(New page: '''Lemma 1:''' Assume <math>f \in L^1</math>. Then <math>\lim_{p \rightarrow 0} \int_X |f|^p d\mu = \mu (\left\{x \in X|f(x) \neq 0 \right\})</math> '''Proof:''' We have <math>\lim_{p ...) |
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'''Lemma 2:''' If <math>f \in L^1(X), f>0 </math> a.e., <math>\psi(t) := \int_X |f|^td\mu </math> is differentiable <math>\forall \ t \in (0,\frac{1}{2}) </math>, and <math>\psi'(t) = \int_X |f|^t \log|f| d\mu</math>. | '''Lemma 2:''' If <math>f \in L^1(X), f>0 </math> a.e., <math>\psi(t) := \int_X |f|^td\mu </math> is differentiable <math>\forall \ t \in (0,\frac{1}{2}) </math>, and <math>\psi'(t) = \int_X |f|^t \log|f| d\mu</math>. | ||
| − | '''Proof:''' <math>\lim_{h \rightarrow 0} \frac{f(t+h)-f(t)}{h} = \lim_{h \rightarrow 0} \int_X \frac{|f|^{t+h}-|f|^t}{h} d\mu = \lim_{h \rightarrow 0} \int_X |f|^\xi \log|f|d\mu</math>, for some <math>\xi\in(t,t+h)</math>, by the Mean Value Theorem. For <math>|f|>1, |f| > |f|^{2\xi} > |f|^\xi(|f|^\xi-1) \geq |f|^\xi log|f| \downarrow |f|^tlog|f| | + | '''Proof:''' <math>\lim_{h \rightarrow 0} \frac{f(t+h)-f(t)}{h} = \lim_{h \rightarrow 0} \int_X \frac{|f|^{t+h}-|f|^t}{h} d\mu = \lim_{h \rightarrow 0} \int_X |f|^\xi \log|f|d\mu</math>, for some <math>\xi\in(t,t+h)</math>, by the Mean Value Theorem. For <math>|f|>1, \frac{|f|}{t} > \frac{|f|}{\xi} > \frac{|f|^{2\xi}}{\xi} > \frac{|f|^\xi(|f|^\xi-1)}{\xi} \geq \frac{|f|^\xi log(|f|^\xi)}{\xi} = |f|^\xi \log{|f|} \downarrow |f|^tlog|f|</math>, and for <math>0<|f|\leq1, 0 \geq |f|^\xi \log|f| \downarrow |f|^tlog|f|</math>, so two applications of MCT allow us to pass the limit inside the integral, yielding the result. <math>\square</math>. |
'''Corollary 3:''' Under the assumptions of Lemma 2, <math>\varphi(t) := \log(\psi(t))</math> is differentiable <math>\forall t \in (0,1)</math>, and <math>\varphi'(t) = ||f||_t^{-t}\int_X |f|^t \log|f| d\mu</math> | '''Corollary 3:''' Under the assumptions of Lemma 2, <math>\varphi(t) := \log(\psi(t))</math> is differentiable <math>\forall t \in (0,1)</math>, and <math>\varphi'(t) = ||f||_t^{-t}\int_X |f|^t \log|f| d\mu</math> | ||
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By the Mean Value Theorem, <math>\lim_{p \rightarrow 0} \frac{\varphi(p)}{p} = \lim_{p\rightarrow 0}\varphi'(q) = \lim_{p\rightarrow 0}||f||_q^{-q}\int_X |f|^q \log|f| d\mu</math>, for some <math>q \in (0,p)</math>. As <math>p \rightarrow 0, q\rightarrow 0</math>, so we can take the limit on the right. | By the Mean Value Theorem, <math>\lim_{p \rightarrow 0} \frac{\varphi(p)}{p} = \lim_{p\rightarrow 0}\varphi'(q) = \lim_{p\rightarrow 0}||f||_q^{-q}\int_X |f|^q \log|f| d\mu</math>, for some <math>q \in (0,p)</math>. As <math>p \rightarrow 0, q\rightarrow 0</math>, so we can take the limit on the right. | ||
| − | We have <math>\lim_{q \rightarrow 0} \int_{f \leq 1} |f|^q \log|f| d\mu = \int_{f \leq 1} \log|f| d\mu </math> by MCT, since <math> 0 > |f|^q \log|f| \downarrow log|f| </math>, and <math>\lim_{q \rightarrow 0} \int_{|f| > 1} |f|^q \log|f| d\mu = \int_{f > 1} \log|f| d\mu </math> by DCT, since <math> | + | We have <math>\lim_{q \rightarrow 0} \int_{f \leq 1} |f|^q \log|f| d\mu = \int_{f \leq 1} \log|f| d\mu </math> by MCT, since <math> 0 > |f|^q \log|f| \downarrow log|f| </math>, and <math>\lim_{q \rightarrow 0} \int_{|f| > 1} |f|^q \log|f| d\mu = \int_{f > 1} \log|f| d\mu </math> by DCT, since for <math>q<\frac{1}{2}, 0 < |f|^q\log|f| < |f|^{\frac{1}{2}}\log|f| < 2|f|^{\frac{1}{2}}(|f|^{\frac{1}{2}}-1)<2|f| \in L^1</math>. Summing yields the result, since we showed <math>\int_{f > 1} \log|f| d\mu < \infty</math> |
By Lemma 1, <math>\lim_{p \rightarrow 0} ||f||_p^{-p} = 1</math>. | By Lemma 1, <math>\lim_{p \rightarrow 0} ||f||_p^{-p} = 1</math>. | ||
Latest revision as of 11:54, 10 July 2008
Lemma 1: Assume $ f \in L^1 $. Then $ \lim_{p \rightarrow 0} \int_X |f|^p d\mu = \mu (\left\{x \in X|f(x) \neq 0 \right\}) $
Proof: We have $ \lim_{p \rightarrow 0} \int_X |f|^p d\mu = \lim_{p \rightarrow 0} \int\limits_{\left\{|f|>1\right\}} |f|^p d\mu + \lim_{p \rightarrow 0} \int\limits_{\left\{0<|f| \leq 1\right\}} |f|^p d\mu $.
