Complex Riesz–Thorin Endpoint Interpolation

Discard null fibers and zero coefficients without changing the classes, and define their omitted contributions to be zero for every z. For the remaining coefficients put $\alpha (z)=p((1-z)\mathrm{/}{p}_0+z\mathrm{/}{p}_1)$ and $a_j(z)=(a_j\mathrm{/}\mathrm{\mid }{a}_j\mathrm{\mid })\exp (\alpha (z)\log \mathrm{\mid }{a}_j\mathrm{\mid })$. Positive coefficient moduli have defined logarithms; at z=theta, $\alpha (\theta )=1$, so $a_j(\theta )=a_j$. Set $f_z={\sum }_j{a}_j(z){\mathbf{1}}_{E_j}$.

The affine alpha is entire; the chain rule and the entire exponential make every $a_j(z)$ entire. For $z=x+it$, the exponential modulus formula gives $\mathrm{\mid }{a}_j(z)\mathrm{\mid }=\exp (p((1-x)\mathrm{/}{p}_0+x\mathrm{/}{p}_1)\log \mathrm{\mid }{a}_j\mathrm{\mid })=\mathrm{\mid }{a}_j{\mathrm{\mid }}^{p((1-x)\mathrm{/}{p}_0+x\mathrm{/}{p}_1)}$. For $0\le x\le 1$ this is bounded by the larger of the two boundary powers. The finite list of coefficients is therefore bounded throughout the strip. At $x=\mathrm{\ell }$ with $\mathrm{\ell }=0$ or $1$, disjointness gives $\mathrm{\parallel }{f}_{\mathrm{\ell }+it}{\mathrm{\parallel }}_{p_{\mathrm{\ell }}}^{p_{\mathrm{\ell }}}={\sum }_j\mathrm{\mid }{a}_j{\mathrm{\mid }}^p\mu (E_j)=\mathrm{\parallel }f{\mathrm{\parallel }}_p^p$, proving the asserted norm formula.

Since $1\mathrm{/}{r}_{\mathrm{\ell }}=1-1\mathrm{/}{q}_{\mathrm{\ell }}$, we have $1\mathrm{/}r=(1-\theta )\mathrm{/}{r}_0+\theta \mathrm{/}{r}_1$. If r is finite, put $\beta (z)=r((1-z)\mathrm{/}{r}_0+z\mathrm{/}{r}_1)$ and $b_k(z)=(b_k\mathrm{/}\mathrm{\mid }{b}_k\mathrm{\mid })\exp (\beta (z)\log \mathrm{\mid }{b}_k\mathrm{\mid })$. The preceding entire-function and modulus calculations apply with r and the b-coefficients, and $\beta (\theta )=1$. For finite $r_{\mathrm{\ell }}$, $\mathrm{\parallel }{g}_{\mathrm{\ell }+it}{\mathrm{\parallel }}_{r_{\mathrm{\ell }}}^{r_{\mathrm{\ell }}}={\sum }_k\mathrm{\mid }{b}_k{\mathrm{\mid }}^r\nu (F_k)=\mathrm{\parallel }g{\mathrm{\parallel }}_r^r$. If $r_{\mathrm{\ell }}=\mathrm{\infty }$, each surviving coefficient has modulus one on that boundary, so the essential maximum is one: a nonzero class has at least one positive-measure surviving fiber.

If $r=\mathrm{\infty }$, the positive weights $1-\theta ,\theta$ and nonnegative reciprocals force $1\mathrm{/}{r}_0=1\mathrm{/}{r}_1=0$, hence $q_0=q_1=1$. The constant family $g_z=g$ is entire coefficientwise, bounded, has $g_{\theta }=g$ and unchanged infinity norm. Identically zero families handle zero classes on either side, including empty or zero-measure spaces, with no logarithm of zero. Thus every asserted branch is established.

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.

Fix theta, put $p=p_{\theta }$ and $q=q_{\theta }$, and denote its finite core bound by K (the stated endpoint bound when theta is an endpoint). Density and countable choice give finite simple $s_n$ for $n\ge 1$ with $\mathrm{\parallel }{s}_n-f{\mathrm{\parallel }}_p<1\mathrm{/}n$ for any fixed $f\in L^p$. The bound $\mathrm{\parallel }T{s}_n-T{s}_m{\mathrm{\parallel }}_q\le K\mathrm{\parallel }{s}_n-s_m{\mathrm{\parallel }}_p$ makes $T{s}_n$ Cauchy. Completeness, with its stated countable-choice hypothesis, gives a limit; define $T_{\theta }f$ to be that limit.

