(Removing all content from page) |
|||
| Line 1: | Line 1: | ||
| + | '''Given:''' <math>f \in L^p, p \geq 1, \int_0^1 f(y)sin(xy) dy = 0</math> | ||
| + | '''Show:''' f=0 a.e. | ||
| + | |||
| + | '''Proof:''' Assume wlog that f(0)=0, and extend to the entire line: <math>f(x) = -f(-x), x\in [-1,0), f(x)=0, |x|>1</math>. | ||
| + | |||
| + | We recall some facts about Fourier transfroms: | ||
| + | |||
| + | We proved on a HW that <math>f, g \in L^1 \Rightarrow \widehat{f*g} = \hat{f}\hat{g}</math>. | ||
| + | |||
| + | This implies that <math>\hat{f} = 0 \Rightarrow f=0</math> a.e. | ||
| + | |||
| + | Let <math>g(x) = \sin(x)\chi_{[-1,1]}</math>. We observe that <math>f*g = 0</math>: | ||
| + | |||
| + | <math>(f*g)(x) = \int_{-1}^1 f(t)\sin(x-t)dt = \int_{-1}^1 f(t)[\sin(x)\cos(t) - \cos(x)\sin(t)] = 0 \ \forall x</math>, | ||
| + | |||
| + | since <math>\int_{-1}^1 f(t)\cos(t) = 0</math> since <math>f(t)\cos(t)</math> is odd, and | ||
| + | |||
| + | <math>\int_{-1}^1 f(t)\sin(t) dt = 2\int_{0}^1 f(t)\sin(t) dt = 0</math> by assumption. | ||
| + | |||
| + | This implies that <math>\hat{f}\hat{g} = 0 \Rightarrow \hat{f} = 0 \Rightarrow f=0</math> a.e. | ||
Revision as of 09:47, 16 July 2008
Given: $ f \in L^p, p \geq 1, \int_0^1 f(y)sin(xy) dy = 0 $
Show: f=0 a.e.
Proof: Assume wlog that f(0)=0, and extend to the entire line: $ f(x) = -f(-x), x\in [-1,0), f(x)=0, |x|>1 $.
We recall some facts about Fourier transfroms:
We proved on a HW that $ f, g \in L^1 \Rightarrow \widehat{f*g} = \hat{f}\hat{g} $.
This implies that $ \hat{f} = 0 \Rightarrow f=0 $ a.e.
Let $ g(x) = \sin(x)\chi_{[-1,1]} $. We observe that $ f*g = 0 $:
$ (f*g)(x) = \int_{-1}^1 f(t)\sin(x-t)dt = \int_{-1}^1 f(t)[\sin(x)\cos(t) - \cos(x)\sin(t)] = 0 \ \forall x $,
since $ \int_{-1}^1 f(t)\cos(t) = 0 $ since $ f(t)\cos(t) $ is odd, and
$ \int_{-1}^1 f(t)\sin(t) dt = 2\int_{0}^1 f(t)\sin(t) dt = 0 $ by assumption.
This implies that $ \hat{f}\hat{g} = 0 \Rightarrow \hat{f} = 0 \Rightarrow f=0 $ a.e.