Now $ \lim_{p \rightarrow 0} \int\limits_{\left\{0 < |f| \leq 1\right\}} |f|^p d\mu = \mu (\left\{x \in X|0<f(x) \leq 1 \right\}) $ by Bounded Convergence Theorem, since $ 0 \leq |f|^p \leq 1 \in L^1 $ when $ |f| \leq 1 $, and $ \mu(X) < \infty $
Similarly, $ \lim_{p \rightarrow 0} \int\limits_{\left\{|f|>1\right\}} |f|^p d\mu = \mu (\left\{x \in X|f(x) > 1 \right\}) $ by MCT, since we're bounded above by $ |f| \in L^1 $, for $ p<1 $. Adding yields the result. $ \square $
Lemma 2: If $ f \in L^1(X), f>0 $ a.e., $ \psi(t) := \int_X |f|^td\mu $ is differentiable $ \forall \ t \in (0,\frac{1}{2}) $, and $ \psi'(t) = \int_X |f|^t \log|f| d\mu $.
Proof: $ \lim_{h \rightarrow 0} \frac{f(t+h)-f(t)}{h} = \lim_{h \rightarrow 0} \int_X \frac{|f|^{t+h}-|f|^t}{h} d\mu = \lim_{h \rightarrow 0} \int_X |f|^\xi \log|f|d\mu $, for some $ \xi\in(t,t+h) $, by the Mean Value Theorem. For $ |f|>1, \frac{|f|}{t} > \frac{|f|}{\xi} > \frac{|f|^{2\xi}}{\xi} > \frac{|f|^\xi(|f|^\xi-1)}{\xi} \geq \frac{|f|^\xi log(|f|^\xi)}{\xi} = |f|^\xi \log{|f|} \downarrow |f|^tlog|f| $, and for $ 0<|f|\leq1, 0 \geq |f|^\xi \log|f| \downarrow |f|^tlog|f| $, so two applications of MCT allow us to pass the limit inside the integral, yielding the result. $ \square $.
Corollary 3: Under the assumptions of Lemma 2, $ \varphi(t) := \log(\psi(t)) $ is differentiable $ \forall t \in (0,1) $, and $ \varphi'(t) = ||f||_t^{-t}\int_X |f|^t \log|f| d\mu $
Proof: Chain rule.
Main Result: Assume $ f>0 $ a.e.
$ \lim_{p \rightarrow 0} \log||f||_p = \lim_{p \rightarrow 0} \frac{\varphi(p)}{p} = \lim_{p \rightarrow 0} \frac{\varphi(p)-\varphi(0)}{p} $, since $ |f|>0 a.e. \Rightarrow |f|^0 = 1 \ a.e. \Rightarrow \varphi(0)=\log(\int_X d\mu) = 0 $, since $ \mu(X)=1 $.
By the Mean Value Theorem, $ \lim_{p \rightarrow 0} \frac{\varphi(p)}{p} = \lim_{p\rightarrow 0}\varphi'(q) = \lim_{p\rightarrow 0}||f||_q^{-q}\int_X |f|^q \log|f| d\mu $, for some $ q \in (0,p) $. As $ p \rightarrow 0, q\rightarrow 0 $, so we can take the limit on the right.
We have $ \lim_{q \rightarrow 0} \int_{f \leq 1} |f|^q \log|f| d\mu = \int_{f \leq 1} \log|f| d\mu $ by MCT, since $ 0 > |f|^q \log|f| \downarrow log|f| $, and $ \lim_{q \rightarrow 0} \int_{|f| > 1} |f|^q \log|f| d\mu = \int_{f > 1} \log|f| d\mu $ by DCT, since for $ q<\frac{1}{2}, 0 < |f|^q\log|f| < |f|^{\frac{1}{2}}\log|f| < 2|f|^{\frac{1}{2}}(|f|^{\frac{1}{2}}-1)<2|f| \in L^1 $. Summing yields the result, since we showed $ \int_{f > 1} \log|f| d\mu < \infty $
By Lemma 1, $ \lim_{p \rightarrow 0} ||f||_p^{-p} = 1 $.
Combining these yields $ \lim_{p \rightarrow 0} \log||f||_p = \int_X \log|f| d\mu $. Finally,
$ \exp(\int_X \log|f| d\mu) = \exp (\lim_{p \rightarrow 0} \log||f||_p)= \lim_{p \rightarrow 0} ||f||_p $ by continuity of the exponential function. This proves the statement if $ |f|>0 $ a.e.
General Case: If f = 0 a.e. the statement is trivially true, using the convention $ \log(0) = -\infty, e^{-\infty} = 0 $.
Now let A denote the set where $ |f| = 0, 0 \leq \mu(A) < 1 $. We define a new measure $ \nu = \frac{\mu}{\mu(A^c)} $. Under this measure and restricting our attention to $ A^c $, $ A^c $ is a probability space, $ |f|>0 $ a.e. on $ A^c $, so we reduce to the special case considered above. As mentioned before, the statement is true trivially on A. This concludes the proof. $ \square $
