Complex Riesz–Thorin Endpoint Interpolation: Examples

Counting measure assigns masses $0,1,1,2$ to the four subsets and is countably additive because a disjoint family has at most two nonempty members. Every complex tuple is a finite simple function and H is complex-linear. The complex Lp conventions give $\mathrm{\parallel }(a,b){\mathrm{\parallel }}_1=\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$, $\mathrm{\parallel }(a,b){\mathrm{\parallel }}_2^2=\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2$, and $\mathrm{\parallel }(a,b){\mathrm{\parallel }}_{\mathrm{\infty }}=\max (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$. Since $\mathrm{\mid }a\pm b\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$, the first operator norm is at most one; the input (1,0) has input norm one and output (1,1) of infinity norm one, so the norm is exactly one.

Expanding with complex conjugates gives $\mathrm{\mid }a+b{\mathrm{\mid }}^2+\mathrm{\mid }a-b{\mathrm{\mid }}^2=2\mathrm{\mid }a{\mathrm{\mid }}^2+2\mathrm{\mid }b{\mathrm{\mid }}^2$, since the two cross terms cancel. Hence $\mathrm{\parallel }H(a,b){\mathrm{\parallel }}_2=\sqrt{2}\mathrm{\parallel }(a,b){\mathrm{\parallel }}_2$, proving the second operator norm exactly. For $1<p<2$, apply F2 with A=1 and B=sqrt(2): $1^{2\mathrm{/}p-1}(\sqrt{2}{)}^{2-2\mathrm{/}p}=2^{1-1\mathrm{/}p}$. The endpoints are the two direct calculations.

At $p=4\mathrm{/}3$, $1\mathrm{/}p=3\mathrm{/}4$, $\theta =2-2\mathrm{/}p=2-3\mathrm{/}2=1\mathrm{/}2$, and $1\mathrm{/}{p}^{\mathrm{\prime }}=1-3\mathrm{/}4=1\mathrm{/}4$, so $p^{\mathrm{\prime }}=4$. The bound in F1 becomes $\mathrm{\parallel }Tf{\mathrm{\parallel }}_4\le A^{1\mathrm{/}2}{B}^{1\mathrm{/}2}\mathrm{\parallel }f{\mathrm{\parallel }}_{4\mathrm{/}3}=\sqrt{AB}\mathrm{\parallel }f{\mathrm{\parallel }}_{4\mathrm{/}3}$. For instance A=B=1 gives coefficient one.

At p=1 the reciprocal target exponent is $1-1=0$, giving infinity and the hypothesis $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{\mathrm{\infty }}\le A\mathrm{\parallel }f{\mathrm{\parallel }}_1$. At p=2 it is $1-1\mathrm{/}2=1\mathrm{/}2$, giving two and $\mathrm{\parallel }Tf{\mathrm{\parallel }}_2\le B\mathrm{\parallel }f{\mathrm{\parallel }}_2$. These direct endpoint statements remain meaningful for A=0 or B=0, while the interior square-root coefficient is zero if either vanishes. No Fourier operator or its endpoint bounds are being presumed.

Use Lebesgue measure on (0,1) and let T be the identity on complex finite simple classes. It is complex-linear and has $\mathrm{\parallel }Tf{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }Tf{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Countable choice supplies the stated earlier Lebesgue-measure result, which gives measure one to (0,1) and measure $1\mathrm{/}n$ to $(0,1\mathrm{/}n)$ for each integer $n\ge 2$.

Define $f_n=n{\mathbf{1}}_{(0,1\mathrm{/}n)}$. It is a finite simple function of finite-measure support. Direct integration gives $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_1=n(1\mathrm{/}n)=1$ and $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2^2=n^2(1\mathrm{/}n)=n$. Its infinity norm is n: n is a pointwise bound, and every smaller nonnegative bound fails on a set of measure $1\mathrm{/}n>0$. Thus an $L^1\to L^{\mathrm{\infty }}$ bound C would require $n=\mathrm{\parallel }T{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_1=C$ for every $n\ge 2$, impossible for finite C.

Every finite simple function on a probability space is integrable; finite sums in its integral show that P is complex-linear. The constant one has every displayed norm equal to one. Hence $\mathrm{\parallel }Pf{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\mid }\int f\mathrm{\mid }\le \int \mathrm{\mid }f\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_1$ by the integral triangle inequality. In particular $P{\mathbf{1}}_X={\mathbf{1}}_X$, so the bound is attained on this input.

Complex Cauchy–Schwarz against the constant one gives $\mathrm{\parallel }Pf{\mathrm{\parallel }}_2=\mathrm{\mid }⟨f,1⟩\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }1{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Every probability space is sigma-finite, with the constant exhaustion X, so F3 applies with A=B=1 to give $\mathrm{\parallel }Pf{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le \mathrm{\parallel }f{\mathrm{\parallel }}_p$ for interior p. The endpoint estimates are the two calculations above.