# prove that the operator is compact.

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.

• What does $X$ stand for? – Mercy King Nov 2 '13 at 16:57
• 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. – Daniel Fischer Nov 2 '13 at 16:58
• 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. – Julien Nov 2 '13 at 17:06
• @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 . – Castaroth Nov 2 '13 at 17:24

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.