Riesz–Thorin estimate on the finite simple core

Fix an interior theta and write p,q for its exponents and r for the conjugate of q. If f is zero as a class, linearity gives Tf=0. Otherwise replace f by $f\mathrm{/}\mathrm{\parallel }f{\mathrm{\parallel }}_p$. For a nonzero finite simple dual test replace g by $g\mathrm{/}\mathrm{\parallel }g{\mathrm{\parallel }}_r$; zero tests already have zero integral. These normalizations are legal for nonzero finite simple classes of finite-measure support. The families in F1 then have input boundary norms one and dual boundary norms one, including the constant family when r=infinity.

Write $f_z={\sum }_j{a}_j(z){\mathbf{1}}_{E_j}$ and $g_z={\sum }_k{b}_k(z){\mathbf{1}}_{F_k}$. Put $h_j=T{\mathbf{1}}_{E_j}$. The endpoint hypothesis puts each $h_j$ in both target endpoint spaces. Endpoint Holder with ${\mathbf{1}}_{F_k}$, which belongs to every conjugate space because its support has finite measure, proves $I_{jk}=\int h_j{\mathbf{1}}_{F_k}d\nu$ finite. Linearity gives $H(z)=\int (T{f}_z)g_{z}d\nu ={\sum }_{j,k}{a}_j(z)b_k(z)I_{jk}$. This is a finite sum of products of entire coefficients, so it is entire and continuous on the closed strip. Their strip bounds and the finite constants $I_{jk}$ give a uniform bound for H on that whole strip.

On each boundary line $\mathrm{\ell }\in \{0,1\}$, Holder and the endpoint operator bound yield $\mathrm{\mid }H(\mathrm{\ell }+it)\mathrm{\mid }\le \mathrm{\parallel }T{f}_{\mathrm{\ell }+it}{\mathrm{\parallel }}_{q_{\mathrm{\ell }}}\mathrm{\parallel }{g}_{\mathrm{\ell }+it}{\mathrm{\parallel }}_{r_{\mathrm{\ell }}}\le M_{\mathrm{\ell }}$. The precise closed-strip three-lines theorem therefore gives $\mathrm{\mid }\int (Tf)g\mathrm{\mid }=\mathrm{\mid }H(\theta )\mathrm{\mid }\le M_0^{1-\theta }{M}_1^{\theta }$. It also applies when an $M_{\mathrm{\ell }}$ vanishes, because theta is interior and both powers are positive.

For arbitrary measure spaces with finite $q_0,q_1$, choose representatives of the finitely many $h_j$ and let $Y_0={\bigcup }_{j,m\ge 1}\{\mathrm{\mid }{h}_j\mathrm{\mid }>1\mathrm{/}m\}$. For each j,m, $m^{-q_0}\nu \{\mathrm{\mid }{h}_j\mathrm{\mid }>1\mathrm{/}m\}\le \int \mathrm{\mid }{h}_j{\mathrm{\mid }}^{q_0}<\mathrm{\infty }$. Thus each level set has finite measure, and their countable union is sigma-finite (finite unions give an increasing exhaustion). All the chosen $h_j$, hence the finite-sum representative of every $T{f}_z$, vanish off $Y_0$. The restricted measure and its measurable sets satisfy the same endpoint bounds.

For any unnormalized test s with $\mathrm{\parallel }s{\mathrm{\parallel }}_r\le 1$, either it is the zero class or scaling the estimate for its norm-one version gives $\mathrm{\mid }\int (Tf)s\mathrm{\mid }\le M_0^{1-\theta }{M}_1^{\theta }$. Every such product is integrable by the endpoint Holder calculation. On sigma-finite Y the membership form of the dual-test lemma now proves $Tf\in L^q$ and bounds its norm. Scaling f back proves the desired estimate. No step assumed intermediate target membership before this test.

Apply steps 1.1, 2.1, 3.1 and 4.1 with dual tests on $Y_0$, extended by zero to Y. Their integrals and norms are unchanged, and membership on $Y_0$ gives membership on Y because Tf vanishes off $Y_0$. The argument used only the finitely many source fibers, not source sigma-finiteness. If $Y_0$ is empty, Tf is zero and the estimate holds directly. The endpoint parameters are exactly the original hypotheses.