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If $f:\mathbb C \longrightarrow \mathbb C$ is continuous and $f$ is analytic off the real axis, then show that $f$ is entire.

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I tried using Morera's theorem but I can't seem to get it right –  user60184 Jan 30 '13 at 4:23
    
what do you mean by off? –  lee Jan 30 '13 at 4:29
    
@lee $f$ is analytic on $\mathbb{C}\backslash\mathbb{R}$ –  Brett Frankel Jan 30 '13 at 4:31
    
@lee as Brett said we know that $f$ is analytic on $\mathbb C$\ $\mathbb R$ and $f$ is continuous everywhere this should be enough to get that $f$ is entire i.e $f$ will be analytic on the real axis. –  user60184 Jan 30 '13 at 4:40
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2 Answers

By continuity, $f$ is bounded on $\{x+iy: -a\leq x\leq a, -\varepsilon \leq y \leq \varepsilon\}$. Then by dominated convergence, $\lim_{y\to 0}\int_{-a}^af(x+iy)\ dx=\int_{-a}^af(x)\ dx$.

Now we can take the integral of $f$ on a cirlce of radius $a$ centered at $0$ as the limit of the integrals over two semi-circles, one in each half-plane, and apply Morera's theorem.

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Can you please explain more, and why it's enough to consider circles only ?? Thanx –  user60184 Jan 30 '13 at 4:47
    
If you look at the proof of Morera's theorem, you'll see that circles suffice. But you could apply the argument above to any contour integral. –  Brett Frankel Jan 30 '13 at 4:52
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Theorem: (Symmetry principle) If $f^+$ and $f^−$ are holomorphic functions in $Ω^+$ and $Ω^−$ respectively, that extend continuously to I and $$f^+(x) = f^−(x)\quad for\;all\;x ∈ I,$$ then the function $f$ defined on $Ω$ by $$f(z) = $$$$f^+ (z)\quad if\quad z ∈ Ω^+ ,\quad f^+(z) = f^−(z)\quad if\quad z ∈ I,\quad f^−(z)\quad if\quad ∈ Ω^−$$ is holomorphic on all of $Ω$.(COMPLEX ANALYSIS Elias M. Stein & Rami Shakarchi page 58)

According to the theorem you have $f^+=f^-=f$ , $Ω^+$ is upper half plain $Ω^-$ is lower half plain and $I=\mathbb{R}$ so $f$ is holomorphic over $\mathbb{C}$ means entire.

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