# Using Cauchy Schwartz Inequality (for Integrals)

## 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.

## Answers and Replies

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.