Riesz-Thorin interpolation theorem

First assume that $f$ and $g$ are finite simple functions of finite support [L1, L2, given, choose] on $X$ and $Y$, respectively, with $g$ chosen from $L^{q_{\theta }^{\mathrm{\prime }}}(\nu )$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_{q_{\theta }^{\mathrm{\prime }}}=1$. Write $f={\sum }_{j=1}^m{a}_j{\mathbf{1}}_{E_j},g={\sum }_{k=1}^{\mathrm{\ell }}{b}_k{\mathbf{1}}_{F_k},$ with the sets $E_j,F_k$ pairwise disjoint and of finite measure.

Define the analytic families [step 1.1, construct, algebra] $f_z:={\sum }_{j=1}^m{\alpha }_j\mathrm{\mid }{a}_j{\mathrm{\mid }}^{p_{\theta }((1-z)\mathrm{/}{p}_0+z\mathrm{/}{p}_1)}{\mathbf{1}}_{E_j},$ $g_z:={\sum }_{k=1}^{\mathrm{\ell }}{\beta }_k\mathrm{\mid }{b}_k{\mathrm{\mid }}^{q_{\theta }^{\mathrm{\prime }}((1-z)\mathrm{/}{q}_0^{\mathrm{\prime }}+z\mathrm{/}{q}_1^{\mathrm{\prime }})}{\mathbf{1}}_{F_k},$ where ${\alpha }_j=a_j\mathrm{/}\mathrm{\mid }{a}_j\mathrm{\mid }$ and ${\beta }_k=b_k\mathrm{/}\mathrm{\mid }{b}_k\mathrm{\mid }$ when the coefficient is nonzero and $0$ otherwise. Then $f_{\theta }=f$ and $g_{\theta }=g$. For real $t$, direct calculation on each simple coefficient gives $\mathrm{\parallel }{f}_{it}{\mathrm{\parallel }}_{p_0}=\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}^{p_{\theta }\mathrm{/}{p}_0},\mathrm{\parallel }{f}_{1+it}{\mathrm{\parallel }}_{p_1}=\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}^{p_{\theta }\mathrm{/}{p}_1},$ and likewise $\mathrm{\parallel }{g}_{it}{\mathrm{\parallel }}_{q_0^{\mathrm{\prime }}}=\mathrm{\parallel }g{\mathrm{\parallel }}_{q_{\theta }^{\mathrm{\prime }}}^{q_{\theta }^{\mathrm{\prime }}\mathrm{/}{q}_0^{\mathrm{\prime }}}=1,\mathrm{\parallel }{g}_{1+it}{\mathrm{\parallel }}_{q_1^{\mathrm{\prime }}}=\mathrm{\parallel }g{\mathrm{\parallel }}_{q_{\theta }^{\mathrm{\prime }}}^{q_{\theta }^{\mathrm{\prime }}\mathrm{/}{q}_1^{\mathrm{\prime }}}=1.$

Put [step 2.1, given, algebra] $\mathrm{\Phi }(z):={\int }_Y(T{f}_z)(y)g_z(y)d\nu (y)\mathrm{.}$ Because $f_z$ and $g_z$ are finite linear combinations of exponentials in $z$, $\mathrm{\Phi }$ is continuous on the closed strip $S=\{z\in \mathbb{C}:0\le \mathrm{Re}z\le 1\}$ and holomorphic on its interior. For real $t$, the endpoint bounds and Holder give $\mathrm{\mid }\mathrm{\Phi }(it)\mathrm{\mid }\le M_0\mathrm{\parallel }{f}_{it}{\mathrm{\parallel }}_{p_0}\mathrm{\parallel }{g}_{it}{\mathrm{\parallel }}_{q_0^{\mathrm{\prime }}}\le M_0\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }},$ $\mathrm{\mid }\mathrm{\Phi }(1+it)\mathrm{\mid }\le M_1\mathrm{\parallel }{f}_{1+it}{\mathrm{\parallel }}_{p_1}\mathrm{\parallel }{g}_{1+it}{\mathrm{\parallel }}_{q_1^{\mathrm{\prime }}}\le M_1\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}\mathrm{.}$

