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Pack narsese -- jmc/weierstrass.md


Vol. LX, No. 10, December 1953



John McCarthy, Princeton University

The following is an especially simple example. It is

f (x) =





where g(x) = 1 + x for −2 ≤ x ≤ 0, g(x) = 1 − x for 0 ≤ x ≤ 2 and g(x) has

period 4.

The function f (x) is continuous because it is the uniform limit of con-

tinuous functions. To show that it is not differentiable, take ∆x = ±2−2k


choosing whichever sign makes x and x + ∆x be on the same linear segment

of g(22k

x). We have

x) = 0 for n > k, since g(22n

x)| = 1.

  1. ∆g(22n
  2. |∆g(22k
  3. |∆ Pk−1 Hence |∆f /∆x| ≥ 2−k22k

    The proof that the present example has the required property is simpler

    which goes to infinity with k.

    x)| ≤ (k−1)max|∆g(22n

    x) has period 4 · 2−2n

    x)| ≤ (k−1)22k−1

    < 2k2−2k−1

    − 2k22k−1

    n=1 g(22n




    than that for any other example the author has seen.

    Weierstrass gave the example F (x) = P∞

    n=0 bn cos(anπx) for b < 1 and

    ab > 1 + 3π/2 which is discussed in Goursat-Hedrick Mathematical Analysis.

    A complete discussion of functions with various singular properties is

    given in Hobson, Functions of a Real Variable, volume II, Cambridge, 1926.

    2006 January note: I was tempted to dig up this 1953 note of mine and put it

    on my web page by reading The Calculus Gallery by William Dunham. This

    excellent book includes the first proofs of a number of important theorems,

    including Weierstrass’s proof that his function has the required properties.

    Dunham’s version of Weierstrass’s proof is six pages of what Dunham de-

    scribes as difficult mathematics. Since my proof is 13 lines of what I consider

    easy math, I decided to copy my old note and discuss it. Dunham recounts

    that the famous mathematicians Hermite, Poincare and Picard all expressed

    themselves as repelled by Weierstrass’s “pathological example”.

    I’m sure

    that by the time I was born in 1927, such functions were no longer regarded

    as repellent. My own opinion is that most everywhere continuous functions,

    in some suitable sense of most, are nowhere differentiable.


  4. To prove a function f differentiable at x, one must show that no matter how ∆x goes to zero, f (x + ∆x)/∆x approaches a limit. To prove f non-

    differentiable at x, one need only find a sequence of values of ∆x for which

    the limit doesn’t exist.

  5. If f (x) is to be represented as the sum of a series of continuous func- tions, it suffices to bound the terms by a suitable positive terms, 2−k in our

    case. Then f is sure to be everywhere continuous. It doesn’t matter how

    fast the successive terms wiggle.

  6. In our case, the terms are 2−kg(22k x). The 22k

    x grows fast enough to

    overcome the 2−k damping.


  7. I would expect P∞ i=1(x) to be nowhere differentiable for most any initial
  8. However, the particular g(x) makes the proof easy, because the peri- odicity kills the higher terms of the series for ∆g(x), and using 22k

    x for the

    argument allows the kth term to go to infinity and dominate the earlier terms

    of the series. It seems lucky in whatever sense luck exists in mathematics.