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Long versus Short Proofs

October 17, 2013 Leave a comment

Proofs are very similar to computer programs. And just like computer programs, there are often many different ways of writing a proof.  Sometimes the difference between proofs is great; one might be based on geometry while another on analysis. The purpose of this short note is to emphasise that different proofs are possible at the “simple” level too.  While this may appear mundane, it is actually an important part of learning mathematics: once you have proved a result, spend a few minutes trying to simplify the proof. It will make the next proof you do easier. The two key messages are as follows.

  • There is no substitute for experience when it comes to deriving proofs.
  • Practicing on simple examples to find the “best” proof is training for being able even just to find a proof of a harder result.

For lack of a better on-the-spot example, consider the following problem. Let f\colon\mathbb{C} \rightarrow \mathbb{C} be an analytic function having a double zero at the origin and whose first derivative has at most linear growth: | f'(x) | \leq \alpha |x| for some \alpha \in \mathbb{R}.  What is the most general form of f?

First Approach

A double zero at the origin means f(x) = x^2 h(x) for some analytic function h. Therefore f'(x) = 2xh(x) + x^2h'(x) and \frac{f'(x)}{x} = 2h(x) + xh'(x).  Since $latex 2h(x) + xh'(x)$ is both analytic and bounded, it must be constant, by Liouville’s theorem.  (Here, all analytic functions are entire because they are defined on the whole of the complex plane.) Solving the differential equation $latex 2h(x) + xh'(x) = c$ by substituting in the power series expansion h(x) = \sum a_i x^i shows that h(x) = \frac{c}2. [The general solution of $latex 2h(x) + xh'(x) = 0$ is h(x) = \beta x^{-2} where \beta is an arbitrary scalar. The only general solution that is analytic is h(x)=0.] The conclusion is that the most general form of f is f(x) = ax^2 for some complex scalar a.

Second Approach

The first approach is unattractive and unnecessarily complicated. Instead of starting with the double zero, start instead with the first derivative having a linear bound. (A standard generalisation of Liouville’s theorem is that an entire function with a linear bound must itself be linear. We will pretend here we do not know this fact.) If f(x) has a double zero at the origin then f'(x) has a zero at the origin, therefore f'(x) = x g(x) for some analytic g(x). The linear bound together with Liouville’s theorem means g(x) is constant, that is f'(x) = 2 a x for some scalar a. Therefore f(x) must equal ax^2 if it is to satisfy both f'(x) = 2 a x and f(0)=0.

What was the Difference?

The first approach expressed f as f(x) = x^2 h(x) while the second approach expressed f' as f'(x) = x g(x). Both approaches resulted in a differential equation, but the second approach resulted in the simpler differential equation f'(x) = 2ax. Underlying this example is that a “change of coordinates” can simplify a differential equation. Although both approaches could be made to work in this simple example, there are situations where some approaches are too difficult to follow through to completion. 

One could argue that because the linear bound constraint is “harder” than the double-zero constraint, one should start with the linear bound constraint and not the double-zero constraint, and therefore be led to the simpler differential equation. Yet the real messages are as stated at the start:

  • There is no substitute for experience when it comes to deriving proofs.
  • Practicing on simple examples to find the “best” proof is training for being able even just to find a proof of a harder result.
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