(New page: a) <math>|h(x)| = \left|\int{f(x-y)g(y)dy}\right| \le \int{|f(x-y)g(y)|dy} \le \left(\int{|f(x-y)|^p dy}\right)^{1/p}\left(\int{|g(y)|^q dy}\right)^{1/q}</math> by Holder's Inequality. Le...) |
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Since <math>f \in L^p</math> and <math>g \in L^q</math> we get <math>|h(x)| \le ||f||_p ||g||_q < \infty</math> | Since <math>f \in L^p</math> and <math>g \in L^q</math> we get <math>|h(x)| \le ||f||_p ||g||_q < \infty</math> | ||
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Latest revision as of 14:59, 22 July 2008
a) $ |h(x)| = \left|\int{f(x-y)g(y)dy}\right| \le \int{|f(x-y)g(y)|dy} \le \left(\int{|f(x-y)|^p dy}\right)^{1/p}\left(\int{|g(y)|^q dy}\right)^{1/q} $ by Holder's Inequality.
Let $ z = y - x $. Then $ \left(\int{|f(x-y)|^p dy}\right)^{1/p} \left(\int{|g(y)|^q}\right)^{1/q} = \left(\int{|f(-z)|^p dz}\right)^{1/p} ||g||_q = ||f||_p ||g||_q $ because $ \int{|f(-x)|} = \int{|f(x)|} $.
Since $ f \in L^p $ and $ g \in L^q $ we get $ |h(x)| \le ||f||_p ||g||_q < \infty $
