Is the Limit of the Integral of a Function Equal to its Maximum Value?

[Hunterelite7]

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 I am sure I need to set the limit to less than or equal to and greater than or equal to the max but i don't quite know how

 

[trambolin]

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...

 

[Hunterelite7]

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

 

[trambolin]

Why is the following true?

<br /> \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)|<br />
This proof lacks only one limiting argument, can you find it?

 

[Hunterelite7]

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

 

[HallsofIvy]

What function and the max of what?

 

[Hunterelite7]

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