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Homework Help: The Maximum of a Nonnegative Function as an Integral

  1. Nov 21, 2008 #1
    The problem statement, all variables and given/known data
    Let f: [a,b] -> R be continuous such that f(x) ≥ 0 for all x in [a,b]. Show that

    [tex]\lim_{n \to \infty} \left( \int_a^b (f(x))^n \, dx \right)^{1/n} = \max\{f(x) : x \in [a,b]\}.[/tex]

    Relevant equations
    The fundamental theorem of calculus and its corollaries.

    The attempt at a solution
    It is easy to show that the limit is less than or equal to the max of f. This doesn't rely on using the fact that f ≥ 0, so somehow this extra piece of info. turns 'less than or equal' into 'equal', but I have failed to determine why. Any tips?
  2. jcsd
  3. Nov 21, 2008 #2


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    Suppose M is the max. Pick any m such that 0<m<M. Define g_m(x) to be 0 if f(x)<m and m if f(x)>=m. The length of the set where g_m(x)=m is greater than zero, right? Call it d_m. g_m(x)<=f(x). Can you show if you put g_m(x) into your limit, you get m?
    Last edited: Nov 21, 2008
  4. Nov 21, 2008 #3
    Heuristically, you might think of the integral as a Riemann sum, that is

    where [itex]\{x_j\}_{j=0,\dots,N}[/itex] is a partition of [a,b] with [itex]x_0=1,x_N=b[/itex]. Assume f assume its maximal value m in the interval [itex][x_{\tilde j-1},x_{\tilde j })[/itex]. Then you can factor out this m to obtain
    m^n\left[x_{\tilde j}-x_{\tilde j -1}+\sum_{j=1,j\neq\tilde j}^N{\frac{\max{\{f(x):x\in[x_{j-1},x_{j}\}}^n}{m}(x_{j}-x_{j-1})}\right]
    if you take take the nth root of this expression you get
    m\left[x_{\tilde j}-x_{\tilde j -1}+\sum_{j=1,j\neq\tilde j}^N{\frac{\max{\{f(x):x\in[x_{j-1},x_{j}\}}^n}{m}(x_{j}-x_{j-1})}\right]^{1/n}
    Can you show that the second factor (for fixed N) converges to 1?
    Can you use this to prove the original question? (You will have to be careful with interchanging the two limit processes involved.)

    Edit: Ok. Dick's method is a lot easier:smile:
  5. Nov 22, 2008 #4
    We have to assume that M > 0 for that to work. We can prove the case M = 0 easily.

    What do you mean by "length of the set"? Do you mean the diameter of {x in [a,b] : g_m(x) = m}? Also if I put g_m(x) into my limit and get m, what then? Are you somehow going to convert m into M?
  6. Nov 22, 2008 #5
    \left[\int_a^b{dx g_m(x)^n}\right]^{1/n}=m
    for all 0<m<M it follows that
    \left[\int_a^b{dx f(x)^n}\right]^{1/n}\geq m
    Letting [itex]m\to M[/itex]...
  7. Nov 22, 2008 #6
    I was just thinking that. Makes sense. Thanks.
  8. Nov 22, 2008 #7
    I'm trying to prove that g_m(x) is integrable. I wanted to do this by showing that g has finitely many points of discontinuity but I'm not sure that is this is the case. Is there an easier way of proving that g_m(x) is integrable?
  9. Nov 22, 2008 #8
    Hmm...I think the best way to show that g_m(x) is integrable is by using the definition together with the fact that f is integrable.
  10. Nov 22, 2008 #9
    Are you talking about Riemann or Lebesgue integrability? If the latter is the case, it's enough to observe that
    g_m(x)=m\chi_{\{f\geq m\}}
    is essentially the indicator function of the set [itex]d_m=\{x\in[a,b]:f(x)\geq m\}[/itex]. It is therefore integrable if [itex]d_m[/itex] is a Borel set, which is obvious from the continuity f. If you mean Riemann integrable the reasoning is essentially the same. Show that [itex]d_m[/itex] is a union of intervals, and so again [itex]g_m[/itex] is a simple step function which you hopefully know is integrable.
  11. Nov 22, 2008 #10


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    By 'length', sure, I meant measure. Why are you worried about such a simple function being integrable? If you are going to get all bothered about it simplify g_m even more. Just pick a point where f(x)>m. Now there's an open interval around x where f(x)>m. Define g_m to be m in that interval and zero otherwise. There. Now it's a step function. With one step.
  12. Nov 22, 2008 #11
    OK. I have produced a solution to my satisfaction. Thanks everybody.
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