# Complex functions with local primitives

Problem from the book: "A Course in Complex Analysis" - Fisher|Lieb .

Show that if a function $f:G\rightarrow \mathbb{C}$ has local primitive, then for every closed triangle $\Delta\subset G$ , we have

$\int_{\partial\Delta}f(z)dz=0$.

My proof: I take the open cover $\mathcal{U}$of $\Delta$ , defined as $\mathcal{U}=\lbrace B(x):x\in \Delta , f_{\mid U(x)} \: has \: primitive\rbrace$, whith $B(x)$ a ball centered in $x\in \Delta$.

Since the triangle $\Delta$ is a compact subset of $\mathbb{C}$ , exists $U_1,U_2,\ldots , U_n\in \mathcal{U}$ such that $\Delta \subseteq\bigcup_{i=1}^nU_i$. Now I can divide $\Delta$ in a finite number of triangles $T_j \subset \Delta$, $j=1,2,\ldots , k$ such that for all $j\in \lbrace 1,2,\ldots , k\rbrace$ we have $T_j\subset U_i$ for some $i\in \lbrace 1,2,\ldots,n\rbrace$. Then

$\int_{\partial\Delta}f(z)dz=\sum_{j=i}^k\int_{\partial T_j}f(z)dz=0$ because $\int_{\partial T_j}f(z)dz=0$ , $\forall j\in \lbrace 1,2,\ldots , k\rbrace$.

My proof is correct?

• I am wary of why this should be true: $\int_{\partial\Delta}f(z)dz=\sum_{j=i}^k\int_{\partial T_j}f(z)dz$. If I was your grader, I would certainly at least need some more justification. – TomGrubb Jan 2 '16 at 0:36
• I take the same orientation for all $\partial T_j$. – Vincenzo Zaccaro Jan 2 '16 at 0:39
• You are still making a leap here. You need to justify in more depth why you can find such triangles, and why the two integrals are equal. – TomGrubb Jan 2 '16 at 0:46
• @Vincenzo: Can you elaborate hypothesis more (second line)? I was considering $f(z)=1/z$ with $G=\mathbb{C}-\{0\}$. – p Groups Jan 2 '16 at 3:48

You certainly have the right idea, but let's tighten up the argument a bit. Fix $\Delta \subset G.$ Claim: There exists $r>0$ such that if $\Delta_1 \subset \Delta$ is a triangle with $\text { diam } \Delta_1 < r,$ then $\int_{\partial \Delta_1} f = 0.$ Proof: If not, then there exist $\Delta_n \subset \Delta$ with $\text { diam } \Delta_n \to 0$ and $\int_{\partial \Delta_n} f \ne 0.$ Let $z_n \in \Delta_n.$ Because $\Delta$ is compact, some subsequence $z_{n_k} \to z_0$ for some $z_0 \in \Delta.$ Now there is a disc centered at $z_0$ where $f$ has a primitive. For large enough $k,\Delta_{n_k}$ is contained in this disc and the corresponding integrals all equal $0,$ contradiction.
Now there is a canonical way to divide $\Delta$ into subtriangles: Simply add the midpoints of the sides of $\Delta$ into the mix to obtain four subtriangles, each of which has diameter equal to $(1/2)\text { diam } \Delta.$ The smaller triangles inherit an orientation from $\partial \Delta$ so that the integral of $f$ over $\partial \Delta$ is the sum of the corresponding integrals with respect to the smaller triangles. This subdivision process can be continued. At the $n$th stage we will have $4^n$ subtriangles, each with diameter equal to $(\text { diam } \Delta)/4^n.$ When the last quantity is $<r,$ the integral over the boundaries of all subtriangles is $0.$ Hence the integral over $\partial \Delta =0.$