Application of Fubini's Theorem - showing an integral-defined operator on $L^p$ outputs measurable functions

Question

I am trying to understand the solution to problem 79 from chapter 7 of Alberto Torchinsky's Problems in Real and Functional Analysis. The problem is:

Let $(X,\mathcal{M}, \mu)$ and $(Y,\mathcal{N},\nu)$ be $\sigma$-finite measure spaces, and $k$ an $\mathcal{M}\otimes\mathcal{N}$-measurable function on $X\times Y$ such that $$\int_{X}|k(x,y)|d\mu(x)\leq A\text{ for }\nu\text{-a.e. }y\in Y$$ and $$\int_{Y}|k(x,y)|d\nu(y)\leq B\text{ for }\mu\text{-a.e. }x\in X$$ for some $A,B<\infty$. Put $Tf(x)=\int_{Y}k(x,y)f(y)d\nu(y),$ and prove that if $1<p<\infty,$ $T$ is well-defined on $L^p(X)$ and with $q$ the conjugate exponent to $p$, $||Tf||_p\leq A^{1/p}B^{1/q}||f||_p.$

The solution is given in the book, and while I can understand the applications of Holder's Inequality and Fubini's Theorem, there is one assertion that I do not follow, which is in bold below. Here is the solution:

Take $f\in L^p(Y)$ and $g\in L^q(X)$ with $||g||_q\leq 1$.

Since $(x,y)\mapsto |k(x,y)||f(y)|^p$ defines a measurable function on $X\times Y,$ we may put $I:=\left(\int_{X\times Y}|k(x,y)||f(y)|^pd(\mu\otimes\nu)(x,y)\right)^{1/p}.$ An application of Tonelli's Theorem then gives $I\leq A^{1/p}||f||_p.$

Similarly $J:=\left(\int_{X\times Y}|k(x,y)||g(x)|^qd(\mu\otimes\nu)(x,y)\right)^{1/q}\leq B^{1/q}||g||_q\leq B^{1/q}.$

Thus by Holder's Inequality we have $$\int_{X\times Y}|k(x,y)||f(y)||g(x)|d(\mu\otimes\nu)(x,y)\leq IJ\leq A^{1/p}B^{1/q}||f||_p<\infty$$

and so by Fubini's Theorem $\int_{Y}k(x,y)f(y)g(x)d\nu(y)$ exists for $\mu$-a.e. $x\in X$ and defines a measurable and integrable function on $X.$ Since $g$ is arbitrary, $Tf(x)=\int_{Y}k(x,y)f(y)d\nu(y)$ is measurable and finite almost everywhere on $X$.

The fact that $Tf\in L^p(X)$ and satisfies $||Tf||_p\leq A^{1/p}B^{1/q}||f||_p$ follows by more applications of Tonelli's Theorem and Holder's Inequality. However I do not understand the assertion in bold, it seems that we introduced the function $g$ purely so that we could apply Holder's inequality and obtain the finiteness necessary to apply Fubini's Theorem, but what is the generalization being made that allows $g$ to disappear so that we can talk about $T$?

Any help is appreciated!

Answer 1

I think I can see now how one might apply Fubini's Theorem to show that $Tf$ is measurable for each $f\in L^p(X).$ For starters the argument in the question without the additional assumption $||g||_q\leq 1$ shows that

$$\int_{X\times Y}|k(x,y)f(y)g(x)|d(\mu\otimes\nu)(x,y)\leq A^{1/p}B^{1/q}||f||_p||g||_q<\infty$$

for every $g\in L^q(X).$ In particular if $\mu(X)<\infty,$ then one can take $g\equiv 1$ to show that $k(x,y)f(y)\in L^1(X\times Y),$ and conclude by Fubini's Theorem that

$$Tf(x)=\begin{cases}\int_{Y}k(x,y)f(y)d\nu(y), &\text{if }k(x,\cdot)f(\cdot)\in L^1(Y)\\ 0, &\text{otherwise} \end{cases},$$

is an $\mathcal{M}$-measurable function on $X,$ noting that $k(x,\cdot)f(\cdot)\in L^1(Y)$ for $\mu$-a.e. $x\in X$ by Fubini's Theorem.

