Prove that $f(z)$ is identically zero.

Question

Let $$G = \{ z\in \mathbb{C}: 0<\arg z<\frac{\pi}{4}\},\hspace{0.2cm}f \in H(G)\cap C(\bar{G})$$
If $f(x) = 0$ for all real numbers $x \in [a,b]$ where $[a,b]$ is a interval in $\mathbb{R}$. Prove that $f(z)$ is identically zero in $G$.

I'm supposed to use Cauchy integral formula but don't know where to start.

Answer 1

We shall assume that $[a,b]\cap \overline{G}\neq\emptyset$ otherwise we cannot say much. Here $$\overline{G}:=\{z\in\mathbb{C}:0\leqslant\arg z\leqslant\frac{\pi}{4}\}$$ For simplicity assume that $a>0$ so that $[a,b]\cap\overline{G}=[a,b]$. Since $f\in H(G)\cap C(\overline{G})$ then for every $x_0\in[a,b]$ there is a neighborhood $N(x_0)$ such that $N(x_0)\cap G\neq\emptyset$ and $f(z)=0$ for every $z\in N(x_0)\cap G$. On the other hand since $f\in H(G)$ implies that for any closed disk $D$ entirely included in $G$ we have by Cauchy's integral formula $$f(a)=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(z)}{z-a}\,dz$$ for every $a\in \text{int} D$ (interior of $D$) and where $\gamma$ is the boundary of the disk $D$. In particular for disks $D$ such that $D\cap N(x_0)\neq\emptyset$ we get that $f(a)=0$ for all $a\in\text{int}D$. You can then pick an element $z_1\in D\setminus N(x_0)$ and construct another disk around $z_1$ call it $D_1$ such that it is entirely inside $G$ and $D_1\setminus D\neq\emptyset$. Again by Cauchy's formula we get $f(a)=0$ for all $a\in D_1$. Then pick an element $z_2\in D_1\setminus D$ and construct another disk $D_2$ such that it is again entirely included in $G$ and satisfies $D_2\setminus D_1\neq\emptyset$. Again apply Cauchy's formula on the interior of $D_2$. Proceeding in this way you can cover all of $G$ with disks where $f(z)$ vanishes identically. This shows that indeed $f(z)=0$ for all $z\in G$.

Answer 2

This is a consequence of Schwarz's reflexion principle and the identity theorem. Let $D$ be the union of $G$, its reflection on the real axis and the interval $[a,b]$ (which I assume is contained in $(0,+\infty)$.) Define on $D$ the function $$ F(z)=\begin{cases}f(z) & \text{if }z\in G,\\ \overline{f(\bar z)} & \text{if }\bar z\in G,\\ 0 & \text{if }z\in [a,b].\\ \end{cases} $$ Clearly $F$ is holomorphic on $G$, on its reflection on the real axis, and continuous on $D$. Using Morera's theorem it is easy to show that in fact $F$ is holomorphic in $D$. Since it vanishes on $[a,b]$, the identity theorem proves that $F$ is identically zero.