| Line 5: | Line 5: | ||
---- | ---- | ||
| − | + | <math>f*g = g*f</math> so WOLOG take <math>p<\infty</math> | |
Let <math>\epsilon>0</math> | Let <math>\epsilon>0</math> | ||
| Line 11: | Line 11: | ||
<math>\exists h \in C_{0}(\mathbb{R}^n)</math> s.t. <math>\left|\left|f-h\right|\right|_{p}<\epsilon</math> | <math>\exists h \in C_{0}(\mathbb{R}^n)</math> s.t. <math>\left|\left|f-h\right|\right|_{p}<\epsilon</math> | ||
| − | <math>h(x)</math> has compact support, so it is uniformly continuous and <math>\exists r</math> s.t. <math>h(x) = 0 \forall x, |x|>r</math>. Uniform continuity implies <math>\exists\delta>0</math> s.t. <math>\left|x-x'\right|<\delta\Rightarrow\left|h(x)-h(x')\right|<\epsilon</math> Using this <math>\delta</math>, let <math>\left|x-x'\right|<\delta</math> | + | <math>h(x)</math> has compact support, so it is uniformly continuous and <math>\exists r</math> s.t. <math>h(x) = 0 \forall x, |x|>r</math>. Uniform continuity implies <math>\exists\delta>0</math> s.t. <math>\left|x-x'\right|<\delta\Rightarrow\left|h(x)-h(x')\right|<\frac{\epsilon}{\mu(|x|\le r)^{1/p}}</math>. |
| + | |||
| + | Using this <math>\delta</math>, let <math>\left|x-x'\right|<\delta</math>. | ||
<math>\left|(f*g)(x)-(f*g)(x')\right| = </math> | <math>\left|(f*g)(x)-(f*g)(x')\right| = </math> | ||
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<math> \le (\left|\left|(f(x-y)-h(x-y))\right|\right|_{p}+\left|\left|(h(x-y)-h(x'-y))\right|\right|_{p}+\left|\left|(h(x'-y)-f(x'-y))\right|\right|_{p}) \left|\left|g\right|\right|_{q}</math> by Minkowski's Inequality | <math> \le (\left|\left|(f(x-y)-h(x-y))\right|\right|_{p}+\left|\left|(h(x-y)-h(x'-y))\right|\right|_{p}+\left|\left|(h(x'-y)-f(x'-y))\right|\right|_{p}) \left|\left|g\right|\right|_{q}</math> by Minkowski's Inequality | ||
| − | <math>\left|\left|(h(x-y)-h(x'-y))\right|\right|_{p} = \left(\int_{|x|\le r}\left|(h(x-y)-h(x'-y))\right|^{p}dx \right)^\frac{1}{p} < \left(\int_{|x|\le r}\epsilon^{p}dx \right)^\frac{1}{p} = | + | <math>\left|\left|(h(x-y)-h(x'-y))\right|\right|_{p} = \left(\int_{|x|\le r}\left|(h(x-y)-h(x'-y))\right|^{p}dx \right)^\frac{1}{p} < \left(\int_{|x|\le r}\frac{\epsilon^{p}}{\mu(|x|\le r)}dx \right)^\frac{1}{p} =\epsilon</math> |
| − | <math> < (\epsilon + | + | <math> < (\epsilon + \epsilon + \epsilon)\left|\left|g\right|\right|_{q}</math> |
Revision as of 12:43, 29 July 2009
Slaughter a horde of pirates to get back to The_Ninja's_Solutions
Prove that $ *:L^{p}(\mathbb{R}^n)\times L^{q}(\mathbb{R}^n)\rightarrow C(\mathbb{R}^n) $ is well defined, if $ 1/p+1/q=1, 1\le p\le\infty $
$ f*g = g*f $ so WOLOG take $ p<\infty $
Let $ \epsilon>0 $
$ \exists h \in C_{0}(\mathbb{R}^n) $ s.t. $ \left|\left|f-h\right|\right|_{p}<\epsilon $
$ h(x) $ has compact support, so it is uniformly continuous and $ \exists r $ s.t. $ h(x) = 0 \forall x, |x|>r $. Uniform continuity implies $ \exists\delta>0 $ s.t. $ \left|x-x'\right|<\delta\Rightarrow\left|h(x)-h(x')\right|<\frac{\epsilon}{\mu(|x|\le r)^{1/p}} $.
Using this $ \delta $, let $ \left|x-x'\right|<\delta $.
$ \left|(f*g)(x)-(f*g)(x')\right| = $
$ = \left|\int_{\mathbb{R}^n}f(x-y)g(y)dy-\int_{\mathbb{R}^n}f(x'-y)g(y)dy\right| $
$ = \left|\int_{\mathbb{R}^n}(f(x-y)-f(x'-y))g(y)dy\right| $
$ \le \left|\left|(f(x-y)-f(x'-y))\right|\right|_{p} \left|\left|g\right|\right|_{q} $ by Holder's Inequality
$ \le (\left|\left|(f(x-y)-h(x-y))\right|\right|_{p}+\left|\left|(h(x-y)-h(x'-y))\right|\right|_{p}+\left|\left|(h(x'-y)-f(x'-y))\right|\right|_{p}) \left|\left|g\right|\right|_{q} $ by Minkowski's Inequality
$ \left|\left|(h(x-y)-h(x'-y))\right|\right|_{p} = \left(\int_{|x|\le r}\left|(h(x-y)-h(x'-y))\right|^{p}dx \right)^\frac{1}{p} < \left(\int_{|x|\le r}\frac{\epsilon^{p}}{\mu(|x|\le r)}dx \right)^\frac{1}{p} =\epsilon $
$ < (\epsilon + \epsilon + \epsilon)\left|\left|g\right|\right|_{q} $
