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This post, currently closed, asks for the intuition behind Liouville's theorem, which states that every bounded entire function is constant; I find the answers unsatisfactory and hand-wavy. I am looking for a more focused, precise description of what is going on. Is there a geometrical understanding of what is going on? I try to visualize a non-constant analytic function, to understand why unboundedness is necessary along some direction, but it isn't working for me.

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    $\begingroup$ Thanks for clarifying what theorem. It is unclear what is unclear to you. Intuition is usually hand-wavy and a rigorous proof is often unintuitive. What exactly is the problem? $\endgroup$
    – Kurt G.
    Commented Jul 7, 2023 at 6:21
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    $\begingroup$ @KurtG. Real functions that are differentiable everywhere don't have this property; why does the complex extension require unboundedness? Clearly, it is encoded in the definition of an entire function. I am trying to pinpoint why the definition of an entire function necessitates this property. Thank you for you advice. $\endgroup$ Commented Jul 7, 2023 at 6:25
  • $\begingroup$ Complex differentiation is only formally equal to real differentiation. Our professor said "let's do $$\lim_{h\to z}\frac{f(z+h)-f(z)}{h}$$ like you learned in calculus but we start from scratch with $z,h\in\mathbb C\,.$" This did not sound very exciting but he was just kidding. Complex differentiation turns out to be very different from real differentiation (Cauchy-Rieman-equations for example) which makes complex analysis one of the most amazing mathematical fields. $\endgroup$
    – Kurt G.
    Commented Jul 7, 2023 at 6:29
  • $\begingroup$ If you want to understand it better, first understand that $\mathbb{C}$-linear mappings (derivatives are those) have a very special structure. This causes holomorphic equations to be harmonic. Then try to study harmonic functions $u:\mathbb{R}^2 \rightarrow \mathbb{R}$, i.e. those that satisfy $\Delta u = 0$. Read for example the chapter 2.2 about Laplace's equation in Evans' PDE book. This might give you a bridge to the real numbers that you need for a better understanding. $\endgroup$ Commented Jul 7, 2023 at 6:34
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    $\begingroup$ Yes. Harmonic and complex differentiable functions satisfy the maximum/minimum (modulus) principle and have many more properties that one does not expect looking only at their terse definitions. $\endgroup$
    – Kurt G.
    Commented Jul 7, 2023 at 8:16

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You will never be able to “truly” visualize what a holomorphic function “looks like” (you can of course visualize various aspects of it, but never in entirety since the graph is embedded in $\Bbb{C}^2$). So, that already halts the first attempt at a geometric intuition. Anyway, the first answer in the linked post is good, and is what I’ll say, hopefully you find it more convincing.

All the nice features of holomorphic functions come from Cauchy’s integral formula. Here’s a broad slogan to keep in mind when doing analysis (this comes from all our experience with analysis): “derivatives bad $\ddot{\frown}$, integrals good $\ddot{\smile}$”. In general, you’d expect that after taking a derivative, your function becomes less well-behaved (that’s certainly the intuition from real analysis (and it is good intuition; don’t throw it out just because you’re in complex analysis)), whereas integrating makes things better, or the very least doesn’t make things worse: for example, if $f:\Bbb{R}\to\Bbb{R}$ is continuous, then $F:\Bbb{R}\to\Bbb{R}$, $F(x)=\int_0^xf(t)\,dt$ is $C^1$. Likewise, if you have some parameter-dependent integral, $F(x)=\int_a^bf(x,t)\,dt$, then things like continuity/smoothness of $f$ imply that of $F$. Even better are integral operations like convolutions, which are used to “smooth out” functions.

With this slogan in mind, let’s go back to holomorphic functions. Cauchy gives us his wonderful integral formula: \begin{align} f(z)&=\frac{1}{2\pi i}\int_{|\zeta-z|=r}\frac{f(\zeta)}{\zeta-z}\,d\zeta. \end{align} The wonderful thing about this formula is that it expresses $f$ in terms of an integral involving $f$. Furthermore, it’s not just any arbitrary integral, it is a kind of convolution, which as I mentioned before has the effect of smoothing things out. Btw, it is precisely this feature of holomorphic functions which gives us the wonderful regularity theorems (once differentiable on an open set implies infinitely differentiable on that set, and in fact locally admits a power series expansion). Ok, you can actually view this more generally as a corollary of elliptic regularity, but I digress.

