Let \(f\) be analytic in the unit disk \(|z|<1\). Assume that there is a constant \(M>0\) such that \[ \int_0^{2\pi}|f'(re^{i\theta})|\,d\theta\le M \qquad(0\le r<1). \] Prove that \[ \int_0^1|f(x)|\,dx<\infty. \]
Solution
Fix \(R\) with \(0<R<1\) and put \(g=f'\).
1. For \(0\le x<R\),\[ f(x)-f(0) = -\frac{R}{2\pi} \int_0^{2\pi} g(Re^{i\theta})e^{i\theta} \log\left(1-\frac{x}{R}e^{-i\theta}\right) \,d\theta. \]
For \(0\le t<R\), Cauchy’s formula on the circle \(\abs z=R\) gives \[ g(t) = \frac1{2\pi} \int_0^{2\pi} g(Re^{i\theta}) \frac{Re^{i\theta}}{Re^{i\theta}-t} \,d\theta. \] Integrate this identity in \(t\) from \(0\) to \(x\). Since \(x<R\), the integrand is continuous on the compact product of integration domains, so the order of integration may be exchanged. With \(s=t/R\), \[ \int_0^x\frac{Re^{i\theta}}{Re^{i\theta}-t}\,dt = R\int_0^{x/R}\frac{ds}{1-se^{-i\theta}} = -Re^{i\theta}\log\left(1-\frac{x}{R}e^{-i\theta}\right). \] For \(0\le s<1\), the number \(1-se^{-i\theta}\) has positive real part, so the logarithm can be taken on the principal branch throughout the integration path.
2. There is an absolute constant \(C\) such that, for every real \(\theta\),\[ \int_0^1\abs{\log(1-se^{-i\theta})}\,ds\le C. \]
For \(0\le s<1\), \[ 1-s\le\abs{1-se^{-i\theta}}\le1+s\le2, \] and the real part of \(1-se^{-i\theta}\) is positive, so its principal argument has absolute value at most \(\pi/2\). Hence \[ \abs{\log(1-se^{-i\theta})} \le -\log(1-s)+\log2+\frac\pi2. \] The right-hand side is integrable on \([0,1)\); for example one may take \[ C=1+\log2+\frac\pi2. \]
3. For every \(0<R<1\),\[ \int_0^R\abs{f(x)-f(0)}\,dx \le \frac{CM}{2\pi}. \]
Taking absolute values in step 1, integrating in \(x\), and using Tonelli’s theorem gives \[ \begin{aligned} \int_0^R\abs{f(x)-f(0)}\,dx &\le \frac{R}{2\pi} \int_0^{2\pi}\abs{g(Re^{i\theta})} \int_0^R \abs{\log\left(1-\frac{x}{R}e^{-i\theta}\right)} \,dx\,d\theta\\ &=\frac{R^2}{2\pi} \int_0^{2\pi}\abs{g(Re^{i\theta})} \int_0^1\abs{\log(1-se^{-i\theta})}\,ds\,d\theta\\ &\le \frac{CR^2}{2\pi} \int_0^{2\pi}\abs{f'(Re^{i\theta})}\,d\theta\\ &\le\frac{CM}{2\pi}, \end{aligned} \] using step 2 and the hypothesis.
4. For every \(0<R<1\),\[ \int_0^R\abs{f(x)}\,dx \le \abs{f(0)}+\frac{CM}{2\pi}. \]
The triangle inequality and step 3 give \[ \int_0^R\abs{f(x)}\,dx \le R\abs{f(0)} +\int_0^R\abs{f(x)-f(0)}\,dx \le \abs{f(0)}+\frac{CM}{2\pi}. \]
5. Therefore\[ \boxed{\int_0^1\abs{f(x)}\,dx<\infty}. \]
4. Taking the limit \(R\uparrow1\) proves the claim.
6. Q.E.D.
Step 5 is the required conclusion.