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Best bound for simple inequality

  1. Oct 11, 2011 #1
    Hello all, the problem I have is the following:

    Suppose [itex]f \in C^1(0,1)[/itex] and [itex]f(0) = 0[/itex], then

    f^2(x) \le \int_0^1 f^2(x) dx,

    but I was wondering if 1 is the best constant for the inequality. In other words, how do I determine the best bound for

    f^2(x) \le K \int_0^1 f^2(x) dx,

    for K positive?
  2. jcsd
  3. Oct 12, 2011 #2
    Yes,1 is the best possible. Try the functions f_n(x)=x^n.
  4. Oct 12, 2011 #3
    I see. Thanks a lot.
  5. Oct 12, 2011 #4
    Let's see if I got it.

    Suppose I want to find the best constant for the inequality

    [tex] \int_0^\mu f(x) dx \le K \int_0^1 f(x) dx,[/tex]

    where [itex]f(x) \in C^1(0,1)[/itex], [itex]f(0) = 0[/itex], [itex]f(x) \ge 0[/itex], and [itex]0 \le \mu \le 1[/itex].


    [tex]f_n(x) = \begin{cases} \frac{n+2}{n+3} x(2\mu -x), &0 < x \le \mu, \\
    \frac{n+2}{n+3} \mu^2 \mbox{sech} [(n+2)(\mu-x)], &\mu < x < 1. \end{cases}[/tex]

    If [itex] s_n = \int_0^\mu f(x) dx[/itex], and [itex]S_n = \int_0^1 f(x) dx[/itex], then

    [tex]\bigl\{s_n\bigr\} \nearrow \frac{2 \mu^3}{3}, \mbox{ and }\bigl\{S_n\bigr\} \searrow \frac{2 \mu^3}{3},[/tex]

    so the best constant is [itex]K = 1[/itex].

    Is the proof correct?
    Last edited: Oct 12, 2011
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