If $f$ is a continuous mapping from $\mathbb C$ to $\mathbb C$ such that $f$ is analytic except $[-1,1]$ then $f$ is entire function

I want to prove this important theorem. Is there anyone who is going to help me with this? How can I use the fact analytic except $[-1,1]$?

  • $\begingroup$ How is it wrong? Is is an important result. $\endgroup$
    – Kavita
    Mar 30, 2017 at 3:31
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    $\begingroup$ @Kavita just to confirm, you mean $[-1,1]=\{a+bi:b=0\land |a|\leq 1\}$? I ask because interval notation isn't particularly common when talking about subsets of $\mathbb{C}$ $\endgroup$ Mar 30, 2017 at 3:37
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    $\begingroup$ Nvm. If it's continuous then I believe the result is true. $\endgroup$ Mar 30, 2017 at 3:38
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    $\begingroup$ Use Morera's theorem. $\endgroup$ Mar 30, 2017 at 3:39
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    $\begingroup$ I've changed the title of the question. Excessively vague titles are discouraged as they make questions hard for people with similar questions to find later. $\endgroup$ Mar 30, 2017 at 3:41

2 Answers 2


Apply Morera's Theorem. enter image description here

Sorry for the ugly picture. Any how, by continuity when the two segments on the polygons come close to $[-1, 1]$ they cancel each other out. Since $f$ is analytic on $\mathbf{C}\backslash [-1, 1]$ then use Cauchy's Theorem on the inner region of the two polygons.


We use Morera's Theorem. It suffices to show that if $T$ is any triangle in $\mathbb{C}$ that crosses $[-1,1]$, $\int_T f\,dz=0$.

The idea is as follows. Let $\gamma$ be the closed oriented curve round $T$ and $p$ be the point where $\gamma$ crosses $[-1,1]$. Choose points $q$ and $r$ near $p$ (one before and one after $p$), and let $\alpha$ be the rectangular curve that goes round $[-1,1]$ connecting $r$ and $q$.

An example is given in the image below.

enter image description here

Here, $[-1,1]$ is represented by the black line, the red triangle is $\gamma$ and $\alpha$ is the green segment. Say $r$ is the point below $[-1,1]$ and $q$ is the point above it.

Since $f$ is analytic outside $[-1,1]$, we have that the integral along the following path is $0$: start from $q$ and go along $\gamma$ until you meet $r$, then go along $-\alpha$ (opposite orientation) until you meet $q$ again. If we call the red segment of this path $\gamma'$, we thus have that

$$\int_{\gamma'}f\,dz=\int_\alpha f\,dz$$

On the other hand, $\int_\gamma f\,dz=\int_{\gamma'}f\,dz+\int_r^qf\,dz=\int_\alpha f\,dz+\int_r^qf\,dz$. In other words, integrating round the red triangleis the same as integrating round the red-green rectangle.

The thing is, we can repeat this construction making the rectangle tighter and tighter round the interval $[-1,1]$. In the end, it will be very close to going forward then backwards (orientation matters for canceling!) the black segment (inside the rectangle). Using the continuity of $f$, we can make this as small as we want so that in the limit $\int_\alpha f\,dz+\int_r^qf\,dz$ is $0$ and hence $\int_\gamma f\,dz$ is also $0$.

Of course, the triangle doesn't have to be like the one in the image above, this is just meant to give you an idea of how it goes.


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