# 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 $$L^2(X,mu)$$.