Liouville's theorem from complex analysis states that a holomorphic function $f(z)$ on the plane that is bounded in magnitude is constant. The usual proof uses the Cauchy integral formula. But this has always struck me as indirect and unilluminating. There is a proof via harmonic function theory, but this also seems to involve an unnecessarily large amount of prior buildup. So one might seek a more direct proof as below.

Assume that $f(z)$ is nonconstant. The fact that $f(z)$ is holomorphic at every point implies that at any given point, there is a direction such that moving in that direction makes $|f(z)|$ larger. But this doesn't prove that $|f(z)|$ is unbounded, because a priori its magnitude could behave like $5 - \frac{1}{|z|}$ or some such thing.

In the case of $f(z) = \frac{1}{P(z)}$ where $P(z)$ is a polynomial, one knows that $|f(z)|$ tends toward $0$ as $|z| \to \infty$ so that there's some closed disk such that if $|f(z)|$ is bounded, then it has a maximum in the interior of the disk, which contradicts the fact that one can always make $f(z)$ larger by moving in a suitable direction. But for general $f(z)$, one doesn't have this argument.

One can try to reason based on the power series expansion of a holomorphic function $f(z)$ that is not a polynomial. Because polynomials are unbounded as $|z| \to \infty$ and grow in magnitude in a way that's proportional to their degree, one might think that a power series, which can be regarded as an infinite degree polynomial, would also be unbounded as $|z| \to \infty$. This is of course false: take $f(z) = \sin(z)$, then as $|z| \to \infty$ along the real axis, $f(z)$ remains bounded. The point is that the dominant term in the partial sums of the power series varies with $|z|$, and that the relevant coefficients change, alternating in sign and tending toward zero rapidly, so that the gain in size corresponding to moving to the next power of $z$ is counterbalanced by the change in coefficient. But there's some direction that one can move in for which $f(z)$ is unbounded: in particular, for $f(z) = \sin(z)$, $f(z)$ is unbounded along the imaginary axis.

This suggests that we write $a_n = s_{n}e^{i \theta_n}$ for the coefficient of $z^n$ in the power series expansion of $f(z)$ and write $z = re^{i \theta}$ (where $s, r > 0$) so that

$$f(z) = \sum_{n = 0}^{\infty} {a_n}z^n = \sum_{n = 0}^{\infty} sr^n e^{n\theta + \theta_n} $$

and try to find a function $\theta = g(r)$ such that $f(z)$ is unbounded as $r \to \infty$ if one takes $\theta = g(r)$.

But I don't know what to do next. Any ideas? Any ideas for other strategies of proving Liouville's theorem that are more direct than the ones using Cauchy's theorem?

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    $\begingroup$ ams.org/mathscinet-getitem?mr=828328 $\endgroup$ – Ian Agol Dec 20 '12 at 19:57
  • $\begingroup$ Thanks Ian. I hadn't seen this before. My first take on the proof is that it's somewhat unnatural even if elementary, but I'll think about it some more. $\endgroup$ – Jonah Sinick Dec 20 '12 at 20:13
  • $\begingroup$ What you are getting at in the penultimate paragraph is essentially Cauchy's estimates. For instance, in Remmert's book there are two proofs of Cauchy's estimates that start similarly to your reasoning. $\endgroup$ – timur Dec 21 '12 at 8:12

I think the most illuminating proof of Liouville's theorem uses Riemann surfaces. Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be a bounded holomorphic function, and set $g(z) = f(1/z)$. Then $g : \mathbb{C} \setminus 0 \rightarrow \mathbb{C}$ is a bounded holomorphic function, so Riemann's removable singularities theorem says that $g$ can be extended over $0$. Translated into the language of Riemann surfaces, this says that $f$ extends to a holomorphic function $F : \mathbb{P}^1 \rightarrow \mathbb{C}$. Since $\mathbb{P}^1$ is compact, $F$ must have a global maximum. However, the maximum modulus principle says that a nonconstant holomorphic function cannot have a local maximum, so $F$ must be constant.

