How to show a sequence of finite-rank operators converges uniformly

Question

Let $G\in L^2(0,1)^2$. Define a linear mapping $A:L^2(0,1)\mapsto L^2(0,1)$ such that
$$
Af(x):=\int_0^1G(x,\xi)f(\xi)d\xi,\quad \forall f\in L^2(0,1).
$$
Since $L^2(0,1)$ is a separable Hilbert space, one can choose a Hilbert basis of $L^2(0,1)$ denoted by $\{e_i\}_{i=1}^{\infty}$. Define finite-rank operators $T_n(n\in\mathbb{N}^{+})$ as follows
$$
T_nf:=\sum\limits_{i=1}^{n}\left\langle f,e_i\right\rangle e_i,\quad f\in L^2(0,1),
$$
where $\left\langle\cdot,\cdot\right\rangle$ denotes the usual inner product in $L^2(0,1)$. How to prove that
$$
\lim\limits_{n\rightarrow\infty}||T_nAT_n-A||=0,
$$
where $||\cdot||$ denotes the operator norm induced by the norm on $L^2(0,1)$.

Answer 1

It is well-known, that $G\in L^2((0,1)\times (0,1))$ implies that $A$ is a compact operator. As we are on a separable Hilbert space by, e.g., Lemma 3.3.c in this paper, all that is left to show that $T_n$ converges strongly to the identity (because then $T_nAT_n\to \operatorname{id}A\operatorname{id}=A$ in norm as you wanted).

Now $T_n$ is said to converge strongly to $\operatorname{id}$ if for all $x\in L^2$, one has $\lim_{n\to\infty}\|T_nx-x\|_{L^2}=0$. This however holds in arbitrary separable infinite-dimensional Hilbert spaces as is easily seen using the Fourier expansion and Parseval's identity.