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A uniform limit of smooth functions need not be differentiable anywhere

Statement

There is a sequence of smooth functions $S_{N} : R \to R$ that converges uniformly on $R$ to a continuous function which is differentiable at no real point. Consequently, uniform convergence does not preserve even first differentiability, despite every approximating function being $C^{\infty}$ .

Facts & Assumptions

Given: The Weierstrass partial sums $S_{N} (x) = \sum_{n = 0}^{N} a^{n} \cos (b^{n} π x)$ .

[L1]

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

[L2]

The partial sums of the classical Weierstrass series converge uniformly to $W_{a, b}$ on $R$ (The classical Weierstrass series converges uniformly to a continuous function).

[L3]

If $0 < a < 1$ , $b > 1$ is an odd integer, and $a b > 1 + 3 π / 2$ , then $W_{a, b}$ is continuous everywhere and differentiable nowhere (Under $a b > 1 + 3 π / 2$ , the classical Weierstrass function is continuous everywhere and differentiable nowhere).

[L4]

Sine and cosine are differentiable on $R$ , and their derivatives are cosine and negative sine (The derivatives of sine and cosine are cosine and minus sine).

[L5]

The chain rule and the sum and scalar-multiple rules compute derivatives of finite sums and affine composites (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , 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 $g (c) \neq 0$ ).

[L6]

Differentiable real functions are continuous (A function differentiable at $c$ is continuous at $c$ ).

[L7]

There is a unique $γ \in (0, 2)$ with $\cos γ = 0$ , and $π = 2 γ$ (Cosine has a smallest positive zero, lying strictly between zero and two, Pi as twice the smallest positive zero of cosine).

Proof

technique · direct

1.1L3L7choosealgebra

Choose $a = 1 / 2$ and $b = 15$ . The integer $15$ is odd, and [L7] gives $π < 4$ , so $1 + 3 π / 2 < 7 < 15 / 2 = a b$ ; all hypotheses of [L3] hold.

1.2L1L4L5L6algebra

Repeated application of [L4] and [L5] shows that every derivative of every finite partial sum $S_{N}$ is a finite linear combination of sine and cosine functions. Those derivatives are continuous by [L6], so each $S_{N}$ is smooth in the sense of [L1].

2.1step 1.2L2

By [L2], the smooth functions from step 1.2 converge uniformly on $R$ to $W_{1 / 2, 15}$ .

2.2step 1.1L3

The parameter check in step 1.1 lets [L3] identify this uniform limit as continuous everywhere and differentiable nowhere.

3.1step 2.1step 2.2∎

Thus the sequence in step 2.1 consists of smooth functions and converges uniformly to the nowhere-differentiable function in step 2.2.

Depends on

Used by

FALSE: every continuous real function is differentiable somewhere False statement

Dependency tree · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jeff Calder, Weierstrass's Non-Differentiable Function, Theorem 1 (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis, Example 9.24 (standard reference, not scraped)