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Our goal in what follows will be that of proving that any holomorphic function is analytic. This is something quite subtle, which cannot be proved with bare hands, and requires lots of preliminaries. Getting to these preliminaries now, our claim is that a lot of useful knowledge, in order to deal with the holomorphic functions, can be gained by further studying the analytic functions, and even the usual polynomials [math]P\in\mathbb C[X][/math].

So, let us further study the polynomials [math]P\in\mathbb C[X][/math], and other analytic functions. We already know from chapter 5 that in the polynomial case, [math]P\in\mathbb C[X][/math], some interesting things happen, because any such polynomial has a root, and even [math]\deg(P)[/math] roots, after a recurrence. Keeping looking at polynomials, with the same methods, we are led to:

Theorem

Any polynomial [math]P\in\mathbb C[X][/math] satisfies the maximum principle, in the sense that given a domain [math]D[/math], with boundary [math]\gamma[/math], we have:

[[math]] \exists x\in\gamma\quad,\quad |P(x)|=\max_{y\in D}|P(y)| [[/math]]

That is, the maximum of [math]|P|[/math] over a domain is attained on its boundary.

Show Proof

In order to prove this, we can split [math]D[/math] into connected components, and then assume that [math]D[/math] is connected. Moreover, we can assume that [math]D[/math] has no holes, and so is homeomorphic to a disk, and even assume that [math]D[/math] itself is a disk. But with this assumption made, the result follows from by contradiction, by using the same arguments as in the proof of the existence of a root, from chapter 5. To be more precise, assume [math]\deg P\geq 1[/math], and that the maximum of [math]|P|[/math] is attained at the center of a disk [math]D=D(z,r)[/math]:

[[math]] |P(z)|=\max_{x\in D}|P(x)| [[/math]]

We can write then [math]P(z+t)\simeq P(z)+ct^k[/math] with [math]c\neq0[/math], for [math]t[/math] small, and by suitably choosing the argument of [math]t[/math] on the unit circle we conclude, exactly as in chapter 5, that the function [math]|P|[/math] cannot have a local maximum at [math]z[/math], as stated.

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A good explanation for the fact that the maximum principle holds for polynomials [math]P\in\mathbb C[X][/math] could be that the values of such a polynomial inside a disk can be recovered from its values on the boundary. And fortunately, this is indeed the case, and we have:

Theorem

Given a polynomial [math]P\in\mathbb C[X][/math], and a disk [math]D[/math], with boundary [math]\gamma[/math], we have the following formulae, with the integrations being the normalized, mass [math]1[/math] ones:

• [math]P[/math] satisfies the plain mean value formula [math]P(x)=\int_DP(y)dy[/math].
• [math]P[/math] satisfies the boundary mean value formula [math]P(x)=\int_\gamma P(y)dy[/math].

Show Proof

As a first observation, the two mean value formulae in the statement are equivalent, by restricting the attention to disks [math]D[/math], having as boundaries circles [math]\gamma[/math], and using annuli and polar coordinates for the proof of the equivalence. As for the formulae themselves, these can be checked by direct computation for a disk [math]D[/math], with the formulation in (2) being the most convenient. Indeed, for a monomial [math]P(x)=x^n[/math] we have:

[[math]] \begin{eqnarray*} \int_\gamma y^ndy &=&\frac{1}{2\pi}\int_0^{2\pi}(x+re^{it})^ndt\\ &=&\frac{1}{2\pi}\int_0^{2\pi}\sum_{k=0}^n\binom{n}{k}x^k(re^{it})^{n-k}dt\\ &=&\sum_{k=0}^n\binom{n}{k}x^kr^{n-k}\frac{1}{2\pi}\int_0^{2\pi}e^{i(n-k)t}dt\\ &=&\sum_{k=0}^n\binom{n}{k}x^kr^{n-k}\delta_{kn}\\ &=&x^n \end{eqnarray*} [[/math]]

Here we have used the following key identity, valid for any exponent [math]m\in\mathbb Z[/math]:

[[math]] \begin{eqnarray*} \frac{1}{2\pi}\int_0^{2\pi}e^{imt}dt &=&\frac{1}{2\pi}\int_0^{2\pi}\cos(mt)+i\sin(mt)dt\\ &=&\delta_{m0}+i\cdot 0\\ &=&\delta_{m0} \end{eqnarray*} [[/math]]

Thus, we have the result for monomials, and the general case follows by linearity.

