Using Cauchy Schwartz Inequality (for Integrals)

  • Thread starter SP90
  • Start date
  • #1
23
0

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.
 

Answers and Replies

  • #2
312
0
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.
 

Related Threads on Using Cauchy Schwartz Inequality (for Integrals)

  • Last Post
Replies
16
Views
7K
Replies
4
Views
637
  • Last Post
Replies
11
Views
1K
Replies
3
Views
6K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
8K
Replies
6
Views
3K
Replies
1
Views
1K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
14
Views
2K
Top