How exactly is this integral finite?

From here,

Koosis is proving the inequality that $$\int_{- \infty}^{\infty}\frac{\log^+|S(x)|}{1+x^2}dx \qquad (1)$$ is finite by using the fact that $$\int_{- \infty}^{\infty}\frac{\log^+|S(x+i)|}{1+x^2}$$ is finite. Reading through he applies this theorem

and uses a so called "Hall of Mirrors" argument by applying this theorem to the half plane $$Im(z) < 1$$ to get for any real number $$\zeta$$ that $$\log|S(\zeta)| \leq \frac{1}{\pi}\int_{- \infty}^{\infty}\frac{\log^+|S(x+i)|}{1+(x-\zeta )^2}dx.$$ Now, I'm not sure how to use this to get $$(1)$$ as the LHS is not exactly what we know to be finite. His mention of Fubini's Theorem makes be think of multiplying both sides by $$\frac{1}{1+\zeta ^2}$$ and integrating over the reals. Would this work? My thought is that the $$\frac{1}{1+(x-\zeta) ^2}$$ factor would mess with this argument. I'm not sure if this is the right way to go. Any suggestions or words of encouragement?

• What’s $\log^+$ ? – MPW Jun 28 at 0:40
• $\log^+ k = max (0, \log k)$ @MPW – Sean Nemetz Jun 28 at 0:54

1 Answer

Alright guys, so Koosis' suggestion makes a lot more sense now after I actually evaluated the integral. We have that since the integral on the RHS of the last inequality above is positive, we can replace $$\log |S(\zeta)|$$ with $$\log^+ |S(\zeta)|$$ and with my suggestion we get that $$\int_{-\infty}^{\infty}\frac{\log^+ |S(\zeta)|}{1+\zeta^2} d\zeta \leq \frac{1}{\pi}\int_{-\infty}^{\infty} \int_{- \infty} ^ {\infty} \frac{\log^+ |S(x +i)|}{(1+\zeta^2)(1+(x-\zeta)^2)}dxd\zeta .$$ By Tonelli's Theorem, we need only consider the integral $$\int_{- \infty} ^ {\infty} \frac{1}{(1+\zeta^2)(1+(x-\zeta)^2)}d\zeta.$$ Which, by the help of a certain computer algebra system (or partial fractions) this evaluates to $$\frac{2\pi}{x^2+4}$$. Hence, $$\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\log^+ |S(\zeta)|}{1+\zeta^2} d\zeta \leq 2\int_{- \infty} ^ {\infty} \frac{\log^+ |S(x +i)|}{x^2+4}dx \leq 2\int_{- \infty} ^ {\infty} \frac{\log^+ |S(x +i)|}{x^2+1}dx < \infty .$$ As desired.