In the general case, write $X$ as a countable disjoint union of measurable sets with finite measure: $X=\bigsqcup_{n\geq 1}A_n,$ $\mu(A_n)<\infty.$ For each $n,$ put $g_n:=\chi_{A_n}\in L^q(X)$. Then because $k(x,y)f(y)g_n(x)\in L^1(X\times Y),$ it follows by Fubini's Theorem that there exists $B_n\in\mathcal{M}$ with $\mu(B_n)=0$ and $k(x,\cdot)f(\cdot)g_n(x)\in L^1(Y)$ for all $x\in X\setminus B_n.$ Put

$$G_n(x)=\begin{cases} \int_{Y}k(x,y)f(y)g_n(x)d\nu(y), &\text{if }x\in X\setminus B_n\\[3ex] 0, &\text{otherwise} \end{cases}=\begin{cases} \int_{Y}k(x,y)f(y)d\nu(y), &\text{if }x\in A_n\setminus B_n\\[3ex] 0, &\text{otherwise} \end{cases}$$

which is a measurable function on $X$ and let $B:=\bigcup_{n\geq 1}B_n,$ noting $\mu(B)=0.$ Then if $x\in X\setminus B,$ there is a unique $n$ such that $x\in A_n\setminus B_n,$ and so we may put $Tf(x)=\int_{Y}k(x,y)f(y)d\nu(y)=G_n(x).$ Really then $Tf=\sum_{n\geq 1}G_n$, which is the pointwise limit of measurable functions and is therefore measurable.

Answer 2

It seems that there is a minor error in the solution of the author on on your transcribe ing of it. The norm of $T$ seems to be $A^{1/q}B^{1/p}$ as opposed to $A^{1/p}B^{1/q}$.

It is a well known result that if $\phi$ is a measurable function in $(Y,\mathcal{F},\nu)$ and such that $\sup_{\|\psi\|_{L_q(\nu)}=1}\Big|\int_Y \phi\psi\,d\nu\Big|=M<\infty$, then $\phi\in L_p(\nu)$ and $\|\phi\|_{L_p(\nu)}=M$.

This is what the author of the text you are reading sets on using to estimate $\|Tf\|_{L_p(\nu)}$: For any $f\in L_p(\mu)$ and $g\in L_q(\nu)$, where $\frac1p+\frac1q=$, ($1<p<\infty$)

\begin{align} \int_Y&|Tf(y)||g(y)|\,\nu(dy)\leq\int_{X\times Y}|k(x,y)||f(x)||g(y)|\,\mu(x)\nu(dy)\tag{0}\label{zero}\\ &=\int_{X\times Y}|k(x,y)|^{1/p}|f(x)|\,|k(x,y)|^{1/q}|g(y)|\,\mu(dx)\nu(dy)\tag{1}\label{one}\\ &\leq\Big(\int_{X\times Y}|k(x,y)||f(x)|^p\,\mu(dx)\nu(dy)\Big)^{1/p}\Big(\int_{X\times Y}|k(x,y)||g(y)|^q\,\mu(dx)\nu(dy)\Big)^{1/q}\tag{2}\label{two}\\ &\leq\Big(\int_X\Big(\int_Y|k(x,y)|\,\nu(dy)\Big)\,|f(x)|^p\mu(dx)\Big)^{1/p}\Big(\int_Y\Big(\int_X|k(x,y)|\,\mu(dx)\Big)\,|g(y)|^q\nu(dy)\Big)^{1/q}\tag{3}\label{three}\\ &\leq B^{1/p}\|f\|_{L_p(\mu)}A^{1/q}\|g\|_{L_q(\nu)} \end{align} In \eqref{zero} and \eqref{one} we apply the triangle inequality and Fubini-Tonelli's theorem. In \eqref{two} we apply Hölder's inequality where the measure involved in integration is $(\mu\otimes\nu)(dx,dy)$. In \eqref{three} we apply the assumptions in the problem.

Hence $\|Tf\|_{L_p(\nu)}\leq A^{1/q}B^{1/p}\|f\|_{L_p(\mu)}$. This shows that $T$ is a bounded operator from $L_p(\nu)$ into $L_p(\nu)$ with norm $\|T\|\leq A^{1/q}B^{1/p}$.