10
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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 $f\equiv g.$ $\endgroup$ – zhw. Aug 28 '15 at 18:47
6
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.