Integral operator is bounded on $L^p$ if it maps $L^p$ to itself

Question

Here is a homework excercise.

Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space,$1\leq p <\infty.$ and suppose that $k:X\times X\rightarrow \mathbb{C}$ is an $\Omega \times \Omega$ measurable function such that for $f\in L^p(\mu)$ we have $k(x,\cdot)f(\cdot)\in L^1(\mu),a.e.x$ and $(Kf)(x)=\int k(x,y)f(y)d{\mu(y)}$ defines an element $Kf$ of $L^p(\mu)$.

Show that $K:L^p(\mu)\rightarrow L^p(\mu)$ is a bounded operator.

I think we can use the Closed Graph Theorem. suppose that $f_n\rightarrow 0,Kf_n\rightarrow g$ in $L^p$,we only need to prove $g=0$ in $L^p$. Since $Kf_n\rightarrow g$ in $L^p$, then without loss of generality, we can get $Kf_n(x)\rightarrow g(x)$ a.e. Then I want to show $Kf_n(x)\rightarrow 0$ a.e. using $f_n\rightarrow 0$ in $L^p$. But I need to prove that $k(x,\cdot)\in L^q(\mu)$ a.e. $x$ ($1/p+1/q=1$).

How to prove $k(x,\cdot)\in L^q(\mu)$ a.e. $x$?

Answer 1

Let $A_x:L^p(\mu)\rightarrow L^1(\mu)$ be defined by $A_xf(y) = k(x,y)f(y)$ whenever $x$ is chosen such that $A_xf\in L^1$. We show that this map is bounded by applying the Closed Graph Theorem. Suppose that $f_n\rightarrow 0$ in $L^p$ and that $A_xf_n\rightarrow g$ in $L^1$. Now by taking a subsequence if necessary we can assume that $f_n\rightarrow 0$ $\mu$-almost everywhere but then $A_xf_n\rightarrow 0$ $\mu$-almost everywhere so that $g = 0$. By the Closed Graph Theorem we therefore conclude that there is a $C_x>0$ such that \begin{equation*} \int_{X}|k(x,y)f(y)|d\mu(y)\leq C_x\left(\int_{X}|f(y)|^pd\mu(y)\right)^{1/p} \end{equation*} for $\mu$-almost every $x\in X$. If we define a linear functional $l_x$ for every $x$ outside the exceptional set via $$l_x(f) = \int_{X}k(x,y)f(y)d\mu(y)$$ we find that this defines a continuous linear functional by the previous reasoning and therefore $l_x\in (L^p)^\ast$. Since the dual of $L^p$ is $L^q$ where $\frac{1}{q}+\frac{1}{p} = 1$ we can find a function $g_x\in L^q(\mu)$ such that $l_x(f) = \int_{X}g_x(y)f(y)d\mu(y)$ but this implies that \begin{equation*} \int_{X}k(x,y)f(y)d\mu(y) = \int_{X}g_x(y)f(y)d\mu(y) \end{equation*} for every $f\in L^p(\mu)$. This is only possible if $k(x,y) = g_x(y)$ for $\mu$-almost every $y\in X$ and $\mu$-almost every $x\in X$. Thus $y\mapsto k(x,y)$ belongs to $L^q$ for $\mu$-almost every $x$ in $X$.