# Real Analysis

• May 3rd 2010, 03:00 PM
agalmam
Real Analysis
For functions f : R --> R and a>0 set fa(x)=a f(x/a).Prove if the family fa is equicontinuous, then f is Lipshitz continuous. fa as a>0.
• May 3rd 2010, 03:02 PM
Drexel28
Quote:

Originally Posted by agalmam
For functions f : R --> R and a>0 set fa(x)=a f(x/a).Prove if the family fa is equicontinuous, then f is Lipshitz continuous. fa as a>0.

Let's see some work. Also, is what you wrote $f_a(x)=af\left(\frac{x}{a}\right)$?
• May 3rd 2010, 03:07 PM
agalmam
Yes thanx, i did not have the tools to do that, and i would appreciate your help.
• May 3rd 2010, 03:08 PM
Drexel28
Quote:

Originally Posted by agalmam
Yes thanx, i did not have the tools to do that, and i would appreciate your help.

What have you tried? Do you have any leads?
• May 3rd 2010, 03:19 PM
agalmam
Yes i acually tried the definition of the equicontinuous, which's for every eps>0 it exist delt>0 st fa and for every x,y in R Ifa(x)-fa(y)I< eps implies Ix-yI<delt. then since fa(x)=a*f(x/a). Then a*If(x/a)-f(y/a)I< eps for Ix-yI<delt and x/a=x' and y/a=y' then If(x')-f(y')I< eps/a and Ix'-y'I<delt/a that's how far i got
• May 3rd 2010, 03:24 PM
agalmam
I can see f is Lipschitz but i could not prove it properly