But he uses the same idea. I admit this is one of only times I think Rudin could be vaulted is something he wrote.
Here is what he does. Define a continuous function on .
Now for .
You may want to verify that .
Now extend the function to all real numbers this way .
Hence his statement: “extend the function so that ".
That says that, for x between -1 and 1, = |x| and that then repeats with period 2. The graph of |x| is two half-lines joining at (0,0). The graph of is a "sawtooth" graph- a series of "up and down" line segments.
You will find a detailed construction and proof that it is everywhere continuous nowhere differentiable here, Google will also give many other explanations.
There exists a real continuous function on the real line which is nowhere differentiable.
(34) φ(x)=|x| (-1≤x≤1)
and extend the definition of φ(x ) to all real x by requiring that
Then for all s and t
(36) | φ(s) – φ(t) | ≤ | s-t |
In particular, φ is continuous on R'. Define
(37) f(x) = Σ( (3/4)^n) φ(x4^n) )
(note: Σ → n=0 to )
Since 0≤φ ≤1, (37) converges uniformly on R'. f is continuous on R'
Now fix a real number x and a positive integer m. Put
(38) m = ± (1/2)4^-m
where the sign is so chosen that no integer lies between 4^m, and 4^m(x + m).
This can be done, since 4^m|m|=1/2. Define
(39) yn=(φ (4^n(x + m)) - φ (x4^n))/ m
When n>m, then 4^n m is an even integer, so that yn=0. When 0 ≤ n ≤ m, (36) implies that
|yn| ≤ 4^n.
Since |ym|=4^m, we conclude that
| (f(x+ m)-f(x))/ m | = | Σ( (3/4)^n) yn |
(note: Σ → n=0 to m)
≥ 3^m - Σ( (3/4)^n)
(note: Σ → n=0 to m-1)
= 1/2(3^m + 1).
As m → , m→ 0. It follows that f is not differentiable at x.