Difference between revisions of "8.3 Old Kiwi" - Rhea

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-'''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.

Latest revision as of 10:53, 16 July 2008