Interpolation

Let $X$ be a measurable space, $\mu$ a measure on $X$. Let $k(x,y)$ be a function on $X^2$ satisfying the following conditions: $$M_1={sup}_x\mu \left(\right|k(x,-)\left|\right)<\mathrm{\infty },M_2={sup}_y\mu \left(\right|k(-,y)\left|\right)<\mathrm{\infty }$$ Consider the operator $\mathrm{Tf}\left(x\right)={\int }_{X}k(x,y)f\left(y\right)\mu \left(y\right)$. Then by Fubini's theorem $\parallel \mathrm{TF}{\parallel }_1\le M_2\parallel f{\parallel }_1$ for $f\in L^1(X,\mu )$. On the otther hand, we have $\parallel \mathrm{Tf}{\parallel }_{\mathrm{\infty }}\le M_1\parallel f{\parallel }_{\mathrm{\infty }}$. Put $S_z=M_2^{z-1}{M}_1^{-z}T$. Then $S_z$ is holomorphic in $z$ and becomes a contraction for $1$-norm when $\Re z=0$, for $\mathrm{\infty }$-norm when $\Re z=1$.

Put $E$ be the quotient of ${ℒ}^1(X,\mu )\cap {ℒ}^{\mathrm{\infty }}(X,\mu )$ by the almost everywhere equality relation, endowed with the norm $\parallel f{\parallel }_{+}=inf(\parallel g{\parallel }_1+\parallel h{\parallel }_{\mathrm{\infty }}:f=g+h)$. Let $F\left(E\right)$ denote the space of holomorphic functions of $\Gamma =\{0\le \Im z\le 1\}$ into $E$, bounded for the $+$-norm. For each $f\in F\left(E\right)$, put $\parallel f\parallel ={sup}_y(\parallel f(\mathrm{iy}){\parallel }_1,\parallel f(1+\mathrm{iy}\left){\parallel }_{\mathrm{\infty }}\right)$. This determins a norm by the maximal value principle. The subset $K_t=\{f\in F(E):f(t)=0\}$ is closed. Let $S_z$ act on $F\left(E\right)$ by $\left(S_{\mathrm{zf}}\right)\left(z\right)=S_{\mathrm{zf}}\left(z\right)$. Then this becomes a contraction and preserves $K_t$. Hence $S_z$ induces a contraction $M_x^{t-1}{M}_1^{\mathrm{tT}}$ on the qotient $E_t=F\left(E\right)/K_t$.

It remains to show that $E+tsimeq{L}^p(X,\mu )$ for $p=\frac{1}{1-t}$. Let $\phi$ be a positive simple function on $X$. Put $f_z=\parallel \phi {\parallel }_p^{1-p(1-z)}{\phi }^{p(1-z)}$. We have $\parallel f_z{\parallel }_1=\parallel \phi {\parallel }_p$ when $\Re z=0$, $\parallel f_z{\parallel }_{\mathrm{\infty }}=\parallel \phi {\parallel }_p$ when $\Re z=1$ and $f_t=\phi$, hence $\parallel \phi {\parallel }_t\le \parallel \phi {\parallel }_p$.

Conversely, let $f$ be an element of $F\left(E\right)$ satisfying $f_t=\phi$, $\psi$ an $L^q$-function for $p^{-`}+q^{-1}=1$. Put $g_z=\parallel \psi {\parallel }_q^{1-\mathrm{qz}}{\psi }^{\mathrm{qz}}$. $g_{1-z}$ is in $F\left(E\right)$. When $\Re z=r$, $f_z$ is of $L^{\frac{1}{1-r}}$-class while $g_z$ is of $L^{\frac{1}{r}}$-class. Thus $H\left(z\right)={\int }_X{f}_{\mathrm{zg}}{}_{\mathrm{zd}}\mu$ is finite and holomorphic in $z$. By the maximal value princple, $\left|H\right(t\left)\right|<{sup}_y\left(\right|H\left(\mathrm{iy}\right)|,|H(1+\mathrm{iy})\left|\right)$. By $\left|H\right(\mathrm{iy}\left)\right|\le \parallel f_{\mathrm{iy}}{\parallel }_1\parallel g_{\mathrm{iy}}{\parallel }_{\mathrm{\infty }}$ and $\left|H\right(1+\mathrm{iy}\left)\right|\le \parallel f_{1+\mathrm{iy}}{\parallel }_{\mathrm{\infty }}\parallel g_{1+\mathrm{iy}}{\parallel }_1$, $\left|H\right(t\left)\right|$ is bounded by $\parallel f_z\parallel \cdot \parallel g_{1-z}\parallel =\parallel \phi {\parallel }_p\parallel \psi {\parallel }_q$. Hence $\phi =f_t$ is of $L^p$ function with norm less than $\parallel f\parallel$. This proves $\parallel \phi {\parallel }_p\le \parallel \phi {\parallel }_t$.