Extremely hard concept

Hunterelite7 · May 4, 2008

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

trambolin · May 4, 2008

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 · May 4, 2008

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

trambolin · May 5, 2008

Why is the following true?

[tex]
\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)|
[/tex]
This proof lacks only one limiting argument, can you find it?

Hunterelite7 · May 5, 2008

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

HallsofIvy · May 5, 2008

What function and the max of what?

Hunterelite7 · May 5, 2008

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

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Extremely hard concept

Hunterelite7

trambolin

Hunterelite7

trambolin

Hunterelite7

HallsofIvy

Hunterelite7

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