Interpolation for the two-by-two Hadamard matrix

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.