# Show that an operator is weakly compact

If $(X,\Omega,\mu)$ is a finite measure space, $k\in L^\infty(X\times X, \Omega\times \Omega,\mu \times \mu)$ , and $K:L^1(\mu)\to L^1(\mu)$ is defined by $$(Kf)(x)=\int k(x,y) f(y) d\mu(y)$$ show that $K$ is weakly compact and $K^2$ is compact.

My attempt: suppose $\{f_n\}$ is a bounded sequence in $L^1(\mu)$, then the sequence $\{(Kf_n,g)\}$ is a bounded sequence in ${\Bbb C}$, where $g\in L^\infty(\mu)$. Thus there is a subsequence $\{f_{n_i}\}$ such that the sequence $\{(Kf_{n_i},g)\}$ converges. so $K$ is weakly compact.

To show that $K^2$ is compact, I show that the sequence $\{K^2f_n\}$ is cauchy.(where $\{Kf_n\}$ is my subsequence in before part) we have $$||K^2f_n - K^2f_m||_1\leq \int \int |k(x,y)(Kf_n(y) - Kf_m(y))|d\mu(y)d\mu(x)$$ $k_x(y):=k(x,y)\in L^\infty(\mu)$, so using before part, for $\epsilon>0$, there is $n_0$ such that for $n,m>n_0$, $\int |k(x,y)(Kf_n(y) - Kf_m(y))|d\mu(y)< \epsilon/\mu(X)$. therefore $$\int \int |k(x,y)(Kf_n(y) - Kf_m(y))|d\mu(y)d\mu(x) < \epsilon$$ which shows $K^2$ is compact.

• @niki.Have you got an correct answer? – David Lee Feb 26 '16 at 13:25

In the first part, you showed that if $\{ f_{n} \}$ is a bounded sequence in $L^{1}_{\mu}$, then $\langle Kf_{n} | g \rangle$ has a convergent subsequence for a $g \in L^{\infty}_{\mu}$, where $\langle\cdot|\cdot\rangle$ denotes the duality pairing. You forgot that the subsequence $\langle Kf_{n} | k_{x}\rangle$ may be different for each $x$ in the second part, and ignored the issue of a.e. with respect to $x$. So you asserted the existence of a subsequence that works for all such $x$. You compounded the problem by assuming that for the common subsequence which you have not shown to exist, that $$|\langle Kf_{n_{j}}|k_{x}\rangle-\langle Kf_{n_{k}}| k_{x}\rangle|\rightarrow 0 \;\;\;\mbox{ as }\; k,j\rightarrow \infty$$ somehow implied uniformity of convergence with respect to $x$ of $$\langle \;|Kf_{n_{j}}-Kf_{n_{k}}|\;|\; |k_{x}|\; \rangle|\rightarrow 0 \;\;\;\mbox{ as }\; k,j\rightarrow \infty.$$
Firstly,we claim that $K(Ball(L^1(\mu)))$ is weakly closed.
Suppose a net $\{f_j\}_{j\in J}\subset Ball(L^1(\mu))$ and $K(f_j)\rightarrow g$ in $L^1(\mu)$. We need to prove there is a $h\in Ball(L^1(\mu)),s.t, g=K(h).\forall \phi\in L^{\infty}(\mu)$, using Fubini theorem, $$\int_{X}(K(f_j)-g)\bar{\phi}dm=\int_{X}f_j(x)\int_{X}k(x,y)\bar{\phi}(y)dm(y)dm(x)-\int_{X}g(y)\bar{\phi}(y)dm(y) ~~~~~~~~(1)$$ It is easy to check that $\int_{X}k(x,y)\bar{\phi}(y)dm(y)\in C(X)$ and $L^1(\mu)\subset {C(X)}^{*}$.
By Alaoglu's theorem, $Ball(L^{1}(\mu))\subset Ball(C(X)^{*})$ which is compact in weak* topology.It follows that there exist a subnet $\{f_i\}_{i\in I}\subset \{f_j\}_{j\in J}$ and $\nu\in Ball(M(X)),s.t. f_i\rightarrow \nu$ in weak* topology of $C(X)^{*}$. Thus, $$\int_{X}f_i\int_{X}k(x,y)\bar\phi d\mu d\mu\rightarrow\int_{X}\bar\phi\int_{X}k(x,y)d\nu d\nu~~~~~~~~~~~~~~~~~(2)$$ Since $\phi$ is arbitrary, combining (1)(2), we get $$g=\int_{X}k(x,y)d\nu$$ Since $\nu\in Ball(M(X))$, using Radon's theorem, $\exists h\in Ball(L^{1}(\mu))$,s.t. $$g=\int_{X}k(x,y)hd\mu=K(h)$$ Secondly, using the same trick as above, one can easily prove $K(Ball(L^{1}(\mu)))$ is weakly compact.