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Using Cauchy Schwartz Inequality (for Integrals)

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose [itex] \int_{-\infty}^{\infty}t|f(t)|dt < K[/itex]

    Using Cauchy-Schwartz Inequality, show that [itex]\int_{a}^{b} \leq K^{2}(log(b)-log(a))[/itex]

    2. Relevant equations

    Cauchy Schwartz: [itex]|(a,b)| \leq ||a|| \cdot ||b||[/itex]

    3. The attempt at a solution

    Taking CS on [itex]L^{2}[/itex] gives us

    [itex] (|\int_{a}^{b}h(t)g(t)dt|)^{2} \leq (\int_{a}^{b}|h(t)|^{2}dt)(\int_{a}^{b}|g(t)|^{2}dt)[/itex]

    Setting [itex]h(t)=\sqrt{t}|f(t)|[/itex] and [itex]g(t)=\frac{1}{\sqrt{t}}[/itex] we get

    [itex](\int_{a}^{b}|f(t)|dt)^{2} \leq (\int_{a}^{b}t|f(t)|^{2}dt)(\int_{a}^{b}\frac{1}{t}dt)[/itex]

    Obviously [itex]\int_{a}^{b}\frac{1}{t}dt[/itex] is [itex]log(b)-log(a)[/itex]

    So we just need to show [itex] \int_{a}^{b}t|f(t)|^{2}dt \leq K^2[/itex]

    Obviously this is true if [itex] \int_{a}^{b}t|f(t)|^{2}dt \leq (\int_{-\infty}^{\infty}t|f(t)|dt)^2[/itex]

    But this is where I get stuck. The limits aren't equal for these two and the Cauchy Schwartz inequality is the wrong way around for them to work.
     
  2. jcsd
  3. Apr 13, 2012 #2
    your problem statement is probably wrong, since if f(x) is even or odd, K can obviously approach 0, the inequality you want to prove need not hold in general.
     
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