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    $\begingroup$ Can you prove Riemann's theorem "with your bare hands" or does it require Cauchy/Morera ? $\endgroup$ – Ralph Furmaniak Dec 20 '12 at 20:24
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    $\begingroup$ Suppose $f(z)$ has a removable singularity at $z=a$ (i.e. is analytic and bounded in a deleted neighbourhood of $z=a$). Then $(z-a)^2 f(z)$, defined to be $0$ at $z=a$, is analytic in this same deleted neighbourhood and also satisfies the Cauchy-Riemann equations at $z=a$. Now all you need is the fact that the derivative of a holomorphic function is holomorphic. $\endgroup$ – Robert Israel Dec 20 '12 at 23:21
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    $\begingroup$ @Yemon Choi : You need very little complex analysis to set up the basics of Riemann surfaces, and in my opinion many of the basic results in the subject are much clearer from this point of view. Another example is the open mapping theorem (and thus the maximum modulus principle) -- this follows easily from the fact that a holomorphic map between Riemann surfaces is (with respect to an appropriate choice of charts) locally equivalent to $z \mapsto z^n$ for some $n$, which can be easily deduced from the holomorphic version of the inverse function theorem. $\endgroup$ – Andy Putman Dec 21 '12 at 2:21
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    $\begingroup$ @Andy: Riemann surfaces are not really needed. Once you observe that $f(1/z)$ extends holomorphically around zero, it is easy to observe that $f$ achieves its supremum, either on $\mathbf C$, or ``at infinity'' (meaning that the extension of $f(1/z)$ achieves a local supremum at $0$). In either case, $f$ has to be constant. $\endgroup$ – ACL Dec 21 '12 at 8:18
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    $\begingroup$ @ACL : In the end, you're still using (openly or secretly) that $\mathbb{P}^1$ is compact. $\endgroup$ – Andy Putman Dec 21 '12 at 17:29

There is a truly elementary proof. Nothing but high school mathematics + the notion of limit is used.

First one proves Cauchy's inequality for polynomials: $$|f(0)|\leq M(r),$$ where $M(r)$ is the maximum of $|f|$ on the circle $|z|=r$. This is proved as follows: let $\epsilon_k=\exp(2\pi ik/n),$ where $n>$ the degree of the polynomial.



because $\sum_k^n(\epsilon_k)^j=0$ for all $j\in[1,n-1]$, by the Geometric Progression formula, so all other terms except the constant term, in the right hand side cancel. Taking absolute values, we obtain the above inequality.

Then by passing to a limit this inequality is true for all entire functions.

Applying this to $(f-f(0))/z$, we obtain $$|f'(0)|\leq (M(r)+|f(0)|)/r,$$ for all entire functions. If the function is bounded, we conclude that $f'(0)=0$. Applying this to $f(z-a)$ we obtain $f'(a)=0$ for all $a$, that is $f$ is constant.

  • $\begingroup$ This is a great proof! I am curious if this idea can be used somewhere else. $\endgroup$ – Dmitri Panov Dec 20 '12 at 21:04
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    $\begingroup$ Actually this is the same proof as the one proposed by Daniel Litt, on this page, only phrased in somewhat different terms. He shows a more general inequality $|a_n|\leq M(r)/r^n$, which is usually called Cauchy's inequality. I learned this trick, replacing Cauchy's integral from the book of Henrici, Applied and computational complex analysis. $\endgroup$ – Alexandre Eremenko Dec 20 '12 at 23:35
  • $\begingroup$ @Alexandre: No worries. I actually came up with this proof independently a few months ago when thinking about $p$-adic entire functions. $\endgroup$ – Daniel Litt Dec 21 '12 at 2:31
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    $\begingroup$ @Alexandre: Doesn't this proof require to know that an entire function on the plane is the sum of its Taylor series at $0$? — a fact which requires (unless I'm mistaken) the full apparatus of the Cauchy formula. $\endgroup$ – ACL Dec 21 '12 at 8:20
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    $\begingroup$ @ACL: From my point of view, this is the DEFINITION of an entire function. The other definition, which you probably have in mind, is much less elementary, and certainly not phrased in high-school terms. And in applications, entire functions usually arise as power series, not as solutions of CR PDE. See, for example, one of the main applications mathoverflow.net/questions/116749. $\endgroup$ – Alexandre Eremenko Dec 21 '12 at 15:02

I beg to differ with Andy and propose Nelson's proof as THE book proof that a bounded analytic function is constant. In fact Nelson's little gem is for harmonic functions, but the proof is incredibly beautiful and, of course, applies to analytic functions by considering their real and imaginary parts.

There are just two ingredients:

1. The mean value theorem: the value of a harmonic function in the plane (or n-space) at some point is the average of the function over any disc centered at that point. We can even take that as definition of a harmonic function if we have a family of "discs" and a measure with which to average.

2. A geometric property of metric discs on the plane: Given any two points $x$ and $y$, for sufficiently large radii $R$, the symmetric difference of the disc of radius $R$ centered at $x$ and the disc of the same radius centered at $y$ has negligible area compared with the area of the discs.

Now the proof goes as follows: take any two points $x$ and $y$ on the plane and choose discs of a very large radius $R$ centered at each of these points. If the harmonic function is bounded, its average over the two discs, that by (2) basically coincide, has to be almost the same. Let $R$ go to infinity and you're done.

Remark/Question Clearly the proof makes sense for a class of metric measure spaces. Has this sort of spaces (symmetric difference between large metric balls having small relative measure) been studied on its own? Normed spaces are in this category.

