How to prove that a smooth function is NOT analytic?

Question

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.

You show that it is not equal to its Taylor series in any neighborhood of $0.$ This goes back to Cauchy I think — zhw., Sep 28 '15 at 01:57
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. — William Stagner, Sep 28 '15 at 01:58

score 16 · Accepted Answer · answered Sep 28 '15 at 02:08

16

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.

answered Sep 28 '15 at 02:08

2

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 http://math.stackexchange.com/questions/63050/every-power-series-is-the-taylor-series-of-some-c-infty-function – peter a g Sep 28 '15 at 02:20
Obvious "typo" in my previous comment - TS is not into, but is onto. – peter a g Sep 28 '15 at 03:58

score 0 · Answer 2 · answered Sep 28 '15 at 01:59

0

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.

answered Sep 28 '15 at 01:59

Aleksandar

1,771

He's asking about smooth non analytic functions. – Sep 28 '15 at 02:01
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@avid19 No, he is asking how to prove if a function is analytic or not, read the question. – Aleksandar Sep 28 '15 at 02:02
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There is more to analysis than complex analysis :). – Count Iblis Sep 28 '15 at 02:02
@CountIblis True. – Aleksandar Sep 28 '15 at 02:02
The OP says smooth at $0.$ I take that to mean $C^\infty$ in a neighborhood of $0.$ – zhw. Sep 28 '15 at 06:18

How to prove that a smooth function is NOT analytic?

2 Answers2