0
$\begingroup$

I was reading the proof of Morera's Theorem on Arfken's book and in it he glossed over one part of the proof.

Here it says that given a function $f(z)$ that is continuous and satisfies $\oint_C \! f(z) \, dz = 0$, integrating it from z_1 to z_2 would result in:

$$F(z_2) - F(z_1) = \int_{z_1}^{z_2} \! f(z) \, dz$$

which results in:

$$\frac{F \left(z_2 \right) - F \left(z_1 \right)}{z_2 - z_1} - f \left(z_1 \right) = \frac{\int_{z_1}^{z_ 2} \, \left[ f \left(t \right) - f \left(z_1 \right) \right] \, dt}{z_2 - z_1}$$

Using the variable $t$ as another complex variable. If we take the limit $z_2 \to z_1$ this gives:

$$ \lim_{z_2 \to z_1} \frac{\int_{z_1}^{z_ 2} \, \left[ f \left(t \right) - f \left(z_1 \right) \right] \, dt}{z_2 - z_1} = 0$$

Since $f \left( t \right)$ is continuous. This is the part where I am confused in since he only remarked that he used the mean value theorem to prove that it is zero. However it is my understanding that using the mean value theorem on the previous equation would result in:

$$\begin{align} lim_{z_2 \to z_1} \frac{\int_{z_1}^{z_ 2} \, \left[ f \left(t \right) - f \left(z_1 \right) \right] \, dt}{z_2 - z_1} & = 0 \\ lim_{z_2 \to z_1} \left[ \frac{\int_{z_1}^{z_2}\, f \left(t \right) \,dt}{z_2 - z_1} - \frac{\int_{z_1}^{z_2} \, f \left(z_1 \right) \, dt}{z_2 - z_1}\right] & = 0 \tag{1}\\ f \left(c \right) - lim_{z_2 \to z_1} f \left(z_1 \right) \frac{z_2-z_1}{z_2-z_1} & = 0 \tag{2}\\ f \left(c \right) - f \left(z_1 \right) & = 0 \\ \end{align}$$

Which wouldn't result in zero unless of course $f \left(c \right) = f \left(z_1 \right)$ which I think is not the case here. Is my application of the mean value theorem incorrect here or is it really $f \left(c \right) = f \left(z_1 \right)$?

$\endgroup$
  • $\begingroup$ $c$ is related to $z_2$. $\endgroup$ – C.Ding Nov 10 '17 at 9:25
  • $\begingroup$ @C.Ding What do you mean? $c = z_2$? $\endgroup$ – Aldon Nov 10 '17 at 9:30
  • $\begingroup$ $\lim_{z_2 \to z_1} \frac{\int_{z_1}^{z_2}f \left(t \right) \,dt}{z_2 - z_1} =\lim_{z_2 \to z_1} f(c_{z_2})$ $\endgroup$ – C.Ding Nov 10 '17 at 9:34
  • $\begingroup$ Can you add some information about what are $f, z_1, z_2, F$? $\endgroup$ – Gribouillis Nov 10 '17 at 9:37
  • $\begingroup$ @Gribouillis $f$ is any function that is continuous and satisfies $\oint_C \! f(z) \, dz = 0$, $z_1, z_2$ are points in the region and $F$ is the antiderivative of $f$. $\endgroup$ – Aldon Nov 10 '17 at 9:41
1
$\begingroup$

The $c$ (In fact, it should be writen as $c_{z_2}$) in the equation (2) is in the line segments $z_1 z_2$, so $c\to z_1$ as $z_2\to z_1$ and $ \lim_{z_2\to z_1} \frac{\int_{z_1}^{z_2}f(t)dt}{z_2-z_1}=\lim_{z_2\to z_1}f(c)= f(z_1).$

$\endgroup$
  • $\begingroup$ Thank you. However I still don't understand how $c$ is related to $z_2$. Is it because from the mean value theorem that $c \in \left[z_1, z_2 \right]$ such that taking the limit $z_2 \to z_1$ would restrict the value of $c$ to $c=z_1=z_2$? $\endgroup$ – Aldon Nov 10 '17 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.