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All the above is very nice, but we can in fact do even better, with a more powerful integration formula. Let us start with some preliminaries. We first have:

Proposition

We can integrate functions [math]f[/math] over curves [math]\gamma[/math] by setting

[[math]] \int_\gamma f(x)dx=\int_a^bf(\gamma(t))\gamma'(t)dt [[/math]]

with this quantity being independent on the parametrization [math]\gamma:[a,b]\to\mathbb C[/math].

Show Proof

We must prove that the quantity in the statement is independent on the parametrization. In other words, we must prove that if we pick two different parametrizations [math]\gamma,\eta:[a,b]\to\mathbb C[/math] of our curve, then we have the following formula:

[[math]] \int_a^bf(\gamma(t))\gamma'(t)dt=\int_a^bf(\eta(t))\eta'(t)dt [[/math]]

But for this purpose, let us write [math]\gamma=\eta\phi[/math], with [math]\phi:[a,b]\to[a,b][/math] being a certain function, that we can assume to be bijective, via an elementary cut-and-paste argument. By using the chain rule for derivatives, and the change of variable formula, we have:

[[math]] \begin{eqnarray*} \int_a^bf(\gamma(t))\gamma'(t)dt &=&\int_a^bf(\eta\phi(t))(\eta\phi)'(t)dt\\ &=&\int_a^bf(\eta\phi(t))\eta'(\phi(t))\phi'(t)dt\\ &=&\int_a^bf(\eta(t))\eta'(t)dt \end{eqnarray*} [[/math]]

Thus, we are led to the conclusions in the statement.

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The main properties of the above integration method are as follows:

Proposition

We have the following formula, for a union of paths:

[[math]] \int_{\gamma\cup\eta}f(x)dx=\int_\gamma f(x)dx+\int_\eta f(x)dx [[/math]]

Also, when reversing the path, the integral changes its sign.

Show Proof

Here the first assertion is clear from definitions, and the second assertion comes from the change of variable formula, by using Proposition 6.24.

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Now by getting back to polynomials, we have the following result:

Theorem

Any polynomial [math]P\in\mathbb C[X][/math] satisfies the Cauchy formula

[[math]] P(x)=\frac{1}{2\pi i}\int_\gamma\frac{P(y)}{y-x}\,dy [[/math]]

with the integration over [math]\gamma[/math] being constructed as above.

Show Proof

This follows by using abstract arguments and computations similar to those in the proof of Theorem 6.23. Indeed, by linearity we can assume [math]P(x)=x^n[/math]. Also, by using a cut-and-paste argument, we can assume that we are on a circle:

[[math]] \gamma:[0,2\pi]\to\mathbb C\quad,\quad \gamma(t)=x+re^{it} [[/math]]

By using now the computation from the proof of Theorem 6.23, we obtain:

[[math]] \begin{eqnarray*} \int_\gamma\frac{y^n}{y-x}\,dy &=&\int_0^{2\pi}\frac{(x+re^{it})^n}{re^{it}}\,rie^{it}dt\\ &=&i\int_0^{2\pi}(x+re^{it})^ndt\\ &=&i\cdot 2\pi x^n \end{eqnarray*} [[/math]]

Thus, we are led to the formula in the statement.

