Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 4

A function is holomorphic on an open set when it is complex differentiable at every point of that set. Explicitly, for every $z\in U$, the limitexists as a finite complex number. The limit is independent of the direction in which the complex increment approaches zero; existence of derivatives only along the real axis would not suffice.

Cauchy's integral theorem says that if $f$ is holomorphic on an open set $U$ and a closed, piecewise smooth contour $\gamma$ is null-homotopic in $U$, thenIn particular, this holds for every closed contour when $U$ is simply connected, and for a simple closed contour whose boundary and enclosed region lie in $U$. For the later arguments it suffices to use the triangular version: the integral around a triangle is zero if the triangle and its interior lie in $U$ and $f$ is holomorphic there.

Both the holomorphicity condition and a suitable condition excluding holes inside the contour matter. Continuity alone, which is the starting hypothesis for this question, does not imply the theorem's conclusion; nor does holomorphicity on an arbitrary punctured open set imply zero integral around every loop.

Morera's theorem states that a continuous function on an open subset of $\mathbb{C}$ is holomorphic if its integral around every triangular contour, with triangle and interior contained in that open set, is zero.

To prove this, work in any open disk $D$ whose closure lies in $U$, and fix $z_0\in D$. Definewhere the segment lies in $D$ by convexity. The zero integral on the triangle with vertices $z_0,z,z+h$ givesContinuity at $z$ implies that the last integral tends to $f(z)$ as $h\to 0$: its difference from $f(z)$ is bounded by ${\sup }_{0\le t\le 1}|f(z+th)-f(z)|$. Therefore $F^{\prime }=f$ on $D$. The function $F$ is holomorphic, and the allowed theorem that a holomorphic derivative is holomorphic implies that $f=F^{\prime }$ is holomorphic on $D$.

Every point of $U$ lies in such a disk, proving the result throughout $U$:The proof is local and does not require $U$ to be simply connected.

For two moving vertices $p(s),q(s)$, the integral along their joining segment isThe integral factor is continuous in $s$ by the allowed fact, and the other factor is continuous because the vertices are continuous. Thus $J_{p,q}$ is continuous, including when its two endpoints coincide.

The integral around the triangle is exactlyEach term is continuous, so their sum is continuous at zero andNo holomorphicity is used here. Compactness of the fixed containing triangle also gives a direct proof of the allowed fact through uniform continuity of $f$.

First take $z\in U_{-}$ and put $w=\bar{z}\in U_{+}$. Reflection symmetry of the function givesHence $f$ is holomorphic on $U_{-}$, with $f^{\prime }(z)=\overline{f^{\prime }(\bar{z})}$ there.

It remains to cross the real axis. Fix a real point $p\in U$ and a small disk about $p$ with closure in $U$. Consider a triangle in a still smaller disk. Intersect its filled interior with the closed upper and lower half-planes and triangulate the resulting convex polygons. This gives finitely many triangles lying on one side of the real axis, possibly with edges on that axis. Degenerate pieces contribute zero.

For an upper piece, translate its three vertices by $i\epsilon$ with $\epsilon >0$. For sufficiently small $\epsilon$ the translated triangle lies entirely in $U_{+}$ and remains in $U$, so its boundary integral is zero by Cauchy's theorem. Part (d), or its segment-integral proof, shows that the integral converges to that around the untranslated piece as $\epsilon \downarrow 0$. Thus the original piece has zero integral. Translate lower pieces by $-i\epsilon$ to obtain the same conclusion in $U_{-}$.

When all the oriented triangle integrals are added, their internal edges cancel, including the segments on the real axis. The original triangle therefore has zero boundary integral. Continuity of $f$ on $U$ now allows Morera's theorem in the disk, proving holomorphicity near $p$. Together with the two half-plane arguments this givesThis is a Morera gluing lemma across a straight line; the original continuity hypothesis is essential to the limiting step.