Limits of integral to find a maximum from [a,b]

[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




3. I am certian the solution is in front of my face because if i can see how to establish the limit is both less than or equal to and greater than or equal to than it would be easy but i can t see how to set the inequalities

 

[lippyka]

up.The solution lies in using the Mean Value Theorem for Integrals. We know that for any continuous function f(x) on the closed interval [0,1], there exists a number c in [0,1] such that \int_0^1 f(x)dx = f(c). Now, consider the limit of the expression \lim_{p \to \infty}\left(\int_0^1 |f(t)|^p dt\right)^{1/p}. By the mean value theorem, this equals |f(c)|^p for some c in [0,1]. Since the power of p is going to infinity, this expression must equal the maximum value of |f(x)| in [0,1].