prove that K is Lp- bounded operator

Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space, $1\leq p <\infty$, and suppose that $k:X\times X\rightarrow \mathbb{F}$ is an $\Omega \times \Omega$ measurable function such that for $f$ in $L^p(\mu)$ and a.e x, $k(x,\cdot)f(\cdot)\in L^1(\mu)$ 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 that this problem is similar with young's inequality. But, I don't know where $k(x,\cdot)f(\cdot)\in L^1(\mu)$ condition applies. How to prove that??

• But, this problem has some different with Theorem 6.18... – asfajaf May 17 '16 at 14:57
• The integrability condition is needed so that the definition of $(Kf)(x)$ makes sense. Also, this is the same question as Integral operator is bounded on a certain condition. – user147263 May 17 '16 at 20:27