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Reprinted from the AMERICAN MATHEMATICAL MONTHLY, \\
Vol. LX, No. 10, December 1953
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AN EVERYWHERE CONTINUOUS NOWHERE DIFFRENTIABLE FUNCTION

John McCarthy, Princeton University


The following is an especially simple example.  It is
\begin{displaymath}
  f(x) = \sum_{n=1}^{\infty}2^{-n}g(2^{2^n}x)
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where $g(x) = 1+x$ for $-2 \leq x \leq 0$, $g(x) = 1 - x$ for $0 \leq
x \leq 2$ and $g(x)$ has period 4.

The function $f(x)$ is continuous because it is the uniform limit of
continuous functions.  To show that it is not differentiable, take
$\Delta x = \pm 2^{-2^k}$, choosing whichever sign makes $x$ and
$x + \Delta x$ be on the same linear segment of $g(2^{2^k}x)$.  We
have

1. $\Delta g(2^{2^n}x) = 0$ for $n > k$, since $g(2^{2^n}x)$ has
period $4\cdot 2^{-2^n}.$

2. $|\Delta g(2^{2^k}x)| =1.$

3. $|\Delta\sum_{n=1}^{k-1}g(2^{2^n}x)| \leq (k-1) \rm{max} |\Delta
g(2^{2^n}x)| \leq (k - 1)2^{2^{k-1}}2^{-2^k} < 2^k2^{-2^{k-1}}.$

Hence $|\Delta f/{\Delta x}| \geq 2^{-k}2^{2^k} - 2^k2^{2^{k-1}}$
which goes to infinity with $k$.

The proof that the present example has the required property is
simpler than that for any other example the author has seen.

Weierstrass gave the example $F(x) = \sum_{n=0}^\infty b^n \cos(a^n\pi 
  x)$ for $b < 1$ and $ab > 1 + 3\pi/2$ which is discussed in
  Goursat-Hedrick \textit{Mathematical Analysis}.

A complete discussion of functions with various singular properties is 
given in Hobson, \textit{Functions of a Real Variable}, volume II,
Cambridge, 1926.

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\noindent
2006 January note: I was tempted to dig up this 1953 note of mine and
put it on my web page by reading \textit{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 describes 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:

1. To prove a function $f$ differentiable at $x$, one must show that no
matter how $\Delta x$ goes to zero, $f(x + \Delta x)/{\Delta x}$
approaches a limit.  To prove $f$ non-differentiable at $x$, one need
only find a sequence of values of $\Delta x$ for which the limit
doesn't exist.

2. If $f(x)$ is to be represented as the sum of a series of continuous 
functions, 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.

3. In our case, the terms are $2^{-k}g(2^{2^k}x)$.  The $2^{2^k}x$
grows fast enough to overcome the $2^{-k}$ damping.

4. I would expect $\sum_{i=1}^\infty(x)$ to be nowhere differentiable
for most any initial $g(x)$.  

5. However, the particular $g(x)$ makes the proof easy, because the
periodicity kills the higher terms of the series for $\Delta g(x)$,
and using $2^{2^k}x$ for the argument allows the $k$th term to go to
infinity and dominate the earlier terms of the series.  It seems lucky 
in whatever sense luck exists in mathematics.




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