Exercise 15.3.5

Proof. (a) We recall the Principle of Analytic Continuation (or Identity Theorem), given in [13,6.1.1] in some larger context:

“ Let $f,g$be analytic in a region (connected open set) $\mathrm{\Omega }\subset ℂ$. Suppose that there is some $a\in \mathrm{\Omega }$, and a sequence ${(z_n)}_{n\in \mathbb{N}}\in {(\mathrm{\Omega }\setminus \{a\})}^{\mathbb{N}}$of points of $\mathrm{\Omega }$distinct of $a$converging to $a\in \mathrm{\Omega }$, such that $f(z_n)=g(z_n)$for all $n\in \mathbb{N}=\{0,1,2,\cdots \}$. Then $f(z)=g(z)$for all $z\in \mathrm{\Omega }$.”

Here $\mathrm{\Omega }=\{z\in ℂ\mid z\ne (m+\mathit{in})\frac{\varpi }{2},m\equiv n\equiv 1(\mathit{mod}2)\}$. Then $\mathrm{\Omega }\supset ℝ$ is open, and path-connected, thus is connected. Suppose that $f,g$ are analytic on $\mathrm{\Omega }$, and $f(x)=g(x)$ for all $x\in ℝ$. Since any point $a$ of $ℝ$ is a limit of some sequence ${(z_n)}_{n\in \mathbb{N}}\in {ℝ}^{\mathbb{N}}$ (for instance $z_n=a+\frac{1}{n+1},n\in \mathbb{N}=\{0,1,2,\cdots \}$), where $z_n\ne a$ for all $n\in \mathbb{N}$. Since $z_n\in ℝ$ for all $n$, $f(z_n)=g(z_n)$. The Principle of Analytic Continuation shows that $f(z)=g(z)$ for all $z\in \mathrm{\Omega }$. (b) If we define $f,g:\mathrm{\Omega }\to ℂ$ by $f(z)={\phi {}^{\prime }}^2(z),g(z)=1-{\phi }^4(z)$ for all $z\in \mathrm{\Omega }$, then $f,g$ are analytic and $f(x)=g(x)$ for all $x\in ℝ$ by Section 15.2. Then part (b) shows that $f(z)=g(z)$ for all $z\in \mathrm{\Omega }$, thus

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