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

In summary, the conversation is discussing a proof for the limit of a certain integral involving a function f(t) as p approaches infinity. The participants suggest using a limiting argument and simplifying the terms by taking the 20th power. The final step involves showing that the function is greater than or equal to the maximum of |f(t)| between [0,1].
  • #1
Hunterelite7
5
0
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
 
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  • #2
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...
 
  • #3
im sorry I am having a hard time folowing yor terminology is there any way to rephrase
 
  • #4
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?
 
  • #5
ok so how do I show the opposite or that the function is greater than or equal to the max
 
  • #6
What function and the max of what?
 
  • #7
the function is just vague f(t) and the max is the maximum of |f(t)| between [0,1]
 

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