The second key feature of Cauchy’s integral formula is the particular choice of “kernel”, i.e the function we integrate $f(\zeta)$ against. This is $\frac{1}{\zeta-z}$. This is such an important thing that it’s often called “the Cauchy kernel”. Anyway, the point is that the function $\frac{1}{\zeta}$ captures some geometric information, namely by integrating it you essentially get the circumference of the unit circle: $\int_{|\zeta|=1}\frac{1}{\zeta}\,d\zeta=2\pi i$; that’s why one has the extra $\frac{1}{2\pi i}$ in Cauchy’s formula.

Now, we understand (or at the very least, appreciate in hindsight) from our slogan that holomorphic functions are very nice and their integrals are very closely related to circles. The precise corollary of this fact is the mean-value property: the value $f(z)$ of a holomorphic function is equal to its average value over any circle centered at $z$. Also, by linearity of integrals, it follows that both the real and imaginary parts of a holomorphic function have the mean-value property. Now, we can certainly visualize the real and imaginary parts separately (as their graphs can be embedded in $\Bbb{R}^3$).

The mean-value property implies that your function cannot grow crazily in all directions. Said another way, the manner in which the function changes values from point to point must be consistent with keeping the mean-value property. Of course, this is still not a completely obvious thing to visualize, but we certainly have examples, e.g. $u(x,y)=ax+by+c$.

So, now more generally speaking, let’s consider a continuous function $f:\Bbb{R}^n\to\Bbb{R}$ which has the mean-value property (in fact if you know about vector-valued integration, then you can replace the target space $\Bbb{R}$ with any real Banach space $V$). I shall now make the strong assumption that $f$ has a finite limit at infinity, meaning there is an $L$ such that for every $\epsilon>0$, there is an $R>0$ such that for all $x\in\Bbb{R}^n$ with $\|x\|>R$, we have $|f(x)-L|<\epsilon.$ Note that this implies $f$ is bounded. The converse is true in this case, but it is an a-posteriori observation after having proved Liouville’s theorem, but since I’m trying to motivate Liouville’s theorem, I hope you don’t mind me making the slightly stronger a-priori assumption.

Now, one has that for any point $p$ and any $r>0$, $f(p)$ is equal to the average value of $f$ over the sphere of radius $r$ centered at $p$: \begin{align} f(p)&=\frac{1}{\sigma(S_r(p))}\int_{S_r(p)}f(x)\,d\sigma(x), \end{align} where $\sigma(S_r(p))$ is the surface area of the sphere of radius $r$ having center $p$. Now, notice what happens when $r$ is very very large. The integrand is approximately equal to $L$, so \begin{align} f(p)&=\frac{1}{\sigma(S_r(p))}\int_{S_r(p)}f(x)\,d\sigma(x)\approx\frac{1}{\sigma(S_r(p))}\int_{S_r(p)}L\,d\sigma=\frac{1}{\sigma(S_r(p))}L\cdot \sigma(S_r(p))=L. \end{align} So, $f(p)$ is approximately equal to $L$, with the approximation getting better as $r\to\infty$. So, we can conclude $f(p)=L$. Since $p$ was arbitrary, it follows $f$ is constantly equal to $L$. Therefore, we’ve shown (modulo me writing out the $\epsilon$-$\delta$ proof) that for a function that has a limit at infinity, if it has the mean-value property, then it is constant.

The more general theorem which I didn’t prove (but can be proved essentially along the same lines, by a more careful estimate of the integrals) is that every bounded continuous function on $\Bbb{R}^n$ which has the mean-value property is constant. This is a purely geometric statement. Therefore, for an entire function $f$, both the real and imaginary parts satisfy the mean-value property, so if we assume boundedness in addition, then the real and imaginary parts are constant, and hence $f$ itself is constant. Now, you may wonder where we used the fact that the domain of $f$ is all of $\Bbb{R}^n$, and that’s a good question. The devil is in the details; we used it to make all our approximations valid.

So, really, you see that once again, underlying all of this is the integral formula which gives us so many nice things, like regularity (due to the convolution), geometric satisfaction (with circles), and also it allows us to invoke many of the powerful theorems/inequalities for integrals (the main one being Holder’s inequality… it may seem like we didn’t use it above, but we’re unconsciously using the very obvious $L^1$-$L^{\infty}$ version of it). And at the end of the day, it reaffirms our slogan that integrals are good (and highly flexible:).

