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I have this question which I`m not sure at all how to finish the proof. Please help me: Let $f$ be an analytic function on $B_1(0)$: For $z \in B_1(0)$; let

$z : [0; 1] \to \mathbb C$; $\gamma(t) = tz$: Show that $F(z) =\int_\gamma f dt$ is an analytic function satisfying $F' = f$. I tried to write the integral in terms of $\gamma$ and this will give a function of 2 variables $z$ and $t$. Should I then try to prove Cauchy-Riemman equations by taking derivative using Leibniz rule? Please help me out...

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Do you know Cauchy's theorem (or a weaker version of it, such as this one for triangles)? – Antonio Vargas Sep 18 '12 at 18:20
do you mean that I can prove it like the proof of Cauchs theorem or use Cauchys theorem here? – Danny Sep 18 '12 at 20:47
Yes, sorry if I was unclear. I meant to ask whether you were allowed to use it or not, since there is a nice proof with it. I'll post it as an answer. – Antonio Vargas Sep 18 '12 at 20:57
up vote 8 down vote accepted

Fix $z$ with $|z| < 1$ and let $h$ be a complex number small enough so that $|z+h| < 1$. Denote by $C(z)$ the straight-line path from $0$ to $z$, so that $C(z) = \gamma$ in your notation.

Consider the difference

$$ F(z+h) - F(z) = \int_{C(z+h)} f(t)\,dt - \int_{C(z)} f(t)\,dt = \int_{C(z+h)-C(z)} f(t)\,dt, $$

where in the last integral the line segment $C(z)$ has the reverse orientation, as in the image below.

enter image description here

If we also integrate twice over the line segment connecting $z$ and $z+h$, once from $z$ to $z+h$ then once from $z+h$ back to $z$, the contributions from these traversals will cancel each other and the integral will remain unchanged.

enter image description here

Since $f$ is analytic in the disk, Cauchy's theorem for triangles tells us that the integral around the triangle we have constructed is $0$, which leaves us with only the integral along the line segment from $z$ to $z+h$. Let's call this line segment $E$.

enter image description here

We have shown that

$$ F(z+h) - F(z) = \int_E f(t)\,dt. $$

Since $f$ is analytic at $z$ it is also continuous at $z$, so when $t$ is close to $z$ we can write

$$ f(t) = f(z) + \epsilon(t), $$

where $\epsilon(t) \to 0$ as $t \to z$. Thus, for $h$ small enough, we have

$$ \begin{align} F(z+h) - F(z) &= \int_E f(z)\,dt + \int_E \epsilon(t)\,dt \\ &= f(z) \int_E dt + \int_E \epsilon(t)\,dt \\ &= f(z)h + \int_E \epsilon(t)\,dt. \tag{1} \end{align} $$

By the triangle inequality we have

$$ \left|\int_E \epsilon(t)\,dt\right| \leq L(E) \max_{t \in E} |\epsilon(t)| = |h| \max_{t \in E} |\epsilon(t)|, $$

where $L(E)$ is the length of the line segment $E$. Since $\epsilon(t) \to 0$ for all $t \in E$ as $h \to 0$, we know that $\max_{t \in E} |\epsilon(t)| \to 0$ as $h \to 0$. Thus, by dividing equation $(1)$ by $h$, we may conclude that

$$ \lim_{h \to 0} \frac{F(z+h)-F(z)}{h} = f(z), $$

which shows that $F' = f$ on $|z| < 1$. Note that $F$ is analytic on $|z| < 1$ since it is differentiable there.

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Thanks a lot for you. I really appreciate it. – Danny Sep 18 '12 at 22:52
My pleasure.${}$ – Antonio Vargas Sep 19 '12 at 0:44
By the way @Danny, you can "accept" an answer you like by clicking the checkmark ✔ below the up/down voting arrows on that answer. This removes the question from the "unanswered" list on the site. – Antonio Vargas Sep 19 '12 at 5:33

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