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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> | ||
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| + | 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| \leq \int|f_t(x-y)-f(x-y)| |g(y)| dy \leq ||f_t-f||_{p}||g||_{q}</math>. | ||
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| + | Since <math>f \in L^{p}, ||f_{t}-f||_{p} \rightarrow 0</math> as <math>t \rightarrow 0</math>. | ||
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| + | This prove <math>h</math> is continuous. | ||
Latest revision as of 15:22, 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| \leq \int|f_t(x-y)-f(x-y)| |g(y)| dy \leq ||f_t-f||_{p}||g||_{q} $.
Since $ f \in L^{p}, ||f_{t}-f||_{p} \rightarrow 0 $ as $ t \rightarrow 0 $.
This prove $ h $ is continuous.
