Using Cauchy Schwartz Inequality (for Integrals)

[SP90]

Homework Statement

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

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

Homework Equations

Cauchy Schwartz: |(a,b)| \leq ||a|| \cdot ||b||

The Attempt at a Solution

Taking CS on L^{2} gives us

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

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

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

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

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

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

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.

 

[sunjin09]

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.