While we’re talking about the flexibility of integrals, I should mention that although the above proof outline of Liouville’s theorem is more or less geometric in nature, you shouldn’t really think of it as a geometric statement (atleast I don’t). Rather I view it as a rigidity/uniqueness theorem “if you constrain your function to have such and such properties, then boom it is automatically equal to ___”. And many of these rigidity properties come up because of the explicit integral representation (with the relatively simple kernel $\frac{1}{\zeta}$), or more abstractly because the function $f$ satisfies a certain PDE, namely the Cauchy-Riemann equations, which happen to be elliptic (and there’s a vast theory of elliptic PDEs). For example, you have generalized versions of Liouville, such as if an entire function grows at-most polynomially (and even this can be refined slightly), then the entire function is itself a polynomial of at most such and such degree. In other words, growth estimates on $f$ imply certain things about certain coefficients of the Laurent/Taylor expansion, so that gives you more information about your $f$, and if you’re clever you can use this new-found information back in the integral and perhaps you can get more information etc. This just illustrates the flexibility of integrals; see Generalized Liouville Theorem for several complex variables for more specific illustrations of this principle.

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  • $\begingroup$ RE "you will never visualise" - for this particular question, couldn't you draw an analogy with harmonic functions $\mathbb{R}^2 \to \mathbb{R}$, which are not so hard to visualise? $\endgroup$ Commented Jul 7, 2023 at 14:23
  • $\begingroup$ @preferred_anon my 6th paragraph, last line ”Now, we can certainly visualize the real and imaginary parts separately (as their graphs can be embedded in $\Bbb{R}^3$).” Anyway, this is still ‘partial information’ in the sense that we can’t see everything “at the same time”. $\endgroup$
    – peek-a-boo
    Commented Jul 7, 2023 at 14:50
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Compact set version of Liuolle's theorem is Maximum Modulus principle.

Maximum modulus principle is a consequence of Cauchy integral formula. See: Theorem 3.6 https://www.maths.tcd.ie/~richardt/414/414-ch3.pdf

$$f(a) = \frac{1}{2 \pi i} \int_{\partial B_r(a)} \frac{f(z)}{z-a} dz$$

We can choose $r$ however small to imply maxmimum modulus principle.

For $r$ small enough Cauchy integral formula can proved because $f(z)$ is analytic in a neighbourhood of $a$ and in turn is a consequence of an obvious formula:

$$1 = \int_{\partial B_r(a)} \frac{1}{z-a} dz$$.

So i would say Maximum modulus principle is a consequence of $f(z)$ being analytic in a neighbourhood of a point $a$.

Intuitively, Cauchy integral formula is true because you can split the contour into a small contour around $a$ with $r \rightarrow 0$ by Cauchy theorem. Since we have singularity at $z = a$ for $\frac{f(z)}{z-a}$ by appealing to intuition of theorem in real analysis like Stokes theorem, integrating in contour in equivalent to 2D - integration of curl in the region inside the contour. Now since there is a singularity the function $\nabla \times \frac{1}{z-a}$ (assuming $f(z)$ is almost constant as $r \rightarrow 0$) acts like a ($\delta(z-a)$) delta function scaled by $f(a)$ and hence while integrating in the inner region this delta function integrates to $1$ there by giving $f(a)$

So you can actually use or derive Cauchy integral formula with any function $g(z)$ whose curl will act like a delta function intuitively i.e., $\nabla \times g(z-a) \rightarrow \delta(z-a)$ as $z \rightarrow a$. Try to make this precise (the difficulty in making this precise is the difference in the contour line integral definition and line integral definition. I think real part of contour integral gives line integral with curve in contour integral replaced by its conjugate) and try to generalize Cauchy integral formula :-) by choosing functions whose curl in the limit will act like delta functions. But by residue theorem any function $g(z)$ such that $\lim_{z \rightarrow a} (z-a) g(z) = 1$ should work in cauchy integral formula i.e.,

$$f(a) = \int_{\partial B_r(a)} f(z) g(z-a) dz$$ where $$\lim_{z \rightarrow a} (z-a) g(z-a) = 1$$

I vaguely feel $1/(z-a)$ is related to a function called Green's function which is a solution of differential equation involving delta function. May be thats the connection.

I hope this helps. Cheers !

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