseteq H$ of $H$, any function $\lambda:\mathbb H \to \mathbb K$, and for every linear operator $A: D(A) \subseteq H \to H$ with $D(A) = \{v \in H \Big| \sum_{h \in \mathbb H}|\lambda_{h}\langle h,v\rangle_{H}|^{2} \lt \infty\}$ an $\forall$ $v \in D(A)$: $Av = \sum_{h \in \mathbb H}\lambda_{h}\langle h,v \rangle h$, it holds that $A\in L(H)$ IFF sup$(\{|\lambda_{h}| \Big |h \in \mathbb H\} \cup \{0\}) \lt \infty$
Now, starting for the proof, suppose assuming that : sup$(\{|\lambda_{h}| \Big |h \in \mathbb H\} \cup \{0\}) \lt \infty$, I need to show $A \in L(H)$.
With this in mind,
$\Big|\Big|Av\Big|\Big|_{H} = \Big|\Big|\sum\limits_{h\in \mathbb H} \lambda_{h} \langle h,v \rangle_{H} h\Big|\Big|_{H} \leq \sum\limits_{h\in \mathbb H} \Big|\lambda_{h}\Big|\Big| \langle h,v \rangle_{H}\Big|\Big|\Big|h\Big|\Big|_{H} \leq \sum\limits_{h\in \mathbb H}\Big|\lambda_{h}\Big|\Big|\Big|h\Big|\Big|_{H}\Big|\Big|v\Big|\Big|_{H}\Big|\Big|h\Big|\Big|_{H} $
[the first inequality beig the usual property for norm and the second one is due to Cauchy-Schwartz],
So now, we have $\Big|\Big|Av\Big|\Big|_{H} \leq \Bigg(\sum\limits_{h\in \mathbb H}\Big|\lambda_{h}\Big|\Big|\Big|h\Big|\Big|^{2}_{H}\Bigg)\Big|\Big|v\Big|\Big|_{H}$
Consequently, $\Big|\Big|A\Big|\Big|_{L(H)} \leq \Bigg(\sum\limits_{h\in \mathbb H}\Big|\lambda_{h}\Big|\Big|\Big|h\Big|\Big|^{2}_{H}\Bigg)$.
So, if we can show that the R.H.S. of last inequality is finite. We are done.
Now, for me, here the trouble arises. Because due to our assumption, we can pull out the $\lambda_{h}$'s outside the summation (dominating by its supremum). Then all I am left out with is : $\sum\limits_{h\in \mathbb H}\Big|\Big|h\Big|\Big|^{2}_{H}$. Which may not be finite. Isn't it ??
Please let me know how to do it from here??
And what about the converse part ??
You are overdoing your estimates. What you have is, because $\mathbb H$ is orthonormal, \begin{align} \|Av\|_{H}^2 &= \left\|\sum\limits_{h\in \mathbb H} \lambda_{h} \langle h,v \rangle_{H} h\Big|\right\|_{H}^2 =\sum_{h\in\mathbb H}|\lambda_h\,\langle h,v\rangle|^2\\ \ \\ &=\sum_{h\in\mathbb H}|\lambda_h|^2\,|\langle h,v\rangle|^2 \leq\sup\{|\lambda_h|^2:\ h\}\,\sum_{h\in\mathbb H}|\langle h,v\rangle|^2\\ \ \\ &=\sup\{|\lambda_h|:\ h\}^2\,\|v\|^2. \end{align}
For the converse, if the supremum is not finite there is a sequence $\{h_n\}$ such that $|\lambda_{h_n}|\to\infty$. Then $$ \|Ah_n\|=|\lambda_{h_n}|\to\infty, $$ and $A$ would not be bounded.