Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\phi$ is analytic in $\mathbb{C}$ and $g$ is continuous on a closed interval $[a,b]$ in $\mathbb{R}$. Let $$f(z)=\int_a^bg(t)\phi(zt)dt.$$ Prove first that, in $\mathbb{C}\,$, $f$ is continuous, and then prove that $f$ is analytic.

share|cite|improve this question
I first tried expanding $\phi$ as a power series but that got me nowhere. – john Jul 4 '12 at 20:02
how can you do this, the integrand does not contain any denominators? – john Jul 4 '12 at 20:12
up vote 1 down vote accepted

Continuity. Only continuity of $\phi$ is needed. Let $z_0\in\mathbb{C}$. We show that $f$ is continuous at $z_0$. The function $\phi(t\,z)$ is continuous as a function of $(t,z)$, and hence uniformly continuous, on $[a,b]\times\{|z-z_0|\le1\}$. Given $\epsilon>0$ there is a $\delta>0$ such that $$ |\phi(t\,z)-\phi(t\,z_0)|\le\frac{\epsilon}{\int_a^b|g(t)|\,dt},\quad t\in[a,b],\quad |z-z_0|\le\delta. $$ Then $$ |f(z)-f(z_0)|\le\int_a^b|g(t)|\,|\phi(z\,t)-\phi(t\,z_0)|\,dt\le\epsilon. $$

Analiticity. We apply Morera's theorem. Given any closed piecewise $C^1$ curve $\gamma$, $$ \int_\gamma f(z)\,dz=\int_a^bg(t)\Bigl(\int_\gamma\phi(t\,z)\,dz\Bigr)\,dt=0. $$ The change in the order of integration is justified by the continuity of $g(t)\,\phi(t\,z)$ on $[a,b]\times\gamma$, and tha analiticity of $\phi$ implies that $\int_\gamma\phi(t\,z)\,dz=0$ for all $t\in[a,b]$.

share|cite|improve this answer
  • We show continuity. Let $x_0\in\Bbb C$ fixed, and $\gamma$ a closed curve such that for each $z$ which satisfies $|z-z_0|\leq 1$ and each $t\in [a,b]$, $|tz-z'|\geq 1$ for all $z'$ in the support of the curve $\gamma$. We use Cauchy integral formula, which gives, for each $z'$ in the interior of the curve $$\phi(z')=\frac 1{2\pi i}\int_{\gamma}\frac{\phi(\xi)}{\xi-z'}d\xi.$$ Applying it to $z'=tz$ and $z'=tz_0$, we get \begin{align} 2\pi i(f(z)-f(z_0))&=\int_a^bg(t)\int_{\gamma}\phi(\xi)\left(\frac 1{\xi-tz}-\frac 1{\xi-tz_0}\right)d\xi dt\\ &=\int_a^b\int_{\gamma}g(t)\phi(\xi)\frac{\xi-tz_0-(\xi-tz))}{(\xi-tz)(\xi-tz_0)}d\xi dt\\ &=\int_a^b\int_{\gamma}g(t)\phi(\xi)t(z-z_0)\frac 1{(\xi-tz)(\xi-tz_0)}d\xi dt. \end{align} We deduce that $$|f(z)-f(z_0)|\leq \sup_{t\in [a,b]}|tg(t)|\sup_{\xi\in\gamma}|\phi(\xi)|\frac 1{2\pi}|z-z_0|,$$ which gives continuity.
  • Now we show that $f$ is holomorphic. We fix $z\in\Bbb C$, and we shall show that $$\lim_{h\to 0}\frac{f(z+h)-f(z)}h-\int_a^btg(t)\phi'(tz)=0.$$ We use again Cauchy's integral formula, applied to the same curve $\gamma$. We consider $|h|$ small enough. Write $A(h):=2\pi i\left(\frac{f(z+h)-f(z)}h-\int_a^btg(t)\phi'(tz)\right)$. Then $$A(h)=\int_a^bg(t)\left(\frac{\phi((z+h)t)-\phi(zt)}h-t\phi'(zt)\right)dt $$ and \begin{align} \frac{\phi((z+h)t)-\phi(zt)}h-t\phi'(zt)&=\int_{\gamma}\phi(\xi)\left(\frac 1{h(\xi-(z+h)t)}-\frac 1{(\xi-zt)h}-\frac t{(\xi-zt)^2}\right)d\xi\\ &=\int_{\gamma}\phi(\xi)\left(\frac{\xi-zt-\xi+zt+th}{h(\xi-(z+h)t)(\xi-zt)}-\frac t{(\xi-zt)^2}\right)d\xi\\ &=\int_{\gamma}\phi(\xi)\left(\frac{t}{(\xi-(z+h)t)(\xi-zt)}-\frac t{(\xi-zt)^2}\right)d\xi\\ &=t\int_{\gamma}\phi(\xi)\frac{\xi-zt-(\xi-(z+h)t)}{(\xi-(z+h)t)(\xi-zt)^2}d\xi\\ &=t\int_{\gamma}\phi(\xi)\frac{\xi-zt-\xi+zt+ht)}{(\xi-(z+h)t)(\xi-zt)^2}d\xi\\ &=ht^2\int_{\gamma}\phi(\xi)\frac 1{(\xi-(z+h)t)(\xi-zt)^2}d\xi. \end{align} We get that $$|A(h)|\leq h\sup_{t\in [a,b]}|t^2g(t)|\sup_{\xi\in\gamma}|\phi(\xi)|.$$
share|cite|improve this answer
An alternative way to show analiticity once continuity is known is to apply Morera's theorem. For any colsed curve $\gamma$ we have $\int_\gamma f(z)dz=\int_a^bg(t)(\int_\gamma \phi(tz)dz)dt=0$. – Julián Aguirre Jul 5 '12 at 9:49
Right. Maybe you can write it in an answer and is more in the spirit of the exercise. – Davide Giraudo Jul 5 '12 at 9:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.