05 - Existence of Primitives Theorem

Existence of Primitives Theorem

Let $U$ be a domain
Let $f:U\to \mathbb{C}$ be a continuous function

If for any closed path $\gamma$ in $U$ has ${\int }_{\gamma }f(z)dz=0$ then

$$f\text{ has a primitive}$$

Proof

Fix $z_0\in U$

For any $z\in U$ define

$$F(z)={\int }_{\gamma }f(z)dz$$

where $\gamma :[a,b]\to U$ with $\gamma (a)=z_0$ and $\gamma (b)=z$

Need to show that $F(z)$ is independent of the choice of $\gamma$
Suppose ${\gamma }_1,{\gamma }_2$ are two paths with ${\gamma }_1(a)={\gamma }_2(a)=z_0$ and ${\gamma }_1(b)={\gamma }_2(b)=z$ then
Let $\gamma ={\gamma }_1\star {\gamma }_2^{-}$ so $\gamma$ is closed so using Properties of integrals along paths

$$0={\int }_{\gamma }f(z)dz={\int }_{{\gamma }_1}f(z)dz+{\int }_{{\gamma }_2^{-}}f(z)dz={\int }_{{\gamma }_1}f(z)dz-{\int }_{{\gamma }_2}f(z)dz$$

Hence

$${\int }_{{\gamma }_1}f(z)dz={\int }_{{\gamma }_2}f(z)dz$$

Need to show that $F$ is differentiable with $F^{\prime }(z)=f(z)$
Fix $w\in U$ and $ϵ>0$ such that $B(w,ϵ)\in U$

Let any path $\gamma :[a,b]\to U$ with $\gamma (a)=z_0$ and $\gamma (b)=w$
Let straight line path $s:[0,1]\to U$ from $w$ to $z$ be given by

$$s(t)=w+t(z-w)$$

If $z\in B(w,ϵ)\subseteq U$ then concatenation of $\gamma$ with $s$ is a path ${\gamma }_1$ from $z_0$ to $z$ so

$$\begin{array}{rl}F(z)-F(w)& ={\int }_{{\gamma }_1}f(z)dz-{\int }_{\gamma }f(z)dz\\ & =\left({\int }_{\gamma }f(z)dz+{\int }_{s}f(z)dz\right)-{\int }_{\gamma }f(z)dz\\ & ={\int }_{s}f(z)dz\end{array}$$

However for $z\ne w$ then

$$\begin{array}{rl}\left|\frac{F(z)-F(w)}{z-w}-f(w)\right|& =\left|\frac{1}{z-w}\left({\int }_0^{1}f(w+t(z-w))(z-w)dt\right)-f(w)\right|\\ & =\left|{\int }_0^1(f(w+t(z-w))-f(w))dx\right|\\ & \le \underset{t\in [0,1]}{\sup }|f(w+t(z-w))-f(w)|\to 0\text{ as }z\to w\end{array}$$

as $f$ is continuous at $w$

Hence $F$ is differentiable at $w$ with derivative

$$F^{\prime }(w)=f(w)$$