Please explain me the equation (35), page number 154 from the book 'Principles of Mathematical Analysis' by Walter Rudin which says,
φ(x+2)=φ(x) for all -1≤x≤1, where φ(x)=|x|
In my earlier edition Rudin uses a different function to prove the same theorem.
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 $\displaystyle f$ on $\displaystyle [0,2]$.
Now for $\displaystyle \left( {\forall x \in \mathbb{R}} \right)\left[ {g(x) = x - 2\left\lfloor {\frac{x}
{2}} \right\rfloor } \right]$.
You may want to verify that $\displaystyle \left( {\forall x \in \mathbb{R}} \right)\left[ {g(x) \in \left[ {0,2} \right]} \right]$.
Now extend the function $\displaystyle f$ to all real numbers this way $\displaystyle \left( {\forall x \in \mathbb{R}} \right)\left[ {\phi (x) = f(g(x))} \right]$.
Hence his statement: “extend the function so that $\displaystyle \phi (x) = \phi (x + 2)$".
Oh! That's completely different from what you said before.
That says that, for x between -1 and 1, $\displaystyle \phi(x)$= |x| and that then repeats with period 2. The graph of |x| is two half-lines joining at (0,0). The graph of $\displaystyle \phi(x)$ is a "sawtooth" graph- a series of "up and down" line segments.
This looks like a construction of the Blacmange function othewise known as the Takagi fractal (or at least a dilated translate of it) which is the usual suspect that is rolled out when one needs an everywhere continuous nowhere differentiable function.
You will find a detailed construction and proof that it is everywhere continuous nowhere differentiable here, Google will also give many other explanations.
CB
Here is the complete theorem, please explain me the equations (34) and (35).
Theorem:
There exists a real continuous function on the real line which is nowhere differentiable.
Proof
Define
(34) φ(x)=|x| (-1≤x≤1)
and extend the definition of φ(x ) to all real x by requiring that
(35)φ(x+2)=φ(x)
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.