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All this is quite interesting, and obviously, we are now into some serious mathematics. Importantly, Theorem 6.22, Theorem 6.23 and Theorem 6.26 provide us with a path for proving the converse of Theorem 6.21. Indeed, if we manage to prove the Cauchy formula for any holomorphic function [math]f:X\to\mathbb C[/math], then it will follow that our function is in fact analytic, and so infinitely differentiable. So, let us start with the following result:

Theorem

The Cauchy formula, namely

[[math]] f(x)=\frac{1}{2\pi i}\int_\gamma\frac{f(y)}{y-x}\,dy [[/math]]

holds for any holomorphic function [math]f:X\to\mathbb C[/math].

Show Proof

This is something standard, which can be proved as follows:

(1) Our first claim is that given [math]f\in H(X)[/math], with [math]f'\in C(X)[/math], the integral of [math]f'[/math] vanishes on any path. Indeed, by using the change of variable formula, we have:

[[math]] \begin{eqnarray*} \int_\gamma f'(x)dx &=&\int_a^bf'(\gamma(t))\gamma'(t)dt\\ &=&f(\gamma(b))-f(\gamma(a))\\ &=&0 \end{eqnarray*} [[/math]]

(2) Our second claim is that given [math]f\in H(X)[/math] and a triangle [math]\Delta\subset X[/math], we have:

[[math]] \int_\Delta f(x)dx=0 [[/math]]

Indeed, let us call [math]\Delta=ABC[/math] our triangle. Now consider the midpoints [math]A',B',C'[/math] of the edges [math]BC,CA,AB[/math], and then consider the following smaller triangles:

[[math]] \Delta_1=AC'B'\quad,\quad\Delta_2=BA'C'\quad,\quad\Delta_3=CB'A'\quad,\quad\Delta_4=A'B'C' [[/math]]

These smaller triangles partition then [math]\Delta[/math], and due to our above conventions for the vertex ordering, which produce cancellations when integrating over them, we have:

[[math]] \int_\Delta f(x)dx=\sum_{i=1}^4\int_{\Delta_i}f(x)dx [[/math]]

Thus we can pick, among the triangles [math]\Delta_i[/math], a triangle [math]\Delta^{(1)}[/math] such that:

[[math]] \left|\int_\Delta f(x)dx\right|\leq 4\left|\int_{\Delta^{(1)}}f(x)dx\right| [[/math]]

In fact, by repeating the procedure, we obtain triangles [math]\Delta^{(n)}[/math] such that:

[[math]] \left|\int_\Delta f(x)dx\right|\leq 4^n\left|\int_{\Delta^{(n)}}f(x)dx\right| [[/math]]

(3) Now let [math]z[/math] be the limiting point of these triangles [math]\Delta^{(n)}[/math], and fix [math]\varepsilon \gt 0[/math]. By using the fact that the functions [math]1,x[/math] integrate over paths up to 0, coming from (1), we obtain the following estimate, with [math]n\in\mathbb N[/math] being big enough, and [math]L[/math] being the perimeter of [math]\Delta[/math]:

[[math]] \begin{eqnarray*} \left|\int_{\Delta^{(n)}}f(x)dx\right| &=&\left|\int_{\Delta^{(n)}}f(x)-f(z)-f'(z)(x-z)dx\right|\\ &\leq&\int_{\Delta^{(n)}}\left|f(x)-f(z)-f'(z)(x-z)\right|dx\\ &\leq&\int_{\Delta^{(n)}}\varepsilon|x-z|dx\\ &\leq&\varepsilon(2^{-n}L)^2 \end{eqnarray*} [[/math]]

Now by combining this with the estimate in (2), this proves our claim.

(4) The rest is quite routine. First, we can pass from triangles to boundaries of convex sets, in a straightforward way, with the same conclusion as in (2), namely:

[[math]] \int_\gamma f(x)dx=0 [[/math]]

Getting back to what we want to prove, namely the Cauchy formula for an arbitrary holomorphic function [math]f\in H(X)[/math], let [math]x\in X[/math], and consider the following function:

[[math]] g(y)=\begin{cases} \frac{f(y)-f(x)}{y-x}&(y\neq x)\\ f'(x)&(y=x) \end{cases} [[/math]]

Now assuming that [math]\gamma[/math] encloses a convex set, we can apply what we found, namely vanishing of the integral, to this function [math]g[/math], and we obtain the Cauchy formula for [math]f[/math].

