Proving this operator is bounded and then calculating $||T||$ and $||T^*||$. This question was left as an exercise in my online course of Functional Analysis and I am struck no this.

Question:  Let $H_1$ and $H_2$ be two Hilbert spaces and let $(e_i)_{i\in \mathbb{N}}\subseteq H_1$ and $(f_i)_{i\in \mathbb{N}}\subseteq H_2$ be two orthonormals. Show that if $a = (a_i)_{i\in \mathbb{N}} \in l^{\infty}( \mathbb{N}),$ then the formula $Tx= \sum_{i=1}^{\infty} a_i <x,e_i> f_i$ defines a bounded operator. $T: H_1 \to H_2$. Calculate $||T||$ and determine $T^*$.

Attempt: Bounded operator is equivalent to Continuous and Linear.
I have shown that $T$ is linear but I am not able to show that it's continuous.
$T(x_1 -x_2) =\sum_{i=0}^{\infty} a_i <x_1 ,e_i>f_i -\sum_{i=0}^{\infty} a_i <x_2 ,e_i>f_i $ . I have to show that $|| T(x_1 -x_2)||< \epsilon$ , when I am given that $||x_1- x_2||<\epsilon$. $|| T(x_1 -x_2) ||= ||\sum_{i=0}^{\infty} a_i <x_1 -x_2, e_i> f_i||$. But I am not able to think on how should I use $||x_1 -x_2||<\epsilon$.
Can you please give a couple of hints?
 A: Firstly, we go to show that $T$ is well-defined, i.e., $Tx\in H_{2}$.
For $n\in\mathbb{N}$, let $S_{n}=\sum_{i=1}^{n}a_{i}\langle x,e_{i}\rangle f_{i}$
be a partial sum. We go to prove that $(S_{n})$ is a Cauchy sequence
in $H_{2}$. It will follow that $\sum_{i=1}^{\infty}a_{i}\langle x,e_{i}\rangle f_{i}$
converges. Since $a\in l^{\infty},$ we may choose $M>0$ such that
$|a_{i}|\leq M$ for all $i$. Let $\varepsilon>0$. Recall that $\sum_{i=1}^{\infty}|\langle x,e_{i}\rangle|^{2}\leq||x||^{2}$
(Basel's inequality). Therefore, there exists $N$ such that $\sum_{i=N}^{\infty}|\langle x,e_{i}\rangle|^{2}<\frac{\varepsilon^{2}}{M^{2}}$.
For any $m,n$ with $N\leq m<n$, we have that
\begin{eqnarray*}
||S_{n}-S_{m}||^{2} & = & ||\sum_{i=m+1}^{n}a_{i}\langle x,e_{i}\rangle f_{i}||^{2}\\
 & = & \sum_{i=m+1}^{n}|a_{i}\langle x,e_{i}\rangle|^{2}\\
 & \leq & M^{2}\sum_{i=m+1}^{n}|\langle x,e_{i}\rangle|^{2}\\
 & \leq & M^{2}\sum_{i=N}^{\infty}|\langle x,e_{i}\rangle|^{2}\\
 & < & \varepsilon^{2}.
\end{eqnarray*}
Hence, $||S_{n}-S_{m}||<\varepsilon$ whenever $m,n\geq N$. This
shows that $(S_{n})$ is a Cauchy sequence in $H_{2}$ and hence it
is convergent.
It is routine the verify that $T$ is linear. Finally, we go to show
that $T$ is bounded. By Parseval's equality and Basel's inequality
have that
\begin{eqnarray*}
||Tx||^{2} & = & \sum_{i=1}^{\infty}|a_{i}\langle x,e_{i}\rangle|^{2}\\
 & \leq & M^{2}\sum_{i=1}^{\infty}|\langle x,e_{i}\rangle|^{2}\\
 & \leq & M^{2}||x||^{2}.
\end{eqnarray*}
This shows that $||Tx||\leq M||x||$. Hence, $T$ is a bounded linear map.
We further show that $||T||=||a||_{\infty}$. If $a=(0,0,0,\ldots)$,
then $Tx=0$ for all $x$ and hence $||T||=0=||a||_{\infty}$. Suppose
that $a\neq(0,0,\ldots)$, then $||a||_{\infty}>0$. In the above,
we choose $M=||a||_{\infty}.$ Then, $||T||\leq||a||_{\infty}$. We
go to show that $||T||\geq||a||_{\infty}$. Recall that $||T||=\sup_{y\in H_{1},||y||=1}||Ty||$.
Note that $||e_{i}||=1$, so $||T||\geq||Te_{i}||=||a_{i}f_{i}||=|a_{i}|$.
This shows that $||T||\geq\sup_{i}|a_{i}|=||a||_{\infty}$.
A: I assume that the scalar field for Hilbert spaces $H_{1}$ and $H_{2}$
is $\mathbb{R}$. (If the scalar field is $\mathbb{C}$, then the
inner product is sesquilinear only, instead of bilinear. The following
proof needs adjustment.) Recall that $T^{*}:H_{2}\rightarrow H_{1}$
is a bounded linear map such that $\langle T^{\ast}y,x\rangle=\langle y,Tx\rangle$
for any $x\in H_{1}$, $y\in H_{2}$. Define $S:H_{2}\rightarrow H_{1}$
by $Sy=\sum_{i=1}^{\infty}a_{i}\langle y,f_{i}\rangle e_{i}$. By
the same reasoning, $S$ is a bounded linear map. We go to show that
$S=T^{\ast}$. Let $x\in H_1$ and $y\in H_2$ be arbitrary.
By direct calculation,
\begin{eqnarray*}
\langle y,Tx\rangle & = & \langle y,\sum_{i=1}^{\infty}a_{i}\langle x,e_{i}\rangle f_{i}\rangle\\
 & = & \sum_{i=1}^{\infty}a_{i}\langle x,e_{i}\rangle\langle y,f_{i}\rangle.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\langle Sy,x\rangle & = & \langle\sum_{i=1}^{\infty}a_{i}\langle y,f_{i}\rangle e_{i},x\rangle\\
 & = & \sum_{i=1}^{\infty}a_{i}\langle y,f_{i}\rangle\langle e_{i},x\rangle.
\end{eqnarray*}
Hence, $\langle Sy,x\rangle=\langle y,Tx\rangle=\langle T^{\ast}y,x\rangle$.
That is, $\langle Sy-T^{\ast}y,x\rangle=0$. Since $x\in H_{1}$ is
arbitary, we may put $x=Sy-T^{\ast}y$ and obtain $||Sy-T^{\ast}y||^{2}=0$.
That is, $Sy=T^{\ast}y$ for any $y\in H_{2}$. It follows that $S=T^{\ast}$.
