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本帖最后由 hbghlyj 于 2023-1-1 22:42 编辑 相关帖子: Morera's Theorem
complex.pdf
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 $\int_\gamma f(z) d z=0$, then $f$ has a primitive.
Proof. Fix $z_0$ in $U$, and for any $z \in U$ set
\[
F(z)=\int_\gamma f(z) d z .
\]
where $\gamma:[a, b] \rightarrow U$ with $\gamma(a)=z_0$ and $\gamma(b)=z$.
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
\[
0=\int_\gamma f(z) d z=\int_{\gamma_1} f(z) d z+\int_{\gamma_2^{-}} f(z) d z,
\]
hence since $\int_{\gamma_2^{-}} f(z) d z=-\int_{\gamma_2} f(z) d z$ we see that $\int_{\gamma_1} f(z) d z=\int_{\gamma_2} f(z) d z$.
Next we claim that $F$ is differentiable with $F^{\prime}(z)=f(z)$. 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 $s(t)=w+t(z-w)$ from $w$ to $z$ is a path $\gamma_1$ from $z_0$ to $z$. It follows that
\begin{aligned}
F\left(z_1\right)-F(w) &=\int_{\gamma_1} f(z) d z-\int_\gamma f(z) d z \\
&=\left(\int_\gamma f(z) d z+\int_s f(z) d z\right)-\int_\gamma f(z) d z \\
&=\int_s f(z) d z .
\end{aligned}But then we have for $z_1 \neq w$
\[
\begin{aligned}
\left|\frac{F\left(z_1\right)-F(w)}{z_1-w}-f(w)\right| &=\left|\frac1{z_1-w}\left(\int_0^1 f(w+t\left(z_1-w\right))\left(z_1-w\right) d t-f(w)\right)\right|\\
&=\left|\int_0^1\left(f\left(w+t\left(z_1-w\right)\right)-f(w)\right) d t\right| \\
& \leq \sup _{t \in[0,1]}\left|f\left(w+t\left(z_1-w\right)\right)-f(w)\right| \\
& \rightarrow 0 \text { as } z_1 \rightarrow w
\end{aligned}
\]
as $f$ is continuous at $w$. Thus $F$ is differentiable at $w$ with derivative $F^{\prime}(w)=f(w)$ as claimed.
注1.
这个定理中$U$适用条件是“连通区域”, 不需要simply connected(单连通), 比如对于$U=\{z\in\Bbb C:2<\abs z<3\},f(z)=\frac1{z-1}-\frac1{z+1}$定理适用.
注2.
给了这个定理的条件, 由Morera's theorem推出 $f$ 是可导的. |
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