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, whereφ(x)=|x|

- March 17th 2009, 01:35 AMmslghlgPrincple of mathematical analysis, by Walter rudin. Need help....
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|** - March 17th 2009, 07:50 AMHallsofIvy
- March 17th 2009, 10:52 AMmslghlg
- March 17th 2009, 10:59 AMPlato
The pages in my edition do not agree with your edition.

Are you working with his proof for the existence of a continuous nowhere differentiable function?

Please give a more detailed discription of the your problem. - March 17th 2009, 07:30 PMmslghlg
- March 18th 2009, 06:13 PMPlato
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 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 ". - March 18th 2009, 07:19 PMmslghlg
- March 18th 2009, 07:43 PMPlato
- March 19th 2009, 08:12 AMHallsofIvy
Oh! That's completely different from what you said before.

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. - March 19th 2009, 08:30 AMCaptainBlack
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 - March 19th 2009, 10:13 AMmslghlg
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.