View source for Non-analytic smooth function

{{Short description|Mathematical functions which are smooth but not analytic}}
In [[mathematics]], [[smooth function]]s (also called infinitely [[Differentiable function|differentiable]] functions) and [[analytic function]]s are two very important types of [[function (mathematics)|functions]]. One can easily prove that any analytic function of a [[real number|real]] [[Argument of a function|argument]] is smooth. The [[converse (logic)|converse]] is not true, as demonstrated with the [[counterexample]] below.

One of the most important applications of smooth functions with [[compact support]] is the construction of so-called [[mollifier]]s, which are important in theories of [[generalized function]]s, such as [[Laurent Schwartz]]'s theory of [[distribution (mathematics)|distribution]]s.

The existence of smooth but non-analytic functions represents one of the main differences between [[differential geometry]] and [[complex manifold|analytic geometry]]. In terms of [[sheaf theory]], this difference can be stated as follows: the sheaf of differentiable functions on a [[differentiable manifold]] is [[fine sheaf|fine]], in contrast with the analytic case.

The functions below are generally used to build up [[partition of unity|partitions of unity]] on differentiable manifolds.

==An example function==
===Definition of the function===
[[Image:Non-analytic smooth function.png|right|frame|The non-analytic smooth function ''f''(''x'') considered in the article.]]
Consider the function

:<math>f(x)=\begin{cases}e^{-\frac{1}{x}}&\text{if }x>0,\\ 0&\text{if }x\le0,\end{cases}</math>

defined for every [[real number]] ''x''.

===The function is smooth===

The function ''f'' has [[Continuous function|continuous]] [[derivative]]s of all orders at every point ''x'' of the [[real line]].  The formula for these derivatives is

:<math>f^{(n)}(x) = \begin{cases}\displaystyle\frac{p_n(x)}{x^{2n}}\,f(x) & \text{if }x>0, \\ 0 &\text{if }x \le 0,\end{cases}</math>

where ''p<sub>n</sub>''(''x'') is a [[polynomial]] of [[Degree of a polynomial|degree]] ''n''&nbsp;&minus;&nbsp;1 given [[recursion|recursively]] by ''p''<sub>1</sub>(''x'')&nbsp;=&nbsp;1 and

:<math>p_{n+1}(x)=x^2p_n'(x)-(2nx-1)p_n(x)</math>

for any positive [[integer]] ''n''.  From this formula, it is not completely clear that the derivatives are continuous at 0; this follows from the [[one-sided limit]] 

:<math>\lim_{x\searrow 0} \frac{e^{-\frac{1}{x}}}{x^m} = 0</math>

for any [[nonnegative]] integer ''m''.

{{Collapse top|title=Detailed proof of smoothness}}

By the [[Exponential function#Formal definition|power series representation of the exponential function]], we have for every [[natural number]] <math>m</math> (including zero)

:<math>\frac1{x^m}=x\Bigl(\frac1{x}\Bigr)^{m+1}\le (m+1)!\,x\sum_{n=0}^\infty\frac1{n!}\Bigl(\frac1x\Bigr)^n
=(m+1)!\,x e^{\frac{1}{x}},\qquad x>0,</math>

because all the positive terms for <math>n \neq m+1</math> are added. Therefore, dividing this inequality by <math>e^{\frac{1}{x}}</math> and taking the [[One-sided limit|limit from above]],

:<math>\lim_{x\searrow0}\frac{e^{-\frac{1}{x}}}{x^m}
\le (m+1)!\lim_{x\searrow0}x=0.</math>

We now prove the formula for the ''n''th derivative of ''f'' by [[mathematical induction]]. Using the [[chain rule]], the [[reciprocal rule]], and the fact that the derivative of the exponential function is again the exponential function, we see that the formula is correct for the first derivative of ''f'' for all ''x''&nbsp;>&nbsp;0 and that ''p''<sub>1</sub>(''x'') is a polynomial of degree 0. Of course, the derivative of ''f'' is zero for ''x''&nbsp;<&nbsp;0.
It remains to show that the right-hand side derivative of ''f'' at ''x''&nbsp;=&nbsp;0 is zero. Using the above limit, we see that

:<math>f'(0)=\lim_{x\searrow0}\frac{f(x)-f(0)}{x-0}=\lim_{x\searrow0}\frac{e^{-\frac{1}{x}}}{x}=0.</math>

