Proving this operator is bounded and then calculating $||T||$ and $||T^*||$.

Question

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?

Answer 1

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 $||T||\geq\sup_{i}|a_{i}|=||a||_{\infty}$.

Answer 2

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 $S=T^{\ast}$.