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A flat smooth real function has no holomorphic extension near zero

Statement refuted

Every smooth real function near $0$ is the restriction of a holomorphic function on some complex neighbourhood of $0$ .

Facts & Assumptions

Given: The function $ψ : R \to R$ defined by $ψ (0) = 0$ and $ψ (x) = \exp (- 1 / x^{2})$ for $x \neq 0$ .

[L1]

For every natural $m$ and every real $a > 0$ , $x^{m} / \exp (a x) \to 0$ as $x \to + \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

[L2]

The real exponential is smooth and every derivative equals the exponential (The exponential function is smooth and $(\exp)^{'} = \exp$ ).

[L3]

The derivative of a real composite is given by the chain rule under its differentiability hypotheses (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L4]

Products of differentiable real functions are differentiable and satisfy the product rule (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L5]

For each positive natural $n$ , the functions $x \mapsto x^{n}$ on $R$ and $x \mapsto x^{- n}$ off $0$ have the usual power-rule derivatives (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[L6]

A differentiable real function is continuous at every point of differentiability (A function differentiable at $c$ is continuous at $c$ ).

[L7]

A function is smooth when it is $C^{k}$ for every natural $k$ (Higher derivatives and the classes $C^{k}$ and $C^{\infty}$ ).

[L8]

For every real $x$ , $\exp (x) > 0$ and $\exp (- x) = 1 / \exp (x)$ (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

[L9]

Every holomorphic function equals its Taylor series throughout the largest centred open disc contained in its domain (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain).

[L10]

If $F (z) = \sum_{n \geq 0} c_{n} (z - a)^{n}$ near $a$ , then $c_{n} = F^{(n)} (a) / n!$ for every natural $n$ (The coefficients of a complex power series are its derivatives at the centre divided by the corresponding factorials).

Counterexample

technique · contradiction

1.1

For $x \neq 0$ , induction using [L2], [L3], [L4], and [L5] gives $ψ^{(n)} (x) = P_{n} (1 / x) \exp (- 1 / x^{2})$ for a real polynomial $P_{n}$ : $P_{0} = 1$ , and differentiating one such expression produces another polynomial in $1 / x$ times the same exponential.

L2L3L4L5givenalgebra

1.2

For every $x \neq 0$ , [L8] gives $ψ (x) = \exp (- 1 / x^{2}) > 0$ .

L8givenalgebra

2.1

Extend each expression in step 1.1 by the value $0$ at $x = 0$ . By [L1], both $P_{n} (1 / x) \exp (- 1 / x^{2})$ and its difference quotient divided by $x$ tend to $0$ as $x \to 0$ from either side. Inductively, every derivative exists at $0$ , equals $0$ , and is continuous there by [L6]; hence $ψ$ is smooth by [L7].

step 1.1L1L6L7

3.1

Suppose, for contradiction, that a holomorphic function $F$ on a complex neighbourhood of $0$ agrees with $ψ$ on a real interval about $0$ . Derivatives along the real axis then give $F^{(n)} (0) = ψ^{(n)} (0) = 0$ for every natural $n$ , and [L10] makes every Taylor coefficient of $F$ at $0$ equal to $0$ .

step 2.1L10assume-contra

4.1

By [L9], $F$ equals that zero Taylor series on a complex disc about $0$ , so $F$ vanishes there.

step 3.1L9

5.1

Every real interval about $0$ contains a nonzero $x$ , where step 1.2 gives $F (x) = ψ (x) > 0$ , contradicting step 4.1. Thus the smooth function $ψ$ has no holomorphic extension to any complex neighbourhood of $0$ .

step 1.2step 4.1discharge-contradiction∎

Depends on

Used by

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Sources

J. K. Hunter, An Introduction to Real Analysis, Example 10.31 and Corollary 10.30 (standard reference, not scraped)
J. Lebl, Guide to Cultivating Complex Analysis, §2.4 (standard reference, not scraped)