The induction step from ''n'' to ''n''&nbsp;+&nbsp;1 is similar. For ''x''&nbsp;>&nbsp;0 we get for the derivative
:<math>\begin{align}f^{(n+1)}(x)
&=\biggl(\frac{p'_n(x)}{x^{2n}}-2n\frac{p_n(x)}{x^{2n+1}}+\frac{p_n(x)}{x^{2n+2}}\biggr)f(x)\\
&=\frac{x^2p'_n(x)-(2nx-1)p_n(x)}{x^{2n+2}}f(x)\\
&=\frac{p_{n+1}(x)}{x^{2(n+1)}}f(x),\end{align}</math>

where ''p''<sub>''n''+1</sub>(''x'') is a polynomial of degree ''n''&nbsp;= (''n''&nbsp;+&nbsp;1)&nbsp;&minus;&nbsp;1. Of course, the (''n''&nbsp;+&nbsp;1)st derivative of ''f'' is zero for ''x''&nbsp;<&nbsp;0. For the right-hand side derivative of ''f''<sup>&nbsp;(''n'')</sup> at ''x''&nbsp;=&nbsp;0 we obtain with the above limit

:<math>\lim_{x\searrow0} \frac{f^{(n)}(x) - f^{(n)}(0)}{x-0} = \lim_{x\searrow0} \frac{p_n(x)}{x^{2n+1}}\,e^{-1/x} = 0.</math>
{{Collapse bottom}}

===The function is not analytic===
As seen earlier, the function ''f'' is smooth, and all its derivatives at the [[origin (mathematics)|origin]] are&nbsp;0. Therefore, the [[Taylor series]] of ''f'' at the origin converges everywhere to the [[zero function]],

:<math>\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty \frac{0}{n!}x^n = 0,\qquad x\in\mathbb{R},</math>

and so the Taylor series does not equal ''f''(''x'') for ''x''&nbsp;>&nbsp;0. Consequently, ''f'' is not [[analytic function|analytic]] at the origin. 

===Smooth transition functions===
[[Image:Smooth transition from 0 to 1.png|right|frame|The smooth transition ''g'' from 0 to 1 defined here.]]
The function

:<math>g(x)=\frac{f(x)}{f(x)+f(1-x)},\qquad x\in\mathbb{R},</math>

has a strictly positive denominator everywhere on the real line, hence ''g'' is also smooth. Furthermore, ''g''(''x'')&nbsp;=&nbsp;0 for ''x''&nbsp;≤&nbsp;0 and ''g''(''x'')&nbsp;=&nbsp;1 for ''x''&nbsp;≥&nbsp;1, hence it provides a smooth transition from the level 0 to the level 1 in the [[unit interval]] <nowiki>[</nowiki>0, 1<nowiki>]</nowiki>. To have the smooth transition in the real interval <nowiki>[</nowiki>''a'', ''b''<nowiki>]</nowiki> with ''a''&nbsp;<&nbsp;''b'', consider the function

:<math>\mathbb{R}\ni x\mapsto g\Bigl(\frac{x-a}{b-a}\Bigr).</math>

For real numbers {{math|''a'' < ''b'' < ''c'' < ''d''}}, the smooth function

:<math>\mathbb{R}\ni x\mapsto g\Bigl(\frac{x-a}{b-a}\Bigr)\,g\Bigl(\frac{d-x}{d-c}\Bigr)</math>

equals 1 on the closed interval <nowiki>[</nowiki>''b'', ''c''<nowiki>]</nowiki> and vanishes outside the open interval (''a'', ''d''), hence it can serve as a [[bump function]].

==A smooth function that is nowhere real analytic==
[[File:Smooth non-analytic function.png|thumb|Approximation of the smooth-everywhere, but nowhere-analytic function mentioned here. This partial sum is taken from {{math|''k'' {{=}} 2<sup>0</sup>}} to 2<sup>500</sup>.|right]]

A more [[Pathological (mathematics)|pathological]] example is an infinitely differentiable function which is not analytic ''at any point''. It can be constructed by means of a  [[Fourier series]] as follows. Define for all <math>x \in \mathbb{R}</math>
:<math>F(x):=\sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}\cos(2^k x)\ .</math>

Since the series <math>\sum_{k\in \mathbb{N}} e^{-\sqrt{2^k}}{(2^k)}^n</math> converges for all <math>n \in \mathbb{N}</math>, this function is easily seen to be of class  C<sup>∞</sup>,  by a standard inductive application of the [[Weierstrass M-test]] to demonstrate [[uniform convergence]] of each series of derivatives.

