In mathematics, smooth functions (also called infinitely differentiable functions) and analytic functions are two very important types of functions. One can easily prove that any analytic function of a real argument is smooth. The converse is not true, with this article constructing a counterexample.

Definition of the function

Consider the function

defined for every real number x.

The function is smooth

The function f has continuous derivatives of all orders in all points x of the real line, given by

where p_n(x) is a polynomial of degree n − 1 given recursively by p₁(x) = 1 and

Outline of proof

The proof, by induction, is based on the fact that for any natural number m including zero,

which implies that all f ⁽ⁿ⁾ are continuous and differentiable in x = 0, because

Detailed proof

By the power series representation of the exponential function, we have for every natural number m (including zero)

=(m+1)!\,x\exp\Bigl(\frac1x\Bigr),\qquad x>0,
because all the positive terms for n ≠ m + 1 are added. Therefore, using the functional equation of the exponential function,

\le (m+1)!\lim_{x\searrow0}x=0.
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 > 0 and that p₁(x) is a polynomial of degree 0. Of course, the derivative of f is zero for x < 0.
It remains to show that the right-hand side derivative of f at x = 0 is zero. Using the above limit, we see that

The induction step from n to n + 1 is similar. For x > 0 we get for the derivative

&=\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}
where p_n+1(x) is a polynomial of degree n = (n + 1) − 1. Of course, the (n + 1)^st derivative of f is zero for x < 0. For the right-hand side derivative of f ⁽ⁿ⁾ at x = 0 we obtain with the above limit

The function is not analytic

As seen earlier, the function f is smooth, and all its derivatives at the origin are 0. Therefore, the Taylor series of f at the origin converges everywhere to the zero function,

and so the Taylor series does not equal f(x) for x > 0. Consequently, f is not analytic at the origin. 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 f 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

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 often in every neighbourhood of the origin.

Uses

For every radius r > 0,

with Euclidean norm ||x|| defines a smooth function on n-dimensional Euclidean space with support in the ball of radius r.

One of the most important applications of smooth functions with compact support is the construction of so-called mollifiers, which are important in theories of generalized functions, like e.g. Laurent Schwartz's theory of distributions.

The existence of smooth but non-analytic functions represents one of the main differences between differential geometry and 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, in contrast with the analytic case.

The functions above are generally used to build up partitions of unity on differentiable manifolds.

See also

• Bump function