Difference between revisions of "8.3 OldKiwi"

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(New page: '''Given:''' <math>f \in L^p([0,1]), p \geq 1, \int_0^1 f(y)\sin{(xy)}dy = 0 \ \forall x \in [0,1]</math> '''Show:''' <math>f=0</math> a.e. '''Proof:''' Since <math>|[0,1]| < \infty, f\i...)

Latest revision as of 09:53, 16 July 2008 (view source)

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~~'''Given:''' <math>f \in L^p([0,1]), p \geq 1, \int_0^1 f(y)\sin{(xy)}dy = 0 \ \forall x \in [0,1]</math>~~

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~~'''Show:''' <math>f=0</math> a.e.~~

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'''Proof:''' Since <math>|[0,1]| < \infty, f\in L^1</math>. By the mean value theorem and DCT, <math>g(x) = \int_0^1 f(y)\sin{(xy)}dy</math> is differentiable with respect to x, and <math>g'(x) = 0 \ \forall x \in (0,1)</math>. Repeating the process, we see that g has derivatives of all orders, all of which are identically zero. Specifically,

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~~<math>\int_0^1 y^{2k}f(y)\sin{(xy)}dy = 0, \ k=0,1,..., \forall x \in (0,1)</math>.~~

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By the uniform continuity of sin(xy), we can extend this to <math>[0,1]</math>, and <math>\int_0^1 y^{2k}f(y)\sin{(y)}dy = 0</math>. Since we have that <math>f(y)\sin{y} \in L^1</math>, Lusin implies given <math>\epsilon > 0 \ \exists F \subset [0,1], |[0,1]-F|<\epsilon</math>, with F compact and <math>f(y)\sin{(y)}</math> continuous relative to F. By Stone-Weierstrass, there exists a sequence of even polynomials such that <math>P_n(x) \rightarrow f(y)\sin{(y)}</math> uniformly on F. (We can choose these polynomials to be even by extending <math>f(y)\sin{y}</math> by even reflection, as in the proof of the Weierstrass Theorem). We obtain from the UCT that

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<math> 0 = \int_F P_n(x)f(y)\sin{(y)}dy = \lim_{n \rightarrow \infty} \int_F P_n(x)f(y)\sin{(y)}dy = \int_F (f(y)\sin{y})^2dy \Rightarrow f(y)\sin{(y)} = 0</math> a.e. on F <math> \Rightarrow f=0 </math> a.e. on F.

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We now consider the sequence of functions <math>f_n = f(y)\chi_{F_{\frac{1}{n}}}</math>, where <math>F_{\frac{1}{n}}</math> is the set obtained above by setting <math>\epsilon = \frac{1}{n}</math>. We have that <math>f_n \rightarrow f</math> in measure, hence a subsequence converges to f uniformly. This implies that <math>f=0</math> a.e.

Difference between revisions of "8.3 OldKiwi" - Rhea

Latest revision as of 09:53, 16 July 2008

Alumni Liaison

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-'''Given:''' <math>f \in L^p([0,1]), p \geq 1, \int_0^1 f(y)\sin{(xy)}dy = 0 \ \forall x \in [0,1]</math>
-'''Show:''' <math>f=0</math> a.e.
-'''Proof:''' Since <math>|[0,1]| < \infty, f\in L^1</math>.  By the mean value theorem and DCT, <math>g(x) = \int_0^1 f(y)\sin{(xy)}dy</math> is differentiable with respect to x, and <math>g'(x) = 0 \ \forall x \in (0,1)</math>.  Repeating the process, we see that g has derivatives of all orders, all of which are identically zero.  Specifically,
-<math>\int_0^1 y^{2k}f(y)\sin{(xy)}dy = 0, \ k=0,1,..., \forall x \in (0,1)</math>.
-By the uniform continuity of sin(xy), we can extend this to <math>[0,1]</math>, and <math>\int_0^1 y^{2k}f(y)\sin{(y)}dy = 0</math>.  Since we have that <math>f(y)\sin{y} \in L^1</math>, Lusin implies given <math>\epsilon > 0 \ \exists F \subset [0,1], |[0,1]-F|<\epsilon</math>, with F compact and <math>f(y)\sin{(y)}</math> continuous relative to F.  By Stone-Weierstrass, there exists a sequence of even polynomials such that <math>P_n(x) \rightarrow f(y)\sin{(y)}</math> uniformly on F.  (We can choose these polynomials to be even by extending <math>f(y)\sin{y}</math> by even reflection, as in the proof of the Weierstrass Theorem).  We obtain from the UCT that
-<math> 0 = \int_F P_n(x)f(y)\sin{(y)}dy = \lim_{n \rightarrow \infty} \int_F P_n(x)f(y)\sin{(y)}dy = \int_F (f(y)\sin{y})^2dy \Rightarrow f(y)\sin{(y)} = 0</math> a.e. on F <math> \Rightarrow f=0 </math> a.e. on F.
-We now consider the sequence of functions <math>f_n = f(y)\chi_{F_{\frac{1}{n}}}</math>, where <math>F_{\frac{1}{n}}</math> is the set obtained above by setting <math>\epsilon = \frac{1}{n}</math>.  We have that <math>f_n \rightarrow f</math> in measure, hence a subsequence converges to f uniformly.  This implies that <math>f=0</math> a.e.