8

Imagine that you are writing a book on the foundations of analysis.

You have already proved that for each $a > 1$ there is a unique function $f_a(x) = a^x$ satisfying the following:

  1. $f_a$ is an isomorphism of ordered groups between $(\mathbb{R},+)$ and $(\mathbb{R}_{+},\cdot)$;
  2. $f_a(1) = a$.

It follows from the monotonicity and bijectivity of $f_a$ that it is continuous.

Now you would like to prove that $f_a$ is differentiable. At this point, you don't know anything about integration, differential equations or power series.

What is the simplest or most elegant way of doing this?

user191371
  • 146
  • Couldn't you use the order topology and say that the isomorphism preserves differentiability of functions and construct a clearly differentiable function in $(\Bbb R , +)$ – Matthew Levy Nov 10 '14 at 15:51
  • @MatthewLevy I may be missing something, but that seems like a circular argument. For the homeomorphism to preserve differentiability, you already need to know that the homeomorphism is a diffeomorphism, which we don't know yet here. Even for $x \mapsto x^3$, this fails. – user191371 Nov 10 '14 at 15:55
  • I'm not sure exactly what you're allowed to know, but you can consider this a special case of the fact that continuous homomorphisms between Lie groups are automatically smooth. http://books.google.com/books?id=gRTDO-wVhj0C&pg=PA49#v=onepage&q&f=false – Tom Nov 10 '14 at 15:58
  • 1
    @Tom I believe that somewhere in the course of that proof, the existence-uniqueness theorem for differential equations must make an appearance. This is at a stage before that. – user191371 Nov 10 '14 at 16:01

1 Answers1

9

Since $f_a$ is a homomorphism, you only need to show differentiability at $0$, for

$$\frac{f_a(x+h) - f_a(x)}{h} = f_a(x)\frac{f_a(h)-1}{h}.$$

Since $f_a$ is convex [you need to show that, of course], you know that

$$\frac{f_a(h) - 1}{h}$$

is monotonically increasing in $h\in \mathbb{R}\setminus \{0\}$, hence the one-sided derivatives

$$D^+f_a(0) = \lim_{h \searrow 0}\frac{f(h)-1}{h},\quad D^-f_a(0) = \lim_{h\nearrow 0} \frac{f_a(h)-1}{h}$$

exist both. So it remains to see that they are equal. But since $f_a$ is a homomorphism we have

\begin{align} D^-f_a(0) &= \lim_{h\searrow 0}\frac{f_a(-h)-1}{-h} = \lim_{h\searrow 0} \frac{\frac{1}{f_a(h)}-1}{-h}\\ &= \lim_{h\searrow 0}\frac{1}{f_a(h)}\cdot\frac{1-f_a(h)}{-h}\\ &= \lim_{h\searrow 0}\underbrace{\frac{1}{f_a(h)}}_{\to 1} \cdot \underbrace{\frac{f_a(h)-1}{h}}_{\to D^{+}f_a(0)}\\ &= D^+f_a(0). \end{align}

Daniel Fischer
  • 206,697
  • And I suppose you would prove convexity by checking that $f((x + y)/2) \leq (f(x) + f(y))/2$, and pointing out that $f$ is continuous, right? – user191371 Nov 10 '14 at 17:11
  • Yes, midpoint convexity plus continuity. Also, we can again shift everything to $0$ since it's a homomorphism, and have $f(2x) = f(x)^2 = \left(1 + (f(x)-1)\right)^2 \geqslant 1 + 2(f(x)-1)$. – Daniel Fischer Nov 10 '14 at 17:15
  • That's a nice proof. Thanks. – user191371 Nov 10 '14 at 17:15