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This is a problem from Complex Variable (Conway's book) 2nd ed.

(Section 4.4) 9. Show that if $f: \mathbb{C}\to\mathbb{C}$ is a continuous function such that $f$ is analytic off $[-1,1]$ then $f$ is an entire function.

I already have a solution by Morera's theorem that split this problem in 5 cases. I think this solution is too long and I'm trying to solve this using a different approach. Any ideas ?

Dr Richard Clare
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    Consider $$g_r(z) = \frac{1}{2\pi i}\int_{\lvert \zeta - 15\rvert = r} \frac{f(\zeta)}{\zeta - z},d\zeta$$ (for $\lvert z-15\rvert < r$). How does $g_r(z)$ depend on $r$? – Daniel Fischer Jul 27 '17 at 17:23
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    If (for fixed $z$) you can see that $g_r(z)$ doesn't depend on $z$, it follows that $f$ is entire, since $g_r(z) = f(z)$ for $r < 14$ by the integral formula. – Daniel Fischer Jul 27 '17 at 17:36
  • Ok. I'm trying to get your point. Why you choose 15 ?, and how can I know that $g_r(z)$ is entire to conclude that if $f(z) = g_r(z)$ is entire ?. Where we use the continuity of f and and the analycity on $\mathbb{C} \setminus [-1,1]$. ? Sorry if my questions seems obvious to you. – Dr Richard Clare Jul 27 '17 at 17:42
  • I know that $f(z) = \frac{1}{2\pi i}\int_{\gamma}\frac{f(\psi)}{\psi - z}$ for some $\gamma(t) = a+ re^{it}, t \in [0,2\pi]$ where $f$ is analytic in $G$ and $\overline{B}(a;r) \subseteq G$. And you choose 15 and 14 because the ball is not touching $[-1,1]$. – Dr Richard Clare Jul 27 '17 at 17:43
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    The $15$ is completely arbitrary. All we need is a centre $z_0$ that lies outside the problematic segment $[-1,1]$, so that we know $g_r(z) = f(z)$ for small $r > 0$ and $\lvert z - z_0\rvert < r$. The integral theorem gives the independence of $g_r(z)$ from $r$ for $r > \lvert z_0\rvert + 1$ (for $z$ so that all considered $g_r$ are defined at $z$), so $h(z) := \lim\limits_{r\to\infty} g_r(z)$ defines an entire function. For $r < \lvert z_0\rvert < 1$, $g_r(z) = f(z)$ where $g_r$ is defined. Since $f$ is continuous and holomorphic outside $[-1,1]$, – Daniel Fischer Jul 27 '17 at 17:56
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    one can move the contour over that segment without changing the integral (for $\lvert z-z_0\rvert$ small enough), so $h = f$ on a neighbourhood of $z_0$. The identity theorem says $h = f$ on $\mathbb{C}\setminus [-1,1]$, and continuity says $h = f$ everywhere. – Daniel Fischer Jul 27 '17 at 17:56

1 Answers1

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Like Daniel Fischer, I used a Cauchy integral formula approach. Let $\gamma (t) = 2e^{it}, 0\le t \le 2\pi.$ For small $\epsilon>0$ let $R_\epsilon=[-1-\epsilon,1+\epsilon]\times [-\epsilon,\epsilon].$ Consider $\partial R_\epsilon$ as a contour in the natural way, oriented clockwise.

Let $z \in D(0,2)\setminus [-1,1].$ Then $z\notin R_\epsilon$ for small $\epsilon >0.$ For such $\epsilon$ we have by Cauchy

$$f(z) = \frac{1}{2\pi i}\left ( \int_{\gamma + R_\epsilon} \frac{f(w)}{w-z}\,dw\right ).$$

Verify that the integral over $R_\epsilon \to 0$ as $\epsilon\to 0$ (here using the continuity of $f$). Thus we have

$$f(z) = \frac{1}{2\pi i}\left ( \int_\gamma \frac{f(w)}{w-z}\,dw\right )$$

for all $z \in D(0,2)\setminus [-1,1].$ But the last integral defines a holomorphic function in all of $D(0,2).$ Simply by continuity, $f$ equals this integral in all of $D(0,2)$ and we're done.

zhw.
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  • I can understand everything except the last step. – Dr Richard Clare Jul 27 '17 at 18:30
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    @RichardClare what do you mean by the last step? – zhw. Jul 27 '17 at 18:38
  • Simply by continuity, f equals this integral in all of D(0,2) and we're done – Dr Richard Clare Jul 27 '17 at 19:02
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    @RichardClare Call the integral $I(z).$ Then $I$ is holomorphic in $D(0,2).$ And we've seen $f(z)= I(z)$ in $D(0,2)\setminus [-1,1].$ The latter set is dense in $D(0,2).$ So you have two continuous functions on $D(0,2)$, $f(z)$ and $I(z),$ that agree on a dense subset of $D(0,2).$ Hence these two functions agree in $D(0,2),$ and therefore $f$ is holomorphic in $D(0,2).$ – zhw. Jul 27 '17 at 19:15
  • Thank you. The only two points that are not very clear is how can I show that the integrals tends to 0 as $R_{\epsilon} \to 0$ and how this function I(Z) is analytic in D(0,2). All other steps are now clear. – Dr Richard Clare Jul 28 '17 at 16:21
  • @RichardClare For the the contour integral over $\partial R_\epsilon,$ note that the "big part" of it is $\int_{-1}^1 f(t+i\epsilon),dt - \int_{-1}^1 f(t-i\epsilon),dt.$ The limit of this is $0$ by the uniform continuity of $f.$ This is really the only place that the nice straight interval $[-1,1].$ enters the picture. The other parts of the contour integral clearly $\to 0.$ I think you must have seen the proof that $I(z)$ is holomorphic in $D(0,2)$ at some point. This is how you show that holomorphic functions have power series! – zhw. Jul 28 '17 at 17:30
  • what parameteriztion of $$R_1 := (-1-\epsilon) +i\epsilon \to (1+\epsilon) +i\epsilon $$ are you using to transform the integral to say $$ \int_{R_1} \dfrac{f(w)}{w-z} = \int_{-1}^1 f(t+i\epsilon)dt ;; ?$$ – alpastor Feb 05 '20 at 17:57
  • @alpastor I don't mention $R_1.$ I mention $\partial R_\epsilon.$ – zhw. Feb 05 '20 at 18:41
  • Sorry, what did you mean by "the big part" of $\partial R_\epsilon$? I thought you meant the top and bottom of $R_\epsilon$. – alpastor Feb 06 '20 at 04:22
  • @alpastor I defined "big part" in my comment. – zhw. Feb 06 '20 at 16:34