1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What does $X$ stand for? $\endgroup$ – Mercy King Nov 2 '13 at 16:57
  • $\begingroup$ Don't bother with weak convergence. Showing that $(Tx_n)$ contains a convergent subsequence for every sequence $(x_n)$ in the unit ball is straightforward enough. $\endgroup$ – Daniel Fischer Nov 2 '13 at 16:58
  • $\begingroup$ It is also easy to prove that this is the norm limit of a sequence of finite rank operators. By the way, your operator is diagonal. $\endgroup$ – Julien Nov 2 '13 at 17:06
  • $\begingroup$ @DanielFischer I want to know how do I find the convergent subsequence, I don't know how to prove that the convergence of those finite rank operators is uniform . $\endgroup$ – Castaroth Nov 2 '13 at 17:24
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How is Bessel's inequality used here? It usually involves sums of squares. $\endgroup$ – Ulrik Nov 10 '17 at 10:20
  • $\begingroup$ @Ulrik maybe he means Cauchy Schwarz? $\endgroup$ – badatmath May 29 at 15:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.