Let $\{e_k\}_{k=1}^\infty$ denote an orthonormal system. Show that $ T\{c_k\}_{k=1}^\infty = \sum_{k=1}^\infty c_ke_k$ is well-defined and bounded

Question

Let $\{e_k\}_{k=1}^\infty$ denote an orthonormal system in $\mathcal{H}$. Show that the following operator $T$ is (i) well-defined and (ii) bounded$$ T \; : \; \ell^2 \rightarrow \mathcal{H}$$ \begin{equation} T\{c_k\}_{k=1}^\infty = \sum_{k=1}^\infty c_ke_k\end{equation}

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My current suggestion:

i) Well-defined: Need to show that $\sum_{k=1}^\infty c_ke_k$ converges. Let $n,m \in \mathbb{N}$, $n>m$ then \begin{align} \| \sum_{k=1}^n c_ke_k-\sum_{k=1}^m c_ke_k\| & = \|\sum_{k=m+1}^n c_ke_k\| \\ & = \sup_{\|v\| = 1} | \langle\sum_{k=m+1}^n c_ke_k,v \rangle | \\ & \leq \sup_{\|v\| = 1} \sum_{k=m+1}^n| c_k \langle e_k,v \rangle | \\ & \leq (\sum_{k=m+1}^n |c_k|^2 )^{1/2} \sup_{\|v\| = 1} (\sum_{k=m+1}^n| \langle e_k,v \rangle |^2)^{1/2} \\ & \leq (\sum_{k=m+1}^n |c_k|^2 )^{1/2} \end{align} The thing that I'm not sure about here, is the last step. My idea is that since both $\|e_k\| = 1$ and $\|v\| = 1$, then their inner product must be 1, but as said I'm not sure. Final argument is then that $\{c_k\} \in \ell^2$ thus it converges.

ii) I guess it is basically the same calculations as above?

Answer 1

Since $\{e_k\}$ is orthonormal, you have $$ \left\|\sum_{k=m+1}^n c_ke_k\right\|^2=\sum_{k=m+1}^n|c_k|^2, $$ and the $\ell^2$ condition then guarantees that your series converges.

As for the second question, not only is $T$ bounded, but it is isometric, with the same computation: $$ \|Tc\|^2=\left\|\sum_{k=1}^\infty c_ke_k\right\|^2=\sum_{k=1}^\infty|c_k|^2=\|c\|^2. $$