converse to the Fundamental Theorem of Calculus

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complex.pdf page 35
Theorem 5.22. If $U$ is a domain (i.e. it is open and path connected) and $f: U \rightarrow$ $\mathbb{C}$ is a continuous function such that for any closed path in $U$ we have $\int_\gamma f(z)\mathrm{\ d}z=0$, then $f$ has a primitive.
Proof. Fix $z_0$ in $U$, and for any $w \in U$ set
\[
F(w)=\int_\gamma f(z)\mathrm{\ d}z .
\]
where $\gamma:[a, b] \rightarrow U$ with $\gamma(a)=z_0$ and $\gamma(b)=w$.
We claim that $F(z)$ is independent of the choice of $\gamma$. Indeed if $\gamma_1, \gamma_2$ are two such paths, let $\gamma=\gamma_1 \star \gamma_2^{-}$be the path obtained by concatenating $\gamma_1$ and the opposite $\gamma_2^{-}$of $\gamma_2$ (that is, $\gamma$ traverses the path $\gamma_1$ and then goes backward along $\gamma_2$). Then $\gamma$ is a closed path and so, using Proposition 5.14 we have
\[
0=\int_\gamma f(z)\mathrm{\ d}z=\int_{\gamma_1} f(z)\mathrm{\ d}z+\int_{\gamma_2^{-}} f(z)\mathrm{\ d}z,
\]
hence since $\int_{\gamma_2^{-}} f(z)\mathrm{\ d}z=-\int_{\gamma_2} f(z)\mathrm{\ d}z$ we see that $\int_{\gamma_1} f(z)\mathrm{\ d}z=\int_{\gamma_2} f(z)\mathrm{\ d}z$.

Next we claim that $F$ is differentiable with $F'(z)=f(z)$. To see this, fix $w \in U$ and $\epsilon>0$ such that $B(w, \epsilon) \subseteq U$ and choose a path $\gamma:[a, b] \rightarrow U$ from $z_0$ to $w$. If $z_1\in B(w, \epsilon) \subseteq U$, then the concatenation of $\gamma$ with the straight-line path $s:[0,1] \rightarrow U$ given by $s(t)=w+t(z_1-w)$ from $w$ to $z_1$ is a path $\gamma_1$ from $z_0$ to $z_1$. It follows that\begin{aligned}
F\left(z_1\right)-F(w) &=\int_{\gamma_1} f(z)\mathrm{\ d}z-\int_\gamma f(z)\mathrm{\ d}z \\
&=\left(\int_\gamma f(z)\mathrm{\ d}z+\int_s f(z)\mathrm{\ d}z\right)-\int_\gamma f(z)\mathrm{\ d}z \\
&=\int_s f(z)\mathrm{\ d}z .
\end{aligned}But then we have for $z_1 \neq w$\begin{aligned}
\left|\frac{F(z_1)-F(w)}{z_1-w}-f(w)\right| &=\left|\frac{\int_0^1 f\big(w+t(z_1-w)\big)\cancel{(z_1-w)}\mathrm{\ d}t}{\cancel{z_1-w}}-f(w) \right|\\
&=\left|\int_0^1\left(f\big(w+t(z_1-w)\big)-f(w)\right)\mathrm{\ d}t\right|\\
&\leq \sup _{t \in[0,1]}\bigl|f\big(w+t(z_1-w)\big)-f(w)\bigr| \\ &\rightarrow 0 \text { as } z_{1} \rightarrow w
\end{aligned}as $f$ is continuous at $w$. Thus $F$ is differentiable at $w$ with derivative $F'(w)=f(w)$ as claimed.

Remark 5.23. Note that any two primitives for a function $f$ differ by a constant: This follows immediately from Corollary 5.21, since if $F_1$ and $F_2$ are two primitives, their difference $\left(F_1-F_2\right)$ has zero derivative.

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lectures14-Earl.pdf page69
Theorem 206 (Antiderivative Theorem) Let $f$ be a function which is holomorphic on a convex domain $U$. Then there exists a holomorphic function $F$ on $U$ such that $F'(z) = f(z)$.
Proof Fix $a \in U$ and for any $z \in U$ let $[a, z]$ denote the oriented line segment from $a$ to $z$. Define
\[
F(z)=\int_{[a, z]} f(w) \mathrm{d} w .
\]
Let $\varepsilon>0$ be such that $D(z, \varepsilon) \subseteq U$ and take $h$ with $|h|<\varepsilon$. By Cauchy's Theorem for triangles
\[
\int_{[a, z]} f(w) \mathrm{d} w+\int_{[z, z+h]} f(w) \mathrm{d} w+\int_{[z+h, a]} f(w) \mathrm{d} w=0
\]
which may be rearranged to
\[
F(z+h)-F(z)=\int_{[z, z+h]} f(w) \mathrm{d} w
\]
So
\[
\left|\frac{F(z+h)-F(z)}{h}-f(z)\right|=\left|\left(\frac{1}{h} \int_{[z, z+h]} f(w) \mathrm{d} w\right)-f(z)\right|=\left|\frac{1}{h} \int_{[z, z+h]}[f(w)-f(z)] \mathrm{d} w\right|
\]
as $\int_{[z, z+h]} k \mathrm{~d} w=k h$ for any constant function $k$ of $w$. Finally, by the Estimation Theorem, we have
\[
\begin{aligned}
\left|\frac{F(z+h)-F(z)}{h}-f(z)\right| &=\left|\frac{1}{h} \int_{[z, z+h]}[f(w)-f(z)] \mathrm{d} w\right| \\
& \leqslant \frac{1}{|h|} \times|h| \times \sup _{w \in[z, z+h]}|f(w)-f(z)| \\
&=\sup _{w \in[z, z+h]}|f(w)-f(z)|
\end{aligned}
\]
which tends to zero as $h → 0$ by the continuity of $f$ at $z$, showing that $F'(z) = f (z)$ as required.