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Reprinted from the AMERICAN MATHEMATICAL MONTHLY,
Vol. LX, No. 10, December 1953
AN EVERYWHERE CONTINUOUS NOWHERE DIFFRENTIABLE
FUNCTION
John McCarthy, Princeton University
The following is an especially simple example. It is
f (x) =
2−ng(22n
x)
∞
X
n=1
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.
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
2−2k
.
.
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.
Remarks:
differentiable at x, one need only find a sequence of values of ∆x for which
the limit doesn’t exist.
case. Then f is sure to be everywhere continuous. It doesn’t matter how
fast the successive terms wiggle.
x grows fast enough to
overcome the 2−k damping.
g(x)
.
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.