# Examples of statements that are true for real analytic functions but false for smooth functions

I'm writing because I don't know the usefulness of real analytic functions. I mean, I know that analyticity is something more respect differentiable ($C^\infty$ function), but I don't have in mind a result, which is true only for real analytic functions, and then become false for $C^\infty$ functions which are not analytic.

• One nice property: If $f,g$ are real analytic in $(a,b)$ and $f=g$ along a sequence of distinct points converging to some $c\in (a,b),$ then $f\equiv g.$ – zhw. Aug 28 '15 at 18:47

zhw points out a nice property of analytic functions:

If $f,g$ are analytic on $(a,b)$ and $f(x_n)=g(x_n)$ for a sequence of distinct points converging to some some $x_0\in(a,b)$, then $f(x)=g(x)$ for all $x\in(a,b)$.

This becomes false if we loosen the restriction of analyticity, as can be seen by considering the functions $$f(x)=\begin{cases}\exp(-x^{-2})&:x>0\\ 0 &:x\leq0\end{cases}\qquad\text{and}\qquad g(x)=\begin{cases}\exp(-x^{-2})&:x\neq0\\0 &:x=0\end{cases}.$$ It isn't hard to check that $f,g\in C^\infty(\Bbb R)$, and clearly $f(x)=g(x)$ for all $x>0$, but the statement above obviously fails whenever $x<0$; hence, it cannot be applied to $C^\infty$ functions.

The Cauchy-Kowalevski Theorem, together with Lewy's example, provides some food for thought.

The CKT is the main result about local existence of a solution of an analytic first order system of PDEs. Roughly speaking, if all the coefficients, the force and the boundary datum are analytic at some point the PDE admits a local solution which is $C^{\infty}$ and also analytic at the given point.

On the other hand, Lewy's example shows that the analogous theorem for smooth functions does not hold: he considered a system of PDEs with polynomial coefficients and was able to prove that there is a $C^{\infty}$ force for which the system has no solution of class $C^2$ on any open set. (Clearly this $f$ cannot be analytic in view of the CKT)

Needless to say, the proof of the CKT (at least the one I am aware of) is very involved and relies heavily on the power series expansion of the analytic functions.