To compare parameters a,b, fix a finite-valued measurable representative $f\in L^{p_a}\cap L^{p_b}$. For $n\ge 1$, set $E_n=\{1\mathrm{/}n\le \mathrm{\mid }f\mathrm{\mid }\le n\}$. This set has finite measure because $n^{-p_a}\mu (E_n)\le \int \mathrm{\mid }f{\mathrm{\mid }}^{p_a}$. On $E_n$, round each real and imaginary component toward zero to a multiple of $1\mathrm{/}{n}^2$, and put $s_n=0$ elsewhere. The rounding has finitely many values because the components are bounded by n; its fibers are measurable intervals, and its support lies in $E_n$. Also $\mathrm{\mid }{s}_n\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$. At a point with f nonzero, it eventually belongs to $E_n$ and the rounding error is at most $\sqrt{2}\mathrm{/}{n}^2$; at a zero of f every $s_n$ is zero. Thus $s_n\to f$ pointwise and $\mathrm{\mid }{s}_n-f{\mathrm{\mid }}^{p_j}\le 2^{p_j}\mathrm{\mid }f{\mathrm{\mid }}^{p_j}$ for j=a,b. Dominated convergence applied to these errors gives simultaneous convergence in both source norms.

If $(t_n{)}_{n\ge 1}$ is a second core approximation converging to f, $\mathrm{\parallel }T{s}_n-T{t}_n{\mathrm{\parallel }}_q\le K(\mathrm{\parallel }{s}_n-f{\mathrm{\parallel }}_p+\mathrm{\parallel }{t}_n-f{\mathrm{\parallel }}_p)\to 0$, so the definition is independent of approximation. Approximate f and g separately; $\alpha s_n+\beta t_n$ approximates $\alpha f+\beta g$, and core linearity gives linearity of the limits. Norm continuity gives $\mathrm{\parallel }{T}_{\theta }f{\mathrm{\parallel }}_q\le K\mathrm{\parallel }f{\mathrm{\parallel }}_p$. Any bounded extension has the same limit on a dense core, proving uniqueness. This includes K=0.

By step 2.1, the one sequence $T{s}_n$ converges in $L^{q_a}$ to $T_{a}f$ and in $L^{q_b}$ to $T_{b}f$. The a.e.-subsequence theorem first gives a subsequence converging a.e. to a representative of $T_{a}f$; apply it again to that subsequence in $L^{q_b}$ to get a further subsequence converging a.e. to $T_{b}f$. Choosing representatives and taking the countable union of their measurable null discrepancies makes the two pointwise limits comparable on one conull set. Uniqueness of complex pointwise limits gives $T_{a}f=T_{b}f$ as classes. The supplier covers q=infinity as well.

If $f_0+f_1=g_0+g_1$ with endpoint components, then $h=f_0-g_0=g_1-f_1$ lies in $L^{p_0}\cap L^{p_1}$. Agreement gives $T_{0}h=T_{1}h$, so $T_0{f}_0+T_1{f}_1=T_0{g}_0+T_1{g}_1$. Componentwise addition and scalar multiplication prove linearity of this sum map.

If $p_0\le p_1$ and $f\in L^{p_{\theta }}$, split $u=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>1\}}$, $v=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le 1\}}$. For $t>0$, $t^s=\exp (s\log t)$; since exp is increasing and $\exp (0)=1$, $\log t$ has the sign of $t-1$. Thus powers increase with the exponent for $t>1$ and decrease for $0<t\le 1$; at $t=0$ all positive powers are zero. On the first set $\mathrm{\mid }f{\mathrm{\mid }}^{p_0}\le \mathrm{\mid }f{\mathrm{\mid }}^{p_{\theta }}$, and on the second $\mathrm{\mid }f{\mathrm{\mid }}^{p_1}\le \mathrm{\mid }f{\mathrm{\mid }}^{p_{\theta }}$. Thus $u\in L^{p_0}\cap L^{p_{\theta }}$ and $v\in L^{p_1}\cap L^{p_{\theta }}$. Pairwise agreement and linearity give $T_{\theta }f=T_{0}u+T_{1}v$. If $p_1<p_0$, reverse the endpoint labels in this split. If they coincide, f is already in both endpoint spaces. This proves the restriction assertion for all theta, including the endpoints.

For $1<p<2$, set $\theta =2-2\mathrm{/}p$. Then $0<\theta <1$, $(1-\theta )\mathrm{/}1+\theta \mathrm{/}2=1\mathrm{/}p$, and $(1-\theta )\mathrm{/}\mathrm{\infty }+\theta \mathrm{/}2=1-1\mathrm{/}p=1\mathrm{/}{p}^{\mathrm{\prime }}$. Thus the core interpolation theorem with endpoints $(1,\mathrm{\infty })$ and $(2,2)$ gives exactly $A^{1-\theta }{B}^{\theta }=A^{2\mathrm{/}p-1}{B}^{2-2\mathrm{/}p}$ and proves membership in $L^{p^{\mathrm{\prime }}}$. Zero A or B is allowed because the interior powers are positive.

At p=1 the target is infinity and at p=2 the target is two, so the asserted estimates are the respective hypotheses. Assuming countable choice, the compatible-extension result applies with finite source endpoint exponents 1 and 2 and the given sigma-finite spaces, providing the claimed unique bounded extensions and agreement.