fa.functional analysis – A sequence of compact operators, weak convergence versus strong convergence

Let $(X,mu)$ be a finite measure space.

We consider two sequences of bounded linear operators ${T_n}_{n=1}^infty$ and ${S_n}_{n=1}^infty$ on $L^2(X,mu)$. We denote by $mathcal{L}$ the space of bounded linear operators on $L^2(X,mu)$.
We assume that ${T_n}_{n=1}^infty$ strongly converges to some $T in mathcal{L}$, and ${S_n}_{n=1}^infty$ weakly converges to $S in mathcal{L}$. Then, it is easy to show that ${T_nS_n}_{n=1}^infty$ weakly converges to $TS$ in $L^2(X,mu)$.

In general, ${T_nS_n}_{n=1}^infty$ does not converge to $TS$ strongly in $L^2(X,mu)$.

Then, we assume that each $T_n$ possesses a non-negative integral kernel. That is, for $n in mathbb{N}$, there exists $p_n=p_n(x,y)colon X times X to (0,infty)$ such that
begin{align*}
T_nf(x)=int_{X}p_n(x,y)f(y),mu(dy),quad f in L^2(X,mu),quad mutext{-a.e. }xin X.
end{align*}

We assume moreover that there exists $C>0$ such that for any $n in mathbb{N}$,
begin{align*}
mutext{-ess.sup}_{x,y in X},p_n(x,y) in (0,C).
end{align*}

Then, each $T_n$ is of Hilbert–Schmidt type (in particular, $T_n$ is a compact operator on $L^2(X,mu)$). Also, in this situation, ${T_n}_{n=1}^infty$ is expected to have better properties.

If $u in L^2(X,mu)$ is bounded and $sup_{n in mathbb{N}}S_nu$ is a bounded function, can we show that
begin{align*}
lim_{n to infty}int_{X}|T_nS_n u-TSu|^2,dmu=0 ,?
end{align*}

This of course implies that $T_nS_nu$ strongly converges to $TSu$ in $L^2(X,mu)$.