Smooth Real Function is not necessarily Analytic

Theorem

Let $f$ be a real function which is smooth on the open interval $\openint a b$.

real function which is analytic real function on the open interval $\openint a b$.

Then it is not necessarily the case that $f$ is also analytic on $\openint a b$.

Proof

Proof by Counterexample

Consider the real function $f: \R \to \R$ defined as:

$\forall x \in \R: \map f x = \begin {cases} \map \exp {\dfrac {-1} {x^2} } & : x \ne 0 \\ 0 & : x = 0 \end {cases}$

$f$ is of differentiability class $C^\infty$ such that the derivatives of all orders equal $0$.

Thus the Taylor series of $f$ at $x = 0$ converges to $f$ at $0$ only.

Hence by definition $f$ is not an analytic real function.

$\blacksquare$

Also see

• Analytic Real Function has Derivatives of All Orders

Sources

• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): analytic