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a)<math>|h(x)| \leq (\int |f(x-y)|^p dy)^{1/p}(\int |g(y)|^q dy)^{1/q} = (\int |f(z)|^p dz)^{1/p}(\int |g(y)|^q dy)^{1/q} \leq ||f||_{p}||g||_{q}</math> | a)<math>|h(x)| \leq (\int |f(x-y)|^p dy)^{1/p}(\int |g(y)|^q dy)^{1/q} = (\int |f(z)|^p dz)^{1/p}(\int |g(y)|^q dy)^{1/q} \leq ||f||_{p}||g||_{q}</math> | ||
| − | b)Define <math>f_{t}(x)=f(x | + | b)Define <math>f_{t}(x)=f(x+t)\frac{}{}</math>, <math>h_{t}(x)=h(x+t)\frac{}{}</math>. |
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| + | We have | ||
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| + | <math>|h_{t}(x) - h(x)| = |\int[f_t(x-y)-f(x-y)]g(y)dy |</math> | ||
Revision as of 14:18, 22 July 2008
a)$ |h(x)| \leq (\int |f(x-y)|^p dy)^{1/p}(\int |g(y)|^q dy)^{1/q} = (\int |f(z)|^p dz)^{1/p}(\int |g(y)|^q dy)^{1/q} \leq ||f||_{p}||g||_{q} $
b)Define $ f_{t}(x)=f(x+t)\frac{}{} $, $ h_{t}(x)=h(x+t)\frac{}{} $.
We have
$ |h_{t}(x) - h(x)| = |\int[f_t(x-y)-f(x-y)]g(y)dy | $
