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| + | [[Category:MA598RSummer2009pweigel]] | ||
| + | [[Category:MA598]] | ||
| + | [[Category:math]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:real analysis]] | ||
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| + | == Problem #7.8, MA598R, Summer 2009, Weigel == | ||
Slaughter a horde of pirates to get back to [[The_Ninja%27s_Solutions]] | Slaughter a horde of pirates to get back to [[The_Ninja%27s_Solutions]] | ||
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---- | ---- | ||
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| + | <math>f*g = g*f</math> so WOLOG take <math>p<\infty</math> | ||
Let <math>\epsilon>0</math> | Let <math>\epsilon>0</math> | ||
<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|<\frac{\epsilon}{\mu(|x|\le r)^{1/p}}</math>. | ||
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| + | 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|\int_{\mathbb{R}^n}f(x-y)g(y)dy-\int_{\mathbb{R}^n}f(x'-y)g(y)dy\right| </math> | ||
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| + | <math> = \left|\int_{\mathbb{R}^n}(f(x-y)-f(x'-y))g(y)dy\right|</math> | ||
| + | |||
| + | <math> \le \left|\left|(f(x-y)-f(x'-y))\right|\right|_{p} \left|\left|g\right|\right|_{q}</math> by Holder'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 | ||
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| + | <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> | ||
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| + | <math> < (\epsilon + \epsilon + \epsilon)\left|\left|g\right|\right|_{q}</math> | ||
| + | |||
| + | <math>\epsilon</math> is arbitrary so <math>f*g</math> is continuous. | ||
| + | |||
| + | ~Ben Bartle | ||
| + | ---- | ||
| + | [[The_Ninja%27s_Solutions|Back to Ninja Solutions]] | ||
| + | |||
| + | [[MA_598R_pweigel_Summer_2009_Lecture_7|Back to Assignment 7]] | ||
| + | |||
| + | [[MA598R_%28WeigelSummer2009%29|Back to MA598R Summer 2009]] | ||
Latest revision as of 04:54, 11 June 2013
Problem #7.8, MA598R, Summer 2009, Weigel
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} $
$ \epsilon $ is arbitrary so $ f*g $ is continuous.
~Ben Bartle
