Does Cauchy's theorem's hold if we only assume boundedness?

Let $$f$$ be a function $$\mathbb C \to \mathbb C$$. I am not assuming $$f$$ is analytic on $$\mathbb C$$, so Cauchy-Goursat does not apply.

Suppose $$\gamma$$ is a simple closed contour, and suppose that the region $$D = {\rm int}(\gamma) \cup \gamma$$ can be approximated arbitrarily by little squares of arbitrarily small size. I would like to know whether $$\oint_\gamma f = 0$$, under the assumption that $$f$$ is continuous and bounded on $$D$$.

I believe the answer is yes. Since the integral around $$\gamma$$ is equivalent to adding up the integrals from all the small squares, it is sufficient to show that the integral around the small squares approaches zero, which trivially follows from the Cauchy ML-inequality if $$f$$ is bounded on $$D$$.

Is this correct? So there is no need to assume analyticity if a continuous function is bounded?

• The trouble is that the integral of a continuous non-analytic function over a loop can be nonzero. Simple example: $f(z)=\overline z$ over the unit circle with centre $0$. – Angina Seng Dec 31 '18 at 22:44
• I do not really get your argumentation. Let's try to repeat it for the interval $[a,b]$ on the line. An integral $\int_a^b f(x)\,dx$ is the sum of integrals over tiny intervals, and since any such integral approaches zero, the total integral is zero for any continuous function $f$. Right? – A.Γ. Dec 31 '18 at 23:14
• @A.Γ. See the answer by Kenny Wong. – DanielWainfleet Jan 1 '19 at 5:07

It's true that, as the square size approaches zero, the contributions from the individual squares tend to zero. But as the square size approaches zero, the number of squares tends towards infinity! For your argument to work, you need some sort of guarantee that the rate at which the integrals on the individual squares are getting smaller is sufficient to offset the rate at which the number of squares is getting bigger. This is where we require holomorphicity, rather than just boundedness.

For a different perspective on this, let's try and construct a rigorous proof in the special case where the original contour $$C$$ is one big square of width $$L$$. Although this isn't the most general case, it's good enough for the purposes of identifying where the proof breaks down.

So let's assume $$\left| \oint_{C} dz \ f(z) \right| = \epsilon.$$ for some positive $$\epsilon > 0$$. We want to derive a contradiction from this assumption.

Well, if the modulus of the integral of $$f$$ around $$C$$ is $$\epsilon$$, then, subdividing the big square into quarters to create four mini-squares of width $$L/2$$, we see that the modulus of the integral of $$f$$ on at least one of these four squares must be greater than or equal to $$\epsilon / 4$$. Continuing subdividing in this way, we see that, for any $$n \geq 0$$, there exists some square $$C_n$$ of width $$L/2^n$$ such that

$$\left| \oint_{C_n} dz \ f(z) \right| \geq \frac{\epsilon}{4^n}.$$

Does this lead to a contradiction? Not if we only assume that $$f$$ is bounded. With $$|f(z) | \leq M$$ (say), the ML inequality gives

$$\left| \oint_{C_n} dz \ f(z)\right| \leq 4 \times \frac{L}{2^n} \times M,$$ and there is no contradiction. Yes, the ML inequality tells us that $$\oint_{C_n} dz \ f(z) \to 0$$ as $$n \to \infty$$. But $$\oint_{C_n} dz \ f(z)$$ does not get small at a fast enough rate so as to contradict $$\left| \oint_{C_n} dz \ f(z) \right|$$ being greater than or equal to $$\epsilon / 4^n$$.

If we make the stronger assumption that $$f$$ is holomorphic, then we can get a tighter bound on the sizes of these integrals, one which does lead to a contradiction. Let $$a$$ be the (unique) point contained inside all of the squares $$C_n$$. Since $$f$$ is differentiable at $$a$$, we have

$$f(z) = f(a) + f'(a)(z - a) + v_a(z)(z - a),$$

where $$v_a(z)$$ is some function that tends to zero as $$z \to a$$. There must therefore exist some $$n \geq 0$$ such that $$|v_a(z)| < \frac \epsilon {8L^2}$$ on $$C_n$$, and hence $$|v_a (z) (z - a)| < \frac \epsilon {8L^2} \times \frac{2L}{2^n}$$ on $$C_n$$. Applying the ML inequality now, and remembering that the integral of the linear function $$f(a) + f'(a)(z - a)$$ around any closed contour is zero, we get the tighter bound,

$$\left| \oint dz \ f(z) \right| = \left| \oint dz \ v_a(z)(z - a)\right| < 4 \times \frac{L}{2^n} \times \frac \epsilon {8L^2} \times \frac{2L}{2^n} = \frac{\epsilon}{4^n}.$$

This bound, obtained by assuming holomorphicity, is much tighter than the bound we got from assuming boundedness alone, and it is tight enough to derive a contradiction.

• Excellent answer.....To the proposer: The counter-example given by Lord Shark the Unknown in a comment to the Q: Let $\gamma=\{z:|z|=1\}$ and $f(z)=\bar z.$ Then $\int_{\gamma}f(z)dz=$ $\int_{t=0}^{2\pi }e^{-it}d(e^{it})=$ $=\int_0^{2\pi}e^{-it}(ie^{it}dt)=$ $\int_0^{2\pi}idt=2\pi i.$ – DanielWainfleet Jan 1 '19 at 5:17

Kenny Wong has nicely explained why your proposed argument is wrong. The conclusion of the argument is also false, and in fact is false in the strongest possible way: the result of Cauchy's theorem is true only for analytic functions and no others. More precisely, the following theorem (known as Morera's theorem) is true:

Theorem: Suppose $$U$$ is an open subset of $$\mathbb{C}$$ and $$f:U\to\mathbb{C}$$ is a continuous function such that $$\int_\gamma f(z)\,dz = 0$$ for any rectifiable (or even just piecewise linear) closed curve $$\gamma$$ in $$U$$. Then $$f$$ is analytic.

To sketch the proof, if $$f$$ satisfies this condition then we can find an antiderivative of $$f$$ by integrating (just like we do with the fundamental theorem of calculus for functions on $$\mathbb{R}$$, except that we need to know the integral on a closed curve is always $$0$$ to be sure that this is well-defined). But the derivative of an analytic function is analytic, so this implies $$f$$ is analytic.