  • $\begingroup$ De gustibus non est disputandum :). $\endgroup$ – Andy Putman Dec 20 '12 at 20:55
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    $\begingroup$ I think what you're looking for is some form of amenability for metric spaces: en.wikipedia.org/wiki/Amenable_group $\endgroup$ – Ian Agol Dec 20 '12 at 21:41
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    $\begingroup$ I agree here. Liouville's Theorem is not really a result about analytic functions, but a result about harmonic functions. The evidence is that it continues to hold for harmonic functions in dimension other than 2. $\endgroup$ – Gerald Edgar Dec 21 '12 at 15:41

Here's a proof via character theory; in some ways it's a recasting of the Cauchy proof, but it has the advantage of being purely algebraic. In particular, this proof will work over any algebraically closed complete valued field, for example (some mild care must be taken over a field of positive characteristic when one takes the limit $k\to\infty$).

Let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a bounded entire function; we wish to show that $a_n=0$ for $n>0$. We use the fact that $$\sum_{\zeta \text{ a $k$-th rooth of unity}} \zeta^n$$ equals $k$ if $k$ divides $n$ and equals $0$ otherwise. So in particular setting $g_n(z)=f(z)/z^n$ we have for $k>n$ that

$$\frac{1}{k}\sum_{\zeta \text{ a $k$-th rooth of unity}} g_n(r\zeta)=a_n+\sum_{m=1}^\infty a_{km+n}r^{km}.$$ The sum on the right tends to zero with $k$, as the $a_{km+n}$ decrease very rapidly by the entireness of $f$. So we have $$a_n=\lim_{k\to\infty}\frac{1}{k}\sum_{\zeta \text{ a $k$-th rooth of unity}} g_n(r\zeta).$$ But taking the limit as $|r|\to \infty$ and using the boundedness of $f$, this is zero for $n>0$.

I say that this is a recasting of the Cauchy integral proof because the character sum we use is essentially a Riemann sum for the integral of $g_n$ around a circular contour. Indeed, one may prove the Cauchy integral formula for circular contours this way.

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    $\begingroup$ The formula for the sum of roots of unity is incorrect: it's $k$ if $k|n$ and 0 if $k$ does not divide $n$. The displayed formula below that is valid only for $k > n$ (you need $n$ to be minimal positive in its congruence class mod $k$). If you want the proof to work over any alg. closed complete valued field then you pay due diligence to characteristic $p$ when indicating which $k$ are used. Mention something about the role of the radius of convergence formula (valid in any complete valued field) in explaining why the coefficients decrease suff. rapidly for an entire function. $\endgroup$ – KConrad Dec 21 '12 at 23:11
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    $\begingroup$ Over the p-adics, I think this "integral" using roots of unity is essentially the Schnirelmann integral. Look at the end of Koblitz's "p-Adic analysis: a short course on recent work". $\endgroup$ – KConrad Dec 21 '12 at 23:14
  • $\begingroup$ Oops you're right--fixed. $\endgroup$ – Daniel Litt Dec 22 '12 at 0:06
  • $\begingroup$ You call the proof "purely algebraic", but it is not since you're taking limits and needing to make estimates to see certain numbers are 0. $\endgroup$ – KConrad Dec 22 '12 at 0:15
  • $\begingroup$ Well, it's pretty algebraic :). What I mean is it doesn't really use any facts from analysis over $\mathbb{C}$, since it works over e.g. $mathbb{C}_p$. $\endgroup$ – Daniel Litt Dec 22 '12 at 0:45

If your definition of holomorphic is that there is an absolutely convergent power series $\sum_{n=0}^{\infty} a_n z^n$ then Parseval's theorem implies that the mean square of the function on the circle of radius $R$ is $\sum_{n=0}^{\infty} |a_n^2| R^{2n}$ and it follows that $a_n=0$ for all $n>0$.

Edit (new approach)

Without loss of generality let's assume that $f(0)=0$ and $f'(0)=0$. Consider the function $f(x)/x$, continuous on all of $\mathbb{C}$. It is analytic on the punctured plane and converges to 0 on its boundary, hence by the maximum principle must be 0.

You can drop the $f'(0)=0$ condition above if you assume instead that $f$ is twice differentiable at 0, so then $f(x)/x$ is holomorphic on all of $\mathbb{C}$ and decreases to 0, so must be identically 0.

This does presuppose the maximum principle.

  • $\begingroup$ This is new to me / interesting, but I'm not sure if it's any more direct than the proof by Cauchy's theorem :) $\endgroup$ – Jonah Sinick Dec 20 '12 at 20:10
  • $\begingroup$ Added a "first principles" proof :-) $\endgroup$ – Ralph Furmaniak Dec 20 '12 at 20:25
  • $\begingroup$ @ Ralph - Thanks for the new proof! $\endgroup$ – Jonah Sinick Dec 20 '12 at 20:41
  • $\begingroup$ I like the second approach, but you have to but in bold that you are requiring only that : $f(z)$ is holomorphic on $\mathbb{C}$ and $f''(0)$ exists, so that $\frac{f(z)-f(0)}{z}$ is holomorphic on $\mathbb{C}$ $\endgroup$ – reuns Oct 28 '16 at 13:57

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