Integral operator on $L^p$ The problem is : (Conway's exercise in 'The Open Mapping and Closed
Graph Theorems' section)
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*Let $(X, \Omega, \mu)$ be a $\sigma$ -finite measure space, $1 \leqslant p<\infty,$ 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 $(K f)(x)=\int k(x, y) f(y) d \mu(y)$ defines an element $K f$ of $L^{p}(\mu) .$ Show that $K: L^{p}(\mu) \rightarrow L^{p}(\mu)$ is a bounded operator.
  

My idea is if $\sup_{x \in X} \int  k(x,y) d\mu(y) \leq C$ and $\sup_{y \in X} \int  k(x,y) d\mu(x) \leq C$, then I can prove that $K$ is a bounded operator.
But I can't justify above my claim. 
How can I show this?
Or, is this method wrong?
 A: Here is a long winded proof (the proof that @MaoWao linked to is more direct) that uses uniform boundedness. It is not as direct as I thought when I wrote my comment above.
The key tools used here are the closed graph theorem and the Banach Steinhaus theorem. The application of Banach Steinhaus comes from a delightful paper "On a converse of the Holder inequality" by E. Leach in the Proceeding of the AMS 7 (1956), pp. 607-608 (link).
Suppose $f_n \to f$ and $K f_n \to g$, we want to show that $g=Kf$. It is sufficient to show that $g(x) = (Kf)(x) $ for ae. $x$.
One minor subtlety to address first: For each $h \in L^p$ there is some set $N_h$ of measure zero such that $y \mapsto k(x,y)h(y)$ is integrable for $x \notin N_h$. Let $N = N_f \cup \bigcup_n N_{f_n} $ and note that $N$ has measure zero. 
Choose $x \notin N$.
Define a linear functional on $L^p$ by $l_x (h) = (K h) (x)$, if we can show that $l_x$ is continuous, then since $f_n \to f$ we have $K f_n(x) = l_x(f_n) \to l_x (f) = K f(x)$ and so $g(x) = (Kf)(x) $.
Let $\lambda(y) = k(x,y)$, then $l_x (h) = \int \lambda(y) h(y) d \mu(y)$.
Since $\lambda$ is measurable and the space is $\sigma$-finite, there is a sequence of bounded measurable functions $\lambda_n$ supported on a set of finite measure such that $|\lambda_n(y)| \le |\lambda_{n+1}(y)| \le |\lambda(y)|$ and $\lambda_n(y) \to \lambda(y)$ for all $y$.
Define $l_n(h) = \int \lambda_n(y) h(y) d \mu(y)$ and
note that for any fixed $h \in L^p$ we have
$|l_n(h)|=|\int \lambda_n(y) h(y) d \mu(y)| \le \int |\lambda_n(y) || h(y)| d \mu(y) \le \int |\lambda(y) || h(y)| d \mu(y) < \infty$.
Banach Steinhaus shows that the collection $l_n$ is uniformly bounded and there is
some $M$ such that $\|l_n\| \le M$ or equivalently, $\int |\lambda_n(y)|^q d \mu(y) \le M^q$. Now the monotone convergence theorem shows that
$\int |\lambda_n(y)|^q d \mu(y) \uparrow \int |\lambda(y)|^q d \mu(y)$ and
so $\int |\lambda(y)|^q d \mu(y) = \|l_x\|^q \le M^q$.