We now show that <math>F(x)</math> is not analytic at any [[dyadic rational]] multiple of π, that is, at any <math>x := \pi \cdot p \cdot 2^{-q}</math> with <math>p \in \mathbb{Z}</math> and <math>q \in \mathbb{N}</math>.  Since the sum of the first <math>q</math> terms is analytic, we need only consider <math>F_{>q}(x)</math>, the sum of the terms with <math>k>q</math>.  For all orders of derivation <math>n = 2^m</math> with <math>m \in \mathbb{N}</math>, <math>m \geq 2</math> and <math>m > q/2</math> we have

:<math>F_{>q}^{(n)}(x):=\sum_{k\in \mathbb{N}\atop k>q} e^{-\sqrt{2^k}} {(2^k)}^n\cos(2^k x)  = \sum_{k\in \mathbb{N}\atop k>q} e^{-\sqrt{2^k}} {(2^k)}^n \ge  e^{-n} n^{2n}\quad  (\mathrm{as}\; n\to \infty)</math>

where we used the fact that <math>\cos(2^k x) = 1</math> for all <math>2^k > 2^q</math>, and we bounded the first sum from below by the term with <math>2^k=2^{2m}=n^2</math>. As a consequence, at any such <math>x \in \mathbb{R}</math>

:<math>\limsup_{n\to\infty} \left(\frac{|F_{>q}^{(n)}(x)|}{n!}\right)^{1/n}=+\infty\, ,</math>
so that the [[radius of convergence]] of the [[Taylor series]] of  <math>F_{>q}</math>  at  <math>x</math>  is  0  by the [[Cauchy-Hadamard theorem#Statement of the theorem|Cauchy-Hadamard formula]]. Since the set of analyticity of  a function  is an open set, and since dyadic rationals are [[Dense set|dense]], we conclude that  <math>F_{>q}</math>, and hence <math>F</math>, is nowhere analytic in <math>\mathbb{R}</math>.

== Application to Taylor series ==
<!--The article on Taylor series links to this section-->
{{main|Borel's lemma}}
For every sequence α<sub>0</sub>, α<sub>1</sub>, α<sub>2</sub>, .&nbsp;.&nbsp;. of real or [[complex number]]s, the following construction shows the existence of a smooth function ''F'' on the real line which has these numbers as derivatives at the origin.<ref>Exercise 12 on page 418 in [[Walter Rudin]], ''Real and Complex Analysis''. McGraw-Hill, New Delhi 1980, {{isbn|0-07-099557-5}}</ref> In particular, every sequence of numbers can appear as the coefficients of the [[Taylor series]] of a smooth function.  This result is known as [[Borel's lemma]], after [[Émile Borel]].

With the smooth transition function ''g'' as above, define

:<math>h(x)=g(2+x)\,g(2-x),\qquad x\in\mathbb{R}.</math>

This function ''h'' is also smooth; it equals 1 on the closed interval <nowiki>[</nowiki>&minus;1,1<nowiki>]</nowiki> and vanishes outside the open interval (&minus;2,2). Using ''h'', define for every natural number ''n'' (including zero) the smooth function

:<math>\psi_n(x)=x^n\,h(x),\qquad x\in\mathbb{R},</math>

which agrees with the [[monomial]] ''x<sup>n</sup>'' on <nowiki>[</nowiki>&minus;1,1<nowiki>]</nowiki> and vanishes outside the interval (&minus;2,2). Hence, the ''k''-th derivative of ''ψ<sub>n</sub>'' at the origin satisfies

:<math>\psi_n^{(k)}(0)=\begin{cases}n!&\text{if }k=n,\\0&\text{otherwise,}\end{cases}\quad k,n\in\mathbb{N}_0,</math>

and the [[boundedness theorem]] implies that ''ψ<sub>n</sub>'' and every derivative of ''ψ<sub>n</sub>'' is bounded. Therefore, the constants

:<math>\lambda_n=\max\bigl\{1,|\alpha_n|,\|\psi_n\|_\infty,\|\psi_n^{(1)}\|_\infty,\ldots,\|\psi_n^{(n)}\|_\infty\bigr\},\qquad n\in\mathbb{N}_0,</math>

involving the [[supremum norm]] of ''ψ<sub>n</sub>'' and its first ''n'' derivatives, are well-defined real numbers. Define the scaled functions

