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:
(i)
$\langle Te_n, v_n \rangle = \langle Tv_n, e_n \rangle \Rightarrow \left \| \langle Te_n, v_n \rangle v_n \right \|^2 = \left \| \langle Tv_n, e_n \rangle e_n \right \|^2 \Rightarrow \sum_{n \in \mathbb{N}} \left \|Te_n \right \| = \sum_{n \in \mathbb{N}} \left \|Tv_n \right \|$
(ii)
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?