I am unable to prove the following lemma of Cauchy's integral theorem for simple closed curve.
Lemma: Let $R$ be a simply connected region. If $f(z)$ be analytic in $R-\{a\}$ and is continuous on $R$ then for all simple closed curves $\gamma$ in $R$ $$\int_\gamma f(z)dz = 0$$
This result really surprised me a lot! Because the point $a$ may lies on $\gamma$ or inside $\gamma$ (If it lies outside $\gamma$ then it is obvious that, integral evaluates to zero due to 'Cauchy's integral theorem for simple closed curve')
However, I have still manged to prove this lemma for the case of triangle. That is when $\gamma$ is parametrized boundary of a triangle contained in $R$.
So from this, certainly I can extend this proof for the case of any closed polygon.
But, I am unable to extend it for the case of any simple closed curve(due to the point $a$ where $f$ is not analytic)
Moreover, I like to know, what if the function $f(z)$ is not analytic at more than one points? Though it continuous at those points. (I mean, can we extend this lemma?)
Please help me.
Best Answer
One way of establishing this -- the approach taken, for example, by Rudin in his book Real and Complex Analysis -- is to prove the following version of Cauchy's theorem for a triangle
Theorem: Let $D$ be a domain (connected open subset) of $\mathbb C$ and let $f \colon D\backslash \{a\} \to \mathbb C$ be a function defined on all but one point of $D$, and assume that $f$ is holmorphic on $D\backslash\{a\}$ and that there is an $r>0$ such that $|f(z)|\leq C_a$ for all $z\in B(a,r)\backslash\{a\}$ where $C_a$ is some positive constant. Then if $T$ is any closed triangular path in $D$,
$$ \int_T f(z)dz =0, $$ whether or not $a$ lies on the path or in its interior.
Proof: One can deduce this fairly easily from the theorem for a triangle where the domain does not contain any ``bad'' points: First one can reduce to the case where $a$ is a vertex of the triangle -- if $a$ is not on the path or in its interior, then the claim follows from the standard result for a triangle. If $a$ lies in the closed region bounded by the triangle, then one first reduces to the case where $a$ is a vertex of the triangle: If $a$ is not already a vertex, join $a$ to all three vertices of the triangle. This will divide the original triangle into three sub-triangles (or two, if $a$ is located on one of the edges of the triangle).
For the case where $a$ is a vertex of the triangle $T$ and label the other two vertices $b$ and $c$, so that for clarity we may write $T(abc)$ in place of $T$. Now introduce points $d$ and $e$ on the edges $ab$ and $ac$ respectively, both very close to $a$. Then
$$ \begin{split} \int_{T[abc]} f(z)dz &= \int_{T(dae)} f(z)dz + \int_{T(dbe)} f(z)dz + \int_{T(dcb)}\\ &= \int_{T(dae)}f(z)dz \end{split} $$ where in going from the first to the second line we used the fact that the last two integrals on the right-hand side of the first line vanish by the ordinary version of Cauchy's theorem for a triangle. (See the attached figure)
But then it follows that $$ 0\leq |\int_{T(abc)} f(z)dz| = |\int_{T(dae)} f(z)dz| \leq \ell(T(dae)).C_a $$ and since $\ell(T(dae)) \to 0$ as $d,e \to a$, it follows that $\int_{T(abc)} f(z)dz=0$ as required.
Once you have this result, by induction the result holds for a function with finitely many "bad points". Moreover, the proof of Cauchy's theorem for a convex domain immediately goes through to the case with ``bad points'', and similarly one deduces Cauchy's Integral Formula. (This is the reason Rudin proves the "bad point" version of the result -- if you want to prove the integral formula, it is natural to consider the function $g(z)=\frac{f(z)-f(z_0)}{z-z_0}$. If $f$ is holomorphic in a domain $D$, then $g(z)$ is also, except perhaps at $z=z_0$, where one only knows $g$ is continuous. If you can still apply Cauchy's theorem for a disk to such functions, the integral formula follows immediately.
To give a bit more detail: we say that $S$ is a star-shaped domain in the complex plane, if there is a point $s_0 \in S$ with the property that the line segment joining it to any other point in $S$ lies entirely inside $S$. (Thus a convex set is start-shaped with respect to any of its points for example). For such a domain, given a function $f$ with the property that its integral around any triangle is zero, the function $F(w) = \int_{[s_0,w]} f(z)dz$ where $[s_0,w]$ is the path $t \mapsto (1-t)s_0 +tw$. Indeed $F(w+h)-F(w)= \int_{[w,w+h]} f(z)dz$ because the integral of $f$ around the triangle with vertices $s_0,w$ and $w+h$ is zero. It is then easy to see that $F'(w)=f(w)$. But then for any closed contour in the star-like domain, $\int_\gamma f(z)dz =0$, since it is the difference in values of the primitive at the start and end points of the path, which for a closed path are the same.
But Cauchy's Integral Formula (for a disk say) readily shows that a holomorphic function is equal to its Taylor series, and since one now has the Integral Formula for any function which is holomorphic except perhaps at one point, it follows that any such function is equal to its Taylor series, and hence is differentiable (and in fact infinitely differentiable).
Note that because $\mathbb C$ is locally convex, we do not need Cauchy's theorem for a simply-connected domain at all, since it suffices to show that the restriction of $f$ to a disk containing $a$ is holomorphic everywhere in that disk in order to deduce that $f$ was in fact differentiable on the entire simply-connected domain. Indeed if $R$ is a simply-connected region and $f$ is holomorphic function defined on $R\backslash S$ where $S$ is a discrete subset if $R$ (that is, the subspace topology on $S$ is the discrete topology) with $f$ bounded near each element of $S$, then $f$ extends uniquely to a holomorphic function on all or $R$.
Update: A variant of the above approach would be to use a Morera-type result: If $f\colon D \to \mathbb C$ is a function (where $D$ is a domain in $\mathbb C$), and $\int_T f(z)dz=0$ for all triangles in $D$, then if we fix $z_0 \in D$ and pick an open disk $B(z_0,r)\subseteq D$, then $f$ has a primitive in the disk $B(z_0,r)$ (that is, there is a function $F \colon B(z_0,r) \to \mathbb C$ such that $F'=f$. But since $F$ is differentiable, it is analytic and hence infinitely differentiable, so that, in particular, $f$ is differentiable.
Thus it follows that if $f$ is holomorphic except perhaps at one point, nearby which it is bounded, to show that $f$ is in fact holomorphic everywhere it suffices to show that $\int_T f(z)dz=0$ for all triangles $T$.