Fix $\delta >0$ and define [step 3.1, construct, algebra] ${\mathrm{\Psi }}_{\delta }(z):=\mathrm{\Phi }(z)(M_0+\delta {)}^{z-1}(M_1+\delta {)}^{-z}\mathrm{.}$ By step 3.1, $\mathrm{\mid }{\mathrm{\Psi }}_{\delta }\mathrm{\mid }$ is at most $1$ on the two boundary lines of the strip. Multiplying once more by $\exp (\epsilon (z^2-1))$ and applying the maximum-modulus principle on large rectangles inside the strip shows that $\mathrm{\mid }{\mathrm{\Psi }}_{\delta }(z)\mathrm{\mid }\le 1$ throughout $S$. Evaluating at $z=\theta$ and letting first $\epsilon \downarrow 0$ and then $\delta \downarrow 0$ yields $\mathrm{\mid }\mathrm{\Phi }(\theta )\mathrm{\mid }\le M_0^{1-\theta }{M}_1^{\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}\mathrm{.}$

Since $\mathrm{\Phi }(\theta )=\int (Tf)gd\nu$, step 4.1 gives [L1, L2, step 4.1, algebra] $\mid \int (Tf)gd\nu \mid \le M_0^{1-\theta }{M}_1^{\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}$ for every unit $g\in L^{q_{\theta }^{\mathrm{\prime }}}(\nu )$ that is finite simple with finite support. By density [L1] and norm recovery [L2], it follows that $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{q_{\theta }}\le M_0^{1-\theta }{M}_1^{\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}$ for every finite simple $f$ of finite support.

Now let $f\in L^{p_{\theta }}(\mu )$. By [L1], choose finite simple functions [L1, L3, step 5.1, algebra] $f_n$ of finite support with $f_n\to f$ in $L^{p_{\theta }}(\mu )$. Step 5.1 makes $(T{f}_n)$ Cauchy in $L^{q_{\theta }}(\nu )$, so [L3] gives a limit $h\in L^{q_{\theta }}(\nu )$ with $\mathrm{\parallel }h-T{f}_n{\mathrm{\parallel }}_{q_{\theta }}\to 0.$ If $(g_n)$ is another such approximating sequence for $f$, then step 5.1 applied to $f_n-g_n$ shows $\mathrm{\parallel }T{f}_n-T{g}_n{\mathrm{\parallel }}_{q_{\theta }}\le M_0^{1-\theta }{M}_1^{\theta }\mathrm{\parallel }{f}_n-g_n{\mathrm{\parallel }}_{p_{\theta }}\to 0,$ so the limit $h$ is independent of the chosen approximation. Define $\tilde{T}f:=h\mathrm{.}$ Passing to the limit in step 5.1 yields $\mathrm{\parallel }\tilde{T}f{\mathrm{\parallel }}_{q_{\theta }}\le M_0^{1-\theta }{M}_1^{\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_{p_{\theta }}\mathrm{.}$

The definition in step 6.1 extends $T$, because a constant approximating [step 6.1, algebra] sequence may be used when $f$ is already finite simple of finite support. Applying step 6.1 to $f+g$ and to $cf$ shows that $\tilde{T}$ is linear, since linearity holds termwise on every approximating sequence. If $S$ is any other bounded linear extension of $T$ to $L^{p_{\theta }}(\mu )$, then for every $f\in L^{p_{\theta }}(\mu )$ and every approximating sequence $(f_n)$ from step 6.1, $\mathrm{\parallel }Sf-\tilde{T}f{\mathrm{\parallel }}_{q_{\theta }}\le \mathrm{\parallel }S(f-f_n){\mathrm{\parallel }}_{q_{\theta }}+\mathrm{\parallel }\tilde{T}(f_n-f){\mathrm{\parallel }}_{q_{\theta }}\to 0,$ so $S=\tilde{T}$.

Steps 5.1, 6.1, and 7.1 prove the interpolated bounded extension theorem.