Non-analytic smooth function

A smooth function that is nowhere real analytic

A more 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 ${\displaystyle x\in \mathbb{R}}$

${\displaystyle F(x):=\sum _{k\in \mathbb{N}}{e}^{-\sqrt{2^k}}\cos (2^{k}x).}$

Since the series ${\displaystyle \sum _{k\in \mathbb{N}}{e}^{-\sqrt{2^k}}{(2^k)}^n}$ converges for all ${\displaystyle n\in \mathbb{N}}$, this function is easily seen to be of class C^∞, by a standard inductive application of the Weierstrass M-test to demonstrate uniform convergence of each series of derivatives.

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

${\displaystyle F_{>q}^{(n)}(x):=\sum _{\genfrac{}{}{0ex}{}{k\in \mathbb{N}}{k>q}}{e}^{-\sqrt{2^k}}{(2^k)}^n\cos (2^{k}x)=\sum _{\genfrac{}{}{0ex}{}{k\in \mathbb{N}}{k>q}}{e}^{-\sqrt{2^k}}{(2^k)}^n\ge e^{-n}{n}^{2n}(\mathrm{a}\mathrm{s}n\to \mathrm{\infty })}$

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

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{\left(\frac{|F_{>q}^{(n)}(x)|}{n!}\right)}^{1/n}=+\mathrm{\infty },}$

so that the radius of convergence of the Taylor series of ${\displaystyle F_{>q}}$ at ${\displaystyle x}$ is 0 by the Cauchy-Hadamard formula. Since the set of analyticity of a function is an open set, and since dyadic rationals are dense, we conclude that ${\displaystyle F_{>q}}$, and hence ${\displaystyle F}$, is nowhere analytic in ${\displaystyle \mathbb{R}}$.

Application to Taylor series

For every sequence α₀, α₁, α₂, . . . of real or complex numbers, the following construction shows the existence of a smooth function F on the real line which has these numbers as derivatives at the origin. 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

${\displaystyle h(x)=g(2+x)g(2-x),x\in \mathbb{R}.}$

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

${\displaystyle {\psi }_n(x)=x^{n}h(x),x\in \mathbb{R},}$

which agrees with the monomial xⁿ on [−1,1] and vanishes outside the interval (−2,2). Hence, the k-th derivative of ψ_n at the origin satisfies

${\displaystyle {\psi }_n^{(k)}(0)=\{\begin{array}{ll}n!& \text{if }k=n,\\ 0& \text{otherwise,}\end{array}k,n\in {\mathbb{N}}_0,}$

and the boundedness theorem implies that ψ_n and every derivative of ψ_n is bounded. Therefore, the constants

${\displaystyle {\lambda }_n=max\{1,|{\alpha }_n|,\Vert {\psi }_n{\Vert }_{\mathrm{\infty }},\Vert {\psi }_n^{(1)}{\Vert }_{\mathrm{\infty }},\dots ,\Vert {\psi }_n^{(n)}{\Vert }_{\mathrm{\infty }}\},n\in {\mathbb{N}}_0,}$

involving the supremum norm of ψ_n and its first n derivatives, are well-defined real numbers. Define the scaled functions

${\displaystyle f_n(x)=\frac{{\alpha }_n}{n!{\lambda }_n^n}{\psi }_n({\lambda }_{n}x),n\in {\mathbb{N}}_0,x\in \mathbb{R}.}$

By repeated application of the chain rule,

${\displaystyle f_n^{(k)}(x)=\frac{{\alpha }_n}{n!{\lambda }_n^{n-k}}{\psi }_n^{(k)}({\lambda }_{n}x),k,n\in {\mathbb{N}}_0,x\in \mathbb{R},}$

and, using the previous result for the k-th derivative of ψ_n at zero,

${\displaystyle f_n^{(k)}(0)=\{\begin{array}{ll}{\alpha }_n& \text{if }k=n,\\ 0& \text{otherwise,}\end{array}k,n\in {\mathbb{N}}_0.}$

It remains to show that the function

${\displaystyle F(x)=\sum _{n=0}^{\mathrm{\infty }}{f}_n(x),x\in \mathbb{R},}$

is well defined and can be differentiated term-by-term infinitely many times. To this end, observe that for every k

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\Vert f_n^{(k)}{\Vert }_{\mathrm{\infty }}\le \sum _{n=0}^{k+1}\frac{|{\alpha }_n|}{n!{\lambda }_n^{n-k}}\Vert {\psi }_n^{(k)}{\Vert }_{\mathrm{\infty }}+\sum _{n=k+2}^{\mathrm{\infty }}\frac{1}{n!}\underset{\le 1}{\underbrace{\frac{1}{{\lambda }_n^{n-k-2}}}}\underset{\le 1}{\underbrace{\frac{|{\alpha }_n|}{{\lambda }_n}}}\underset{\le 1}{\underbrace{\frac{\Vert {\psi }_n^{(k)}{\Vert }_{\mathrm{\infty }}}{{\lambda }_n}}}<\mathrm{\infty },}$

where the remaining infinite series converges by the ratio test.

Application to higher dimensions

For every radius r > 0,

${\displaystyle {\mathbb{R}}^n\ni x\mapsto {\mathrm{\Psi }}_r(x)=f(r^2-\Vert x{\Vert }^2)}$

with Euclidean norm ||x|| defines a smooth function on n-dimensional Euclidean space with support in the ball of radius r, but ${\displaystyle {\mathrm{\Psi }}_r(0)>0}$.

Complex analysis

This pathology cannot occur with differentiable 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 > 0 to the complex plane, that is, the function

${\displaystyle \mathbb{C}\setminus \{0\}\ni z\mapsto e^{-\frac{1}{z}}\in \mathbb{C},}$

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.