On the compactness of the square of a finite double norm integration operator $T$ on $L^1 (\mu)$.

Question

Good morning everyone, I have read in S. P. Eveson, Compactness Criteria for Integral Operators in L∞ and L1 Spaces, Proceedings of the American Mathematical Society 123, 1995, 3709-3716 :

"If $(S, \Sigma, \mu)$ is a positive measure space, $k: S \times S \rightarrow \mathbb{R}$ is a measurable function, and $p, q > 1$ with $p^{-1} + q^{-1} = 1$, one may define the `double norm' of $k$ on $L^p(S, \Sigma, \mu)$ by

$$ \|k\|^p = \int_S \left( \int_S |k(s, t)|^q \, d\mu(t) \right)^{p/q} \, d\mu(s). $$

If this double-norm is finite, then the integral operator induced by $k$ is a compact transformation on $L^p(S, \Sigma, \mu)$ (see Zaanen, "Linear Analysis" [6, Chapter 11, §2, Example D]).

One may make the obvious generalisations of the double-norm to the spaces $L^1(S, \Sigma, \mu)$ and $L^\infty(S, \Sigma, \mu)$, but in these cases integral operators of finite double-norm, although bounded, are not necessarily compact. In the case of an integral operator $T$ of finite double-norm on $L^1(S, \Sigma, \mu)$, $T^2$ turns out to be compact ([6, Chapter 11, §2, Example E])."

I would like to have the proof that $T^2$ is compact in this case, and eventually if we consider that the $L^1$ spaces are different: operator from $L^1(S, \Sigma, \mu)$ to $L^1(S', \Sigma', \mu')$. But Zaanen's book is a bit old and difficult to access. Does someone have an idea of the proof or another reference?

Thank you very much for your help,

Answer 1

The proof that I know relies on the Eberlein-Smulain theorem, and to be more precise, on the Dunford-Pettois theorem. The issue is that to determine compactness of operators on Banach spaces, it is often easier to look weak compactness. In $L_1$ weak-compactness is related to uniform integrability, which is not a trivial sort of result.

Here I present a solution based on a series of problems that I worked on in my graduate student days.

In Zaanen's, the assumption is that $(S,\mathcal{S},\mu)$ is separable, that is, there is accountable family of integrable sets $\mathcal{C}$ such that $\sigma(\mathcal{C})=\mathscr{S}$ and that $S=\bigcup_nC_n$ for some $C_n\in\mathcal{C}$. This in particular, implies that $L_1(\mu)$ (in fact any $L_p(\mu)$ with $1\leq p<\infty$) is separable. This, in the case $1<p<\infty$, implies that any ball $B(0;a)$ in $L_p(\mu)$ is metrizable in $\sigma(L_p(\mu),L_q(\mu))$.

For the case $p=1$ we only assume $\sigma$-finiteness. Thus, it is enough to consider the case where $\mu$ is finite. The main assumption on $T$ is that $C=\int_X\|T(x,\cdot)\|_{L_\infty(\mu)}\,\mu(dx)<\infty$. Now, consider the ball $B(0;a)\subset L_1(\mu)$. Then $$ |Tf(x)|\leq\|T(x,\cdot)\|_{L_\infty(\mu)}\|f\|_{L_1(\mu)}$$ and so $\mathcal{F}=\{Tf:f\in B(0;a)\}$ is uniform integrable, that is,

• $\mathcal{F}$ is bounded in $L_1(\mu)$ and
• For any $\varepsilon>0$, there is $\delta>0$ such that $\mu(A)<\delta$ implies that $$\sup_{h\in\mathcal{F}}\int_A|h|\,d\mu<\varepsilon. $$

Since $L_1(\mu)$ is finite (or $\sigma$-finte), $(L_1(\mu))^*=L_\infty(\mu)$. By the Dunford-Pettis theorem, $\mathcal{F}$ has compact closure in the weak topology $\sigma(L_1(\mu),L_\infty(\mu))$, and by the Eberlein-Smulian theorem, $\mathcal{F}$ is in fact $\sigma(L_1(\mu),L_\infty(\mu))$-sequentially compact. Hence, for any sequence $(h_n:n\in\mathbb{N})\subset\mathcal{F}$, there is a subsequence $h_{n_k}$ and a function $h_0$ such that $h_{n_k}\xrightarrow{k\rightarrow\infty}h_0$ in $\sigma(L_1(\mu),L_\infty(\mu))$.

We may assume without loss of generality that for all $x\in S$, $T_x:y\mapsto T(x,y)$ is in $L_\infty(\mu)$. Let $(f_n:n\in\mathbb{N})\subset B(0;a)$ and let $h_n=Tf_n$. By the arguments above, there is a subsequence $h_{n_k}$ and a function $h_0\in L_1(\mu)$ such that $$g_k(x):=T^2f_{n_k}(x)=\int_S T(x,y) h_{n_k}(y)\,\mu(dy)\xrightarrow{k\rightarrow\infty}\int_S T(x,y) h_0(y)\,\mu(dy)=: g_0(x)$$ Since $T$ maps $L_1(\mu)$ into $L_1(\mu)$, $g_0,g_k\in :_1(\mu)$ for all $k\in\mathbb{N}$. Furthermore, outside a set of $\mu$-measure $0$, $$|g_k(x)-g_0(x)|\leq \|T(x,\cdot)\|_{L_\infty(\mu)}\|h_{n_k}-h_0\|_{L_1(\mu)}\leq (a+\|h_0\|_1)\|T(x,\cdot)\|_{L_\infty(\mu)}\in L_1(\mu)$$ By dominated convergence $$\|g_k-g_0\|_{L_1(\mu)}\xrightarrow{k\rightarrow\infty}0$$ All this shows that $T^2(B(0;a))$ is sequentially compact in $L_1(\mu)$ and thus compact; therefore, $T^2$ is a compact operator.

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Comments:

• The key ingredients of the proof is the Dunford-Pettis theorem that states that in a $\sigma$--finite measure space $(S,\mathscr{S},\mu)$, a set $K\subset L_1(\mu)$ is $\sigma(L_1(\mu),L_\infty(\mu))$ compact (and also sequentially compact) iff the set $K$ is uniformly integrable.

• The book of Zaanen proves a version of Dunford-Pettis theorem as a Lemma in the presentation of Example E in Chapter 11, §2. It also provides an example, due to von Neumann, that $T$ may fail to be compact when $p=1$.

• For a treatment of the Dunford-Pettis theorem, based on the Eberlien-Smulian theorem one can check Bogachev, V. I., Measure Theory, Vol I, Springer-Verlag, Berlin Heidelberg, 2007, pp. 285-286.

• @PhoemueX provides here a much simpler example of an operator $T$ of the type described above in which $T$ which fails to be compact: Fix $g\in L^+_1((\mathbb{R},\mathscr{B}(\mathbb{R}),\lambda)\setminus\{0\}$, where $\lambda$ is Lebesgue measure. Define $$T(x,y)=g(x)\sum_{n\geq0}\mathbb{1}_{(n,n+1]}(y)e^{inx}$$ Then $\|T(x,\cdot)\|_\infty\leq g(x)\in L_1$. Notice that $$T\mathbb{1}_{(n,n+1]}(x)=g(x) e^{in x}, \qquad n\in\mathbb{Z}_+$$ converges to $0$ in $\sigma(L_1(\lambda),L_\infty(\lambda))$ (by virtue of the Riemann-Lebegue lemma). As $\|\mathbb{1}_{(n,n+1]}\|_1=1$ and $\|T\mathbb{1}_{(n,n+1]}\|_1=\|g\|_1>0$, $T$ cannot be compact.