(5) Finally, the extension to general curves is standard, and standard as well is the discussion of what exactly happens at [math]x[/math], in the above proof. See Rudin .

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As a main application of the Cauchy formula, we have:

Theorem

The following conditions are equivalent, for a function [math]f:X\to\mathbb C[/math]:

• [math]f[/math] is holomorphic.
• [math]f[/math] is infinitely differentiable.
• [math]f[/math] is analytic.
• The Cauchy formula holds for [math]f[/math].

Show Proof

This is routine from what we have, the idea being as follows:

[math](1)\implies(4)[/math] is non-trivial, but we know this from Theorem 6.27.

[math](4)\implies(3)[/math] is something trivial, because we can expand the series in the Cauchy formula, and we conclude that our function is indeed analytic.

[math](3)\implies(2)\implies(1)[/math] are both elementary, known from Theorem 6.21.

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As another application of the Cauchy formula, we have:

Theorem

Any holomorphic function [math]f:X\to\mathbb C[/math] satisfies the maximum principle, in the sense that given a domain [math]D[/math], with boundary [math]\gamma[/math], we have:

[[math]] \exists x\in\gamma\quad,\quad |f(x)|=\max_{y\in D}|f(y)| [[/math]]

That is, the maximum of [math]|f|[/math] over a domain is attained on its boundary.

Show Proof

This follows indeed from the Cauchy formula. Observe that the converse is not true, for instance because functions like [math]\bar{x}[/math] satisfy too the maximum principle. We will be back to this later, when talking about harmonic functions.

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As before with polynomials, a good explanation for the fact that the maximum principle holds could be that the values of our function inside a disk can be recovered from its values on the boundary. And fortunately, this is indeed the case, and we have:

Theorem

Given an holomorphic function [math]f:X\to\mathbb C[/math], and a disk [math]D[/math], with boundary [math]\gamma[/math], the following happen:

• [math]f[/math] satisfies the plain mean value formula [math]f(x)=\int_Df(y)dy[/math].
• [math]f[/math] satisfies the boundary mean value formula [math]f(x)=\int_\gamma f(y)dy[/math].

Show Proof

As usual, this follows from the Cauchy formula, with of course some care in passing from integrals constructed as in Proposition 6.24 to integrals viewed as averages, which are those that we refer to, in the present statement.

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Finally, as yet another application of the Cauchy formula, which is something nice-looking and conceptual, we have the following statement, called Liouville theorem:

Theorem

An entire, bounded holomorphic function

[[math]] f:\mathbb C\to\mathbb C\quad,\quad |f|\leq M [[/math]]

must be constant. In particular, knowing [math]f\to0[/math] with [math]z\to\infty[/math] gives [math]f=0[/math].

Show Proof

This follows as usual from the Cauchy formula, namely:

[[math]] f(x)=\frac{1}{2\pi i}\int_\gamma\frac{f(y)}{y-x}\,dy [[/math]]

Alternatively, we can view this as a consequence of Theorem 6.30, because given two points [math]x\neq y[/math], we can view the values of [math]f[/math] at these points as averages over big disks centered at these points, say [math]D=D_x(R)[/math] and [math]E=D_y(R)[/math], with [math]R \gt \gt 0[/math]:

[[math]] f(x)=\int_Df(z)dz\quad,\quad f(y)=\int_Ef(z)dz [[/math]]

Indeed, the point is that when the radius goes to [math]\infty[/math], these averages tend to be equal, and so we have [math]f(x)\simeq f(y)[/math], which gives [math]f(x)=f(y)[/math] in the limit.

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General references

Banica, Teo (2024). "Calculus and applications". arXiv:2401.00911 [math.CO].