# Princple of mathematical analysis, by Walter rudin. Need help....

• Mar 17th 2009, 01:35 AM
mslghlg
Princple 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 -1x≤1, where φ(x)=|x|
• Mar 17th 2009, 07:50 AM
HallsofIvy
Quote:

Originally Posted by mslghlg
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 -1x≤1, where φ(x)=|x|

I don't have that book on hand so I can't look that up to see if you have quoted it wrong, but it clearly is not true. If, for example, if x= 1, then |x+2|= 3 while |x|= 1.
• Mar 17th 2009, 10:52 AM
mslghlg
Quote:

Originally Posted by HallsofIvy
I don't have that book on hand so I can't look that up to see if you have quoted it wrong, but it clearly is not true. If, for example, if x= 1, then |x+2|= 3 while |x|= 1.

Thats what troubling me, however you can never expect a mistake in Rudin books. Here x is bounded
$|x|<=1$.
• Mar 17th 2009, 10:59 AM
Plato
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?
• Mar 17th 2009, 07:30 PM
mslghlg
Quote:

Originally Posted by Plato
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?

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)

• Mar 18th 2009, 06:13 PM
Plato
Quote:

Originally Posted by mslghlg
Theorem:
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 (35) φ(x+2)=φ(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 $f$ on $[0,2]$.
Now for $\left( {\forall x \in \mathbb{R}} \right)\left[ {g(x) = x - 2\left\lfloor {\frac{x}
{2}} \right\rfloor } \right]$
.

You may want to verify that $\left( {\forall x \in \mathbb{R}} \right)\left[ {g(x) \in \left[ {0,2} \right]} \right]$.
Now extend the function $f$ to all real numbers this way $\left( {\forall x \in \mathbb{R}} \right)\left[ {\phi (x) = f(g(x))} \right]$.
Hence his statement: “extend the function so that $\phi (x) = \phi (x + 2)$".
• Mar 18th 2009, 07:19 PM
mslghlg
Quote:

Originally Posted by Plato
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 $f$ on $[0,2]$.
Now for $\left( {\forall x \in \mathbb{R}} \right)\left[ {g(x) = x - 2\left\lfloor {\frac{x}
{2}} \right\rfloor } \right]$
.

You may want to verify that $\left( {\forall x \in \mathbb{R}} \right)\left[ {g(x) \in \left[ {0,2} \right]} \right]$.
Now extend the function $f$ to all real numbers this way $\left( {\forall x \in \mathbb{R}} \right)\left[ {\phi (x) = f(g(x))} \right]$.
Hence his statement: “extend the function so that $\phi (x) = \phi (x + 2)$".

Please Please let me know the complete theorem that you know-"There exists a real continuous function on the real line which is nowhere differentiable"
• Mar 18th 2009, 07:43 PM
Plato
Quote:

Originally Posted by mslghlg
Please Please let me know the complete theorem that you know-"There exists a real continuous function on the real line which is nowhere differentiable"

Sorry but I have no idea what Rudin does in your edition of that text.
You may quote the exact context of the theorem.
Otherwise, I am of no more use to you.
• Mar 19th 2009, 08:12 AM
HallsofIvy
Quote:

Originally Posted by mslghlg
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)

Oh! That's completely different from what you said before.

That says that, for x between -1 and 1, $\phi(x)$= |x| and that then repeats with period 2. The graph of |x| is two half-lines joining at (0,0). The graph of $\phi(x)$ is a "sawtooth" graph- a series of "up and down" line segments.
• Mar 19th 2009, 08:30 AM
CaptainBlack
Quote:

Originally Posted by mslghlg
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)

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
• Mar 19th 2009, 10:13 AM
mslghlg
Quote:

Originally Posted by Plato
Sorry but I have no idea what Rudin does in your edition of that text.
You may quote the exact context of the theorem.
Otherwise, I am of no more use to you.

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 0φ ≤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.