:<math>f_n(x)=\frac{\alpha_n}{n!\,\lambda_n^n}\psi_n(\lambda_n x),\qquad n\in\mathbb{N}_0,\;x\in\mathbb{R}.</math>

By repeated application of the [[chain rule]],

:<math>f_n^{(k)}(x)=\frac{\alpha_n}{n!\,\lambda_n^{n-k}}\psi_n^{(k)}(\lambda_n x),\qquad k,n\in\mathbb{N}_0,\;x\in\mathbb{R},</math>

and, using the previous result for the ''k''-th derivative of ''ψ<sub>n</sub>'' at zero,

:<math>f_n^{(k)}(0)=\begin{cases}\alpha_n&\text{if }k=n,\\0&\text{otherwise,}\end{cases}\qquad k,n\in\mathbb{N}_0.</math>

It remains to show that the function

:<math>F(x)=\sum_{n=0}^\infty f_n(x),\qquad x\in\mathbb{R},</math>

is well defined and can be differentiated term-by-term infinitely many times.<ref>See e.g. Chapter V, Section 2, Theorem 2.8 and Corollary 2.9 about the differentiability of the limits of sequences of functions in {{Citation
  | last = Amann
  | first = Herbert
  | last2 = Escher
  | first2 = Joachim
  | title = Analysis I
  | place = Basel
  | publisher = [[Birkhäuser Verlag]]
  | year = 2005
  | pages = 373–374
  | isbn = 3-7643-7153-6}}</ref> To this end, observe that for every ''k''

:<math>\sum_{n=0}^\infty\|f_n^{(k)}\|_\infty
\le \sum_{n=0}^{k+1}\frac{|\alpha_n|}{n!\,\lambda_n^{n-k}}\|\psi_n^{(k)}\|_\infty
+\sum_{n=k+2}^\infty\frac1{n!}
\underbrace{\frac1{\lambda_n^{n-k-2}}}_{\le\,1}
\underbrace{\frac{|\alpha_n|}{\lambda_n}}_{\le\,1}
\underbrace{\frac{\|\psi_n^{(k)}\|_\infty}{\lambda_n}}_{\le\,1}
<\infty,</math>

where the remaining infinite series converges by the [[ratio test]].

== Application to higher dimensions ==
[[File:Mollifier Illustration.svg|right|thumb|280px|The function Ψ<sub>1</sub>(''x'') in one dimension.]]
For every radius ''r''&nbsp;>&nbsp;0,

:<math>\mathbb{R}^n\ni x\mapsto \Psi_r(x)=f(r^2-\|x\|^2)</math>

with [[Euclidean norm]] ||''x''|| defines a smooth function on ''n''-dimensional [[Euclidean space]] with [[support (mathematics)|support]] in the [[ball (mathematics)|ball]] of radius ''r'', but <math>\Psi_r(0)>0</math>.

==Complex analysis==
This pathology cannot occur with differentiable [[complex analysis|functions of a complex variable]] rather than of a real variable. Indeed, all [[holomorphic functions are analytic]], so that the failure of the function ''f'' defined in this article to be analytic in spite of its being infinitely differentiable is an indication of one of the most dramatic differences between real-variable and complex-variable analysis.

Note that although the function ''f'' has derivatives of all orders over the real line, the [[analytic continuation]] of ''f'' from the positive half-line ''x''&nbsp;>&nbsp;0 to the [[complex plane]], that is, the function

:<math>\mathbb{C}\setminus\{0\}\ni z\mapsto e^{-\frac{1}{z}}\in\mathbb{C},</math>

has an [[essential singularity]] at the origin, and hence is not even continuous, much less analytic. By the [[great Picard theorem]], it attains every complex value (with the exception of zero) infinitely many times in every neighbourhood of the origin.

==See also==
* [[Bump function]]
* [[Fabius function]]
* [[Flat function]]
* [[Mollifier]]

==Notes==
{{reflist}}

==External links==
* {{planetmath reference|urlname=InfinitelydifferentiableFunctionThatIsNotAnalytic|title=Infinitely-differentiable function that is not analytic}}

[[Category:Smooth functions]]
[[Category:Articles containing proofs]]