Consider a linear operator $T$ given by $$T(f)(x)=\int_{Y}K(x,y)f(y)d\nu(y),\qquad x\in X.$$
Let $1\le p_{1}$, $q_{1}\le\infty$, $p_{0}=1$, $q_{1}=\infty$, $\frac{1}{p_{1}}+\frac{1}{p_{1}'}=\infty$ and suppose that $K(x,y)$ satisfies $$\begin{aligned} A:=\sup_{x\in X}\left(\int_{Y}|K(x,y)|^{p'_{1}}d\nu(y)\right)^{1/p'_{1}}<\infty, \\ B:=\sup_{y\in Y}\left(\int_{X}|K(x,y)|^{q_{0}}d\mu(x)\right)^{1/q_{0}}<\infty. \end{aligned}$$
I want to show that $T$ extends a bounded operator from $L^{p_{\theta}}(Y)$ into $L^{q_{\theta}}(X)$ for $\frac{1}{p_{\theta}}=1-\theta+\frac{\theta}{p_{1}}$, $q_{\theta}=\frac{q_{0}}{1-\theta}$, $0\le\theta\le 1$, and determine its operator norm.
Of course, I have to utilise the Riez-Thorin interpolation theorem. The second equation gives that $T$ maps $L^{q_{0}}$ to $L^{q_{0}}$ with bound $B$. What about the first equation? I don't really have the intuition for it. If $p_{1}'$ were equal to 1, then it would map $L^{\infty}$ to $L^{\infty}$ with bound $A$.
You need the result often called "Minkowski's Inequality for Integrals".
Say $k_y(x)=K(x,y)$. Then $$Tf=\int_Yk_yf(y)\,d\nu(y),$$so$$||Tf||_{q_0}\le\int_Y||k_y||_{q_0}|f(y)|\,d\nu(y)\le B||f||_1=B||f||_{p_0}.$$
Or if you don't buy that use duality (as in the proof of Minkowski for integrals): Say $g\in L^{q_0'}$. It's enough to show $$\left|\int_XTf(x)g(x)\,d\mu(x)\right|\le B||g||_{q_0'}||f||_1,$$which follows from Fubini.