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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.

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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

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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 $t\in[a,b]$.

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  • 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 $$|A(h)|\leq h\sup_{t\in [a,b]}|t^2g(t)|\sup_{\xi\in\gamma}|\phi(\xi)|.$$
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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

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