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## Homework Statement

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]

## Homework Equations

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

## 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.