Let $H$ be a Hilbert space over $\mathbb C$, and $\{f_j\}$ a orthonormal set in $H$. Let $t_j\in \mathbb C$ such that $\displaystyle \lim_{n\to \infty} t_j =0$ i.e $(t_j)_{j\in \mathbb N}\in c_0$. Show that the operator $T:H\to H$ defined by:
$Tx=\sum t_j (f_j \cdot x)f_j$ is compact.
It's easy to prove if $X$ is a reflexive space, then an operator in compact if and only if carries weakly convergent into norm convergent sequences(and it's enough the case weakly convergent to zero). If $H$ is a Hilbert space, then it's reflexive.
Let $x_j \to 0$ weakly. I don't know how to do this problem, because I can bound $||Tx||\le ||x|| \sum |t_j|$ but the right side could be infinite.
Best Answer
For each $n\in \mathbb{N}$, define $$ T_n(x) = \sum_{j=1}^n t_j (f_j\cdot x)f_j $$ Then, $T_n$ is finite rank, and hence compact. Now consider $$ \|T(x)-T_n(x)\| \leq \sum_{j=n+1}^{\infty} |t_j||(f_j\cdot x)| \leq u_n\|x\| $$ by Bessel's inequality, where $$ u_n = \sup_{j\geq n} |t_j| $$ Now, $$ \lim u_n = \limsup |t_n| = 0 $$ and hence for $\epsilon > 0$, there is $N_0 \in \mathbb{N}$ such that $u_n < \epsilon$ for all $n\geq N_0$, in which case $$ \|T-T_n\| < \epsilon \quad\forall n\geq N_0 $$ Hence, $T$ is the limit of finite rank operators, and is hence compact.