Spectral cutoff dual tests and the exact interpolation lower bound

Let $P_N=E_T([1,N])$, $h\in K$, and ${\varphi }_N(z)=⟨F(z),T^{\bar{z}}{P}_{N}h⟩$. Since $T^{\bar{z}}{P}_{N}h$ is norm-antiholomorphic (the logarithm is bounded on the cutoff), this is scalar holomorphic inside and continuous on $S$. By Cauchy–Schwarz its modulus is bounded on all $S$ by $N\mathrm{\parallel }h\mathrm{\parallel }{\sup }_S\mathrm{\parallel }F(z){\mathrm{\parallel }}_K$. At $z=it$ its modulus is at most $M_0\mathrm{\parallel }h\mathrm{\parallel }$, since imaginary powers are unitary. At $z=1+it$, the self-adjoint multiplier product gives ${\varphi }_N(1+it)=⟨TF(1+it),T^{-it}{P}_{N}h⟩$, so the bound is $M_1\mathrm{\parallel }h\mathrm{\parallel }$. All scalar three-lines hypotheses, including boundedness throughout the strip rather than just the boundary, have now been verified.

Three-lines yields $\mathrm{\mid }⟨T^{\theta }{P}_{N}F(\theta ),h⟩\mathrm{\mid }\le M_0^{1-\theta }{M}_1^{\theta }\mathrm{\parallel }h\mathrm{\parallel }$. Taking $h$ to be the normalized vector on the left when it is nonzero gives $\mathrm{\parallel }{T}^{\theta }{P}_{N}F(\theta )\mathrm{\parallel }\le M_0^{1-\theta }{M}_1^{\theta }$. The squares of these norms increase to ${\int }_{[1,\mathrm{\infty })}{\lambda }^{2\theta }d(E_T{)}_{F(\theta )}$. Their uniform bound makes this integral finite; the exact Borel-calculus domain formula proves the membership and inequality. Taking the infimum over $F$ gives the lower inequality; the upper inequality from the profile lemma proves equality, and hence definiteness and completeness of this interpolation norm. No density assertion about analytic functions and no preexisting operator interpolation theorem is used.