For class we are exploring an example of a function that is smooth at $x=0$ but not analytic in any open interval centered at $0$. My question is, how can one prove that a function is not analytic? I am unaware of what tools are available to do so.

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    $\begingroup$ You show that it is not equal to its Taylor series in any neighborhood of $0.$ This goes back to Cauchy I think $\endgroup$ – zhw. Sep 28 '15 at 1:57
  • $\begingroup$ I don't know any other way than just checking the definition: compute the taylor series around $x=0$, and given $\epsilon > 0$, check if there is a point in $(x-\epsilon, x+\epsilon)$ where the Taylor series does not converge to the value of the function. $\endgroup$ – William Stagner Sep 28 '15 at 1:58

An explicit example might help:

$$f(x) =\begin{cases}e^{-1/x} \text{ for } x >0 \\ 0 \text{ for } x \leq 0\end{cases}$$

This is smooth but not analytic at $x=0$. Note that $f^n(0)=0$ for all $n$, so the Taylor series at $x=0$ is just $0$, which is clearly not $f(x)$ for any neighborhood.

However if you don't have smoothness, or even continuity, you don't have analyticity.

So ways you can tell is by if it's continuous/differentiable/smooth. If it IS smooth, you can check to see if it actually equals its Taylor series.

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    $\begingroup$ As an addendum: It is an interesting fact that the map TS (for taylor series) : $\{$ smooth functions $\}$ to $\{$ sequence of real numbers $\}$ : $$ g \to \{ g^{(n)} (0)/n!\} $$ is onto - i.e., any sequence of real numbers can be realized as the taylor series (convergent or otherwise) of a smooth function (smooth in some neighborhood of $0$). This is a theorem of Emile Borel,. Anyways, the "TS" function is neither into (your example), nor onto. See also math.stackexchange.com/questions/63050/… $\endgroup$ – peter a g Sep 28 '15 at 2:20
  • $\begingroup$ Obvious "typo" in my previous comment - TS is not into, but is onto. $\endgroup$ – peter a g Sep 28 '15 at 3:58

Analyticity implies differentiability which implies continuity. If a function is not continous or differentiable then it is not analytic.

Also, if you split a function, $f(z)$ into $f(x+iy)=u(x,y)+iv(x,y)$ and,

$u_x \neq v_y$ and/or $u_y \neq -v_x$ then the function is not analytic.

These are known as the Cauchy-Riemann equations and if they are not satisfied then the function is not analytic.

  • $\begingroup$ He's asking about smooth non analytic functions. $\endgroup$ – user223391 Sep 28 '15 at 2:01
  • $\begingroup$ @avid19 No, he is asking how to prove if a function is analytic or not, read the question. $\endgroup$ – Aleksandar Sep 28 '15 at 2:02
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    $\begingroup$ There is more to analysis than complex analysis :). $\endgroup$ – Count Iblis Sep 28 '15 at 2:02
  • $\begingroup$ @CountIblis True. $\endgroup$ – Aleksandar Sep 28 '15 at 2:02
  • $\begingroup$ The OP says smooth at $0.$ I take that to mean $C^\infty$ in a neighborhood of $0.$ $\endgroup$ – zhw. Sep 28 '15 at 6:18

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