Inmathematics,smooth functions (also called infinitelydifferentiable functions) andanalytic functions are two very important types offunctions. One can easily prove that any analytic function of arealargument is smooth. Theconverse is not true, as demonstrated with thecounterexample below.

One of the most important applications of smooth functions withcompact support is the construction of so-calledmollifiers, which are important in theories ofgeneralized functions, such asLaurent Schwartz's theory ofdistributions.

The existence of smooth but non-analytic functions represents one of the main differences betweendifferential geometry andanalytic geometry. In terms ofsheaf theory, this difference can be stated as follows: the sheaf of differentiable functions on adifferentiable manifold isfine, in contrast with the analytic case.

The functions below are generally used to build uppartitions of unity on differentiable manifolds.

Consider the function

defined for everyreal numberx.

The functionf hascontinuousderivatives of all orders at every pointx of thereal line. The formula for these derivatives is

wherep_n(x) is apolynomial ofdegreen − 1 givenrecursively byp₁(x) = 1 and

${\displaystyle p_{n+1}(x)=x^2{p}_n^{\prime }(x)-(2nx-1)p_n(x)}$

for any positiveintegern. From this formula, it is not completely clear that the derivatives are continuous at 0; this follows from theone-sided limit

${\displaystyle \underset{x\searrow 0}{lim}\frac{e^{-\frac{1}{x}}}{x^m}=0}$

for anynonnegative integerm.

As seen earlier, the functionf is smooth, and all its derivatives at theorigin are 0. Therefore, theTaylor series off at the origin converges everywhere to thezero function,

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

and so the Taylor series does not equalf(x) forx > 0. Consequently,f is notanalytic at the origin.

The function

${\displaystyle g(x)=\frac{f(x)}{f(x)+f(1-x)},x\in \mathbb{R},}$

has a strictly positive denominator everywhere on the real line, henceg is also smooth. Furthermore,g(x) = 0 forx ≤ 0 andg(x) = 1 forx ≥ 1, hence it provides a smooth transition from the level 0 to the level 1 in theunit interval [0, 1]. To have the smooth transition in the real interval [a,b] witha < b, consider the function

${\displaystyle \mathbb{R}\ni x\mapsto g(\frac{x-a}{b-a}).}$

For real numbersa <b <c <d, the smooth function

${\displaystyle \mathbb{R}\ni x\mapsto g(\frac{x-a}{b-a})g(\frac{d-x}{d-c})}$

equals 1 on the closed interval [b,c] and vanishes outside the open interval (a,d), hence it can serve as abump function.

A morepathological example is an infinitely differentiable function which is not analyticat any point. It can be constructed by means of aFourier 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 theWeierstrass M-test to demonstrateuniform convergence of each series of derivatives.

We now show that${\displaystyle F(x)}$ is not analytic at anydyadic 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 theradius of convergence of theTaylor series of${\displaystyle F_{>q}}$ at${\displaystyle x}$ is 0 by theCauchy-Hadamard formula. Since the set of analyticity of a function is an open set, and since dyadic rationals aredense, we conclude that${\displaystyle F_{>q}}$, and hence${\displaystyle F}$, is nowhere analytic in${\displaystyle \mathbb{R}}$.

For every sequence α₀, α₁, α₂, . . . of real orcomplex numbers, the following construction shows the existence of a smooth functionF on the real line which has these numbers as derivatives at the origin.^[1] In particular, every sequence of numbers can appear as the coefficients of theTaylor series of a smooth function. This result is known asBorel's lemma, afterÉmile Borel.

With the smooth transition functiong as above, define

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

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

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

which agrees with themonomialxⁿ on [−1,1] and vanishes outside the interval (−2,2). Hence, thek-th derivative ofψ_n at the origin satisfies

and theboundedness theorem implies thatψ_n and every derivative ofψ_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 thesupremum norm ofψ_n and its firstn 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 thechain 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 thek-th derivative ofψ_n at zero,

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.^[2] To this end, observe that for everyk

where the remaining infinite series converges by theratio test.

For every radiusr > 0,

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

withEuclidean norm ||x|| defines a smooth function onn-dimensionalEuclidean space withsupport in theball of radiusr, but${\displaystyle {\mathrm{\Psi }}_r(0)>0}$.

This pathology cannot occur with differentiablefunctions of a complex variable rather than of a real variable. Indeed, allholomorphic functions are analytic, so that the failure of the functionf 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 functionf has derivatives of all orders over the real line, theanalytic continuation off from the positive half-linex > 0 to thecomplex plane, that is, the function

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

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

• Bump function
• Fabius function
• Flat function
• Mollifier

• "Infinitely-differentiable function that is not analytic".PlanetMath.

• Smooth functions

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