I'm trying to solve the following exercise, I think I solved the first two points but I'm not able to solve the third:

Let $(H, \langle \cdot, \cdot \rangle)$ be a separable Hilbert space and let $T$ be a linear bounded operator from $H$ to itself. Let $\{v_n\}_{n \in \mathbb{N}}$ and $\{e_n\}_{n \in \mathbb{N}}$ be two orthonormal basis of $H$. Prove that

(i) $\sum_{n \in \mathbb{N}} \left \|Te_n \right \| = \sum_{n \in \mathbb{N}} \left \|Tv_n \right \|$

(ii) if $$ \sum_{n \in \mathbb{N}} \left \|Te_n \right \|^2 < \infty \quad \quad \quad (1)$$ then $T$ is compact

(iii) Let $H= L^2([0,1])$ and let $k \in L^2([0,1]^2)$ s.t. $k(t,s)=k(s,t)$ a.e. in $[0,1]^2$, then $$ Ku(t) = \int_0^t k(t,s)u(s)ds$$ is a linear bounded operator, it is symmetric and satisfies (1).

My attempt:


$\langle Te_n, v_n \rangle = \langle Tv_n, e_n \rangle \Rightarrow \left \| \langle Te_n, v_n \rangle v_n \right \|^2 = \left \| \langle Tv_n, e_n \rangle e_n \right \|^2 \Rightarrow \sum_{n \in \mathbb{N}} \left \|Te_n \right \| = \sum_{n \in \mathbb{N}} \left \|Tv_n \right \|$


Since $H$ is reflexive it is enough to show that $T$ is weak-strong continuous. Let $\{b_n \}_{n \in \mathbb{N}} \subset H$ s.t. $b_n \rightharpoonup b \in H$. w.l.o.g. I can assume $a_n \rightharpoonup 0$ (it is enough to take $a_n:=b_n-b$).

I want to show $\left \|Ta_n \right \| \to 0$:

$$ \left \|Ta_n \right \|^2 = \sum_{j \in \mathbb{N}} |\langle Ta_n, e_j \rangle |^2 = \sum_{j \in \mathbb{N}} |\langle a_n, Te_j \rangle |^2$$ Consider the sequence of sequences of real numbers, $x_n^{(j)} = \{\{\langle Ta_n, e_j \rangle\}_{j \in \mathbb{N}}\}_{n \in \mathbb{N}}$: it is a subset of $l^2$ because $\left \|x_n \right \|^2 _{l^2} = \sum_{j \in \mathbb{N}} |\langle a_n, Te_j \rangle |^2 \le \sum_{j \in \mathbb{N}} \left \|a_n \right \|^2 \left \|Te_j \right \|^2 \le \sup_{n \in \mathbb{N}} \left \|a_n \right \|^2 \sum_{j \in \mathbb{N}} \left \|Te_j \right \|^2 = M \sum_{j \in \mathbb{N}} \left \|Te_j \right \|^2 < \infty$ and $x_n^{(j)} < M \left \|Te_j \right \| \quad \forall j \in \mathbb{N}$

Then by dominated convergence we have:

$$\lim_{n \to \infty} \left \|Ta_n \right \|^2 = \lim_{n \to \infty} \sum_{j \in \mathbb{N}} |\langle a_n, Te_j \rangle |^2 = \sum_{j \in \mathbb{N}} (\lim_{n \to \infty} |\langle a_n, Te_j \rangle |^2)=0$$ because $a_n$ converges weakly.

(iii) About symmetry, boundedness and linearity I know what to do. About proving (1), I don't know if I can assume that such an operator is compact (I know it is) but even if I do, I maybe can use that (1) is the same of showing

$$\sum_{n \in \mathbb{N}} \lambda_n^2 < \infty$$

where $\{\lambda_n\}_{n \in \mathbb{N}}$ is the sequence of eigenvalues of $K$. I know $\lambda_n \to 0$ but it is not enough... any hint?


Condition (i) is false without squares. For instance, consider orthonormal bases $\{e_n\}$, and let $v_1=\frac1{\sqrt2}(e_1+e_2)$, $v_2=\frac1{\sqrt2}(e_1-e_2)$, $v_n=e_n$ for $n\geq 3$. Let $T$ be the orthogonal projection onto the span of $e_1$, i.e. $Tx=\langle x,e_1\rangle\,e_1$. Then $$ \sum_n\|Te_n\|=\|Te_1\|=1,\ \ \ \sum_n\|Tv_n\|=\|Tv_1\|+\|Tv_2\|=\frac1{\sqrt2}+\frac1{\sqrt2}=\sqrt2. $$ The equality that holds is with squares, because $$ \sum_n\|Te_n\|^2=\sum_n\langle Te_n,Te_n\rangle=\sum_n\langle T^*Te_n,e_n\rangle=\text{Tr}(T^*T)=\sum_n\langle T^*Tv_n,v_n\rangle=\sum_n\|Tv_n\|^2. $$ I cannot really make sense of your attempt at (i).

For part (ii), weak-strong continuity implies finite-rank, and in an infinite dimensional Hilbert space there are compact operators which are not finite rank. For instance, let $T$ be given by $Te_n=\frac1n\,e_n$ and extend linearly. Then $$ \sum_n\|Te_n\|^2=\sum_n\frac1n<\infty, $$ but $T$ is not finite-rank. What you can do from your condition $(1)$ is to show that $T$ is a limit of finite-rank operators, and thus compact. For this, let $P_n$ be the orthogonal projection onto the span of $e_1,\ldots,e_n$. Then $TP_n$ is finite-rank, and if $x=\sum_nx_ne_n$, then \begin{align} \|(T-TP_n)x\|&=\left\|\sum_m x_mTe_m-\sum_{m=1}^nx_nTe_m \right\| =\left\| \sum_{m=n+1}^\infty x_mTe_m\right\|\\ \ \\ &\leq \sum_{m=n+1}^\infty |x_m|\,\|Te_m\| \leq\left(\sum_{m=n+1}^\infty|x_m|^2\right)^{1/2}\,\left(\sum_{m=n+1}^\infty\|Te_m\|^2\right)^{1/2}\\ \ \\ &\leq\|x\|\,\left(\sum_{m=n+1}^\infty\|Te_m\|^2\right)^{1/2}\\ \ \\ \end{align} Now condition $(1)$ guarantees that the right-hand-side goes to zero as $n\to\infty$ (uniformly!). So $\|T-TP_n\|\to0$, and $T$ is compact.

For $(iii)$, you have, for an orthonormal basis $\{e_n\}$, $$ Ke_n(t)=\int_0^1k(t,s)\,e_n(s)\,ds $$ \begin{align} \sum_n\|Ke_n\|^2 &=\sum_n\int_0^1|Ke_n(t)|^2\,dt\\ \ \\ &=\int_0^1\sum_n|Ke_n(t)|^2\,dt\\ \ \\ &=\int_0^1\sum_n\left|\int_0^t k(t,s)\,e_n(s)\,ds \right|^2\,dt\\ \ \\ &=\int_0^1\sum_n|\langle k(t,\cdot),e_n\rangle|^2\,dt\\ \ \\ &=\int_0^1\|k(t,\cdot)\|_2^2\,dt=\int_0^1\int_0^1|k(t,s)|^2\,ds\,dt\\ \ \\ &=\|k\|_2^2<\infty. \end{align} The exchange of the integral and the sum is justified because the integrand in positive. And the fact that $k(t,\cdot)\in L^2[0,1]$ a.e. follows from the fact that $k\in L^2([0,1]^2)$.


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