# Extremely hard concept

I am trying to prove that the Limit as p approaches infinity of {integral from 0 to 1[|f(t)|^p dt]}^(1/p) is in fact equal to the max of |f(x)| between [0,1].

Any suggestions im sure I need to set the limit to less than or equal to and greater than or equal to the max but i dont quite know how

## Answers and Replies

One hint.... What is the most contributing term when you take the n th power of each element and sum them up,

keep it simple and start with taking two elements a,b and take the power of 20th...

im sorry im having a hard time folowing yor terminology is there any way to rephrase

Why is the following true?

$$\left(\int_0^1{|f(t)|^p dt}\right)^{1/p} \leq \left(\int_0^1{\underbrace{(\max{|f(t)|})^p}_{const} dt}\right)^{1/p}=\max|f(t)|\int_0^1{dt}=\max|f(t)|$$
This proof lacks only one limiting argument, can you find it?

ok so how do I show the opposite or that the function is greater than or equal to the max

HallsofIvy
Science Advisor
Homework Helper
What function and the max of what?

the function is just vague f(t) and the max is the maximum of |f(t)| between [0,1]