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| + | a) Prove that <math>\lim_{n\rightarrow\infty} \|f\|_p = \|f\|_{\infty}</math> | ||
| + | Let <math>M = \|f\|_{\infty}</math> | ||
| + | If <math>M^{\prime} < M</math>, then <math>\omega(M') = \left|{x\mid f(x) > M^{\prime}}\right| = |A|>0.</math> | ||
| + | |||
| + | <math>\|f\|_p \geq \left(\int_A|f|^p\right)^{\frac{1}{p}} \geq M^{\prime}|A|^{\frac{1}{p}}</math> | ||
| + | |||
| + | As <math>p \rightarrow \infty, |A|^{\frac{1}{p}} \rightarrow 1 | ||
| + | |||
| + | \Rightarrow \liminf_{p\rightarrow\infty}\|f\|_p \geq M^{\prime}</math> | ||
| + | so <math>\liminf_{p\rightarrow\infty}\|f\|_p \geq M</math> | ||
| + | |||
| + | <math>\|f\|_p \leq \left(\int_X M^p\right)^{\frac{1}{p}} = M|X|^{\frac{1}{p}}</math>, where <math>X</math> is our probability space. | ||
| + | |||
| + | <math>\limsup_{p \rightarrow \infty}\|f\|_p \leq M</math> | ||
| + | |||
| + | <math>\lim_{p \rightarrow \infty}\|f\|_p = M</math> | ||
| + | |||
| + | |||
| + | b) Prove that <math>\lim_{n\rightarrow\infty} \frac{\int_X|f|^{n+1}d\mu}{\int_X|f|^nd\mu} = \|f\|_{\infty}</math> | ||
| + | |||
| + | <math>\int_X|f|^{n+1} \leq \|f\|_{\infty}\int_X|f|^n</math>, by Holder's Inequality | ||
| + | |||
| + | <math>\Rightarrow | ||
| + | \frac{\int_X|f|^{n+1}}{\int_X|f|^{n}} \leq \|f\|_{\infty} | ||
| + | |||
| + | \limsup{\frac{\int_X|f|^{n+1}}{\int_X|f|^{n}}} \leq \|f\|_{\infty}</math> | ||
| + | |||
| + | Consider <math>\varphi(t)= t^{\frac{n+1}{n}}</math> and notice it is convex so | ||
| + | <math>\frac{\int_X|f|^{n+1}}{\int_X|f|^{n}} \geq \frac{\left(\int_X|f|^{n}\right)^{\frac{n+1}{n}}}{\int_X|f|^{n}} = \left(\int_X|f|^{n}\right)^{\frac{1}{n}} \nearrow \|f\|_{\infty}</math> | ||
| + | |||
| + | <math>\liminf\frac{\int_X|f|^{n+1}}{\int_X|f|^{n}} \geq \|f\|_{\infty}</math> | ||
| + | |||
| + | Thus <math>\lim_{n\rightarrow\infty} \frac{\int_X|f|^{n+1}d\mu}{\int_X|f|^nd\mu} = \|f\|_{\infty}</math> | ||
| + | |||
| + | c) Are the results above true for any finite measure space? For any measure space? | ||
| + | |||
| + | Any finite measure space is equivalent to a probability space hence the above parts hold. However, on an infinite measure space consider the constant function <math>f(x)=c</math>. We find that <math>\|f\|_{\infty} =c</math>, but <math>f(x) \notin L^p</math> for <math>0 \leq p < \infty</math> | ||
Latest revision as of 13:00, 11 July 2008
a) Prove that $ \lim_{n\rightarrow\infty} \|f\|_p = \|f\|_{\infty} $
Let $ M = \|f\|_{\infty} $ If $ M^{\prime} < M $, then $ \omega(M') = \left|{x\mid f(x) > M^{\prime}}\right| = |A|>0. $
$ \|f\|_p \geq \left(\int_A|f|^p\right)^{\frac{1}{p}} \geq M^{\prime}|A|^{\frac{1}{p}} $
As $ p \rightarrow \infty, |A|^{\frac{1}{p}} \rightarrow 1 \Rightarrow \liminf_{p\rightarrow\infty}\|f\|_p \geq M^{\prime} $ so $ \liminf_{p\rightarrow\infty}\|f\|_p \geq M $
$ \|f\|_p \leq \left(\int_X M^p\right)^{\frac{1}{p}} = M|X|^{\frac{1}{p}} $, where $ X $ is our probability space.
$ \limsup_{p \rightarrow \infty}\|f\|_p \leq M $
$ \lim_{p \rightarrow \infty}\|f\|_p = M $
b) Prove that $ \lim_{n\rightarrow\infty} \frac{\int_X|f|^{n+1}d\mu}{\int_X|f|^nd\mu} = \|f\|_{\infty} $
$ \int_X|f|^{n+1} \leq \|f\|_{\infty}\int_X|f|^n $, by Holder's Inequality
$ \Rightarrow \frac{\int_X|f|^{n+1}}{\int_X|f|^{n}} \leq \|f\|_{\infty} \limsup{\frac{\int_X|f|^{n+1}}{\int_X|f|^{n}}} \leq \|f\|_{\infty} $
Consider $ \varphi(t)= t^{\frac{n+1}{n}} $ and notice it is convex so $ \frac{\int_X|f|^{n+1}}{\int_X|f|^{n}} \geq \frac{\left(\int_X|f|^{n}\right)^{\frac{n+1}{n}}}{\int_X|f|^{n}} = \left(\int_X|f|^{n}\right)^{\frac{1}{n}} \nearrow \|f\|_{\infty} $
$ \liminf\frac{\int_X|f|^{n+1}}{\int_X|f|^{n}} \geq \|f\|_{\infty} $
Thus $ \lim_{n\rightarrow\infty} \frac{\int_X|f|^{n+1}d\mu}{\int_X|f|^nd\mu} = \|f\|_{\infty} $
c) Are the results above true for any finite measure space? For any measure space?
Any finite measure space is equivalent to a probability space hence the above parts hold. However, on an infinite measure space consider the constant function $ f(x)=c $. We find that $ \|f\|_{\infty} =c $, but $ f(x) \notin L^p $ for $ 0 \leq p < \infty $
