Measure Theory/L^p spaces

< Measure Theory

Recall that an $\mathcal{L}^p$ space is defined as $\mathcal{L}^p(X)=\{f:X\to\mathbb{C}:f\text{ is measurable,}\int_X|f|^pd\mu<\infty\}$

Jensen's inequality

Let $(X,\Sigma,\mu)$ be a probability measure space.

Let $f:X\to\mathbb{R}$ , $f\in\mathcal{L}^1$ be such that there exist $a,b\in\mathbb{R}$ with $a<f(x)<b$

If $\phi$ is a convex function on $(a,b)$ then,

$\displaystyle\phi\left(\int_Xfd\mu\right)\leq\int_X\phi\circ fd\mu$

Proof

Let $t=\displaystyle\int_Xfd\mu$ . As $\mu$ is a probability measure, $a<t<b$

Let $\beta=\sup\{\frac{\phi(t)-\phi(s)}{t-s}:a<s<t<b\}$

Let $t<u<b$ ; then $\beta\leq\displaystyle\frac{\phi(u)-\phi(t)}{u-t}$

Thus, $\displaystyle\frac{\phi(t)-\phi(s)}{t-s}\leq\frac{\phi(u)-\phi(t)}{u-t}$ , that is $\phi(t)-\phi(s)\leq\beta(t-s)$

Put $s=f(x)$

$\phi\left(\int_Xfd\mu\right)-\phi f(x)+\beta\left(f(x)-\int_Xfd\mu\right)\geq 0$ , which completes the proof.

Corollary

Putting $\phi(x)=e^x$ ,
$\displaystyle e^{(\int_Xfd\mu)}\leq\int_Xe^fd\mu$

If $X$ is finite, $\mu$ is a counting measure, and if $f(x_i)=p_i$ , then
$\displaystyle e^{\left(\frac{p_1+\ldots+p_n}{n}\right)}\leq\frac{1}{n}\left(e^{p_1}+e^{p_2}+\ldots+e^{p_n}\right)$

For every $f\in\mathcal{L}^p$ , define $\|f\|_p=\left(\int_X|f|^pd\mu\right)^{\frac{1}{p}}$

Holder's inequality

Let $1<p,q<\infty$ such that $\displaystyle\frac{1}{p}+\frac{1}{q}=1$ . Let $f\in\mathcal{L}^p$ and $g\in\mathcal{L}^q$ .

Then, $fg\in\mathcal{L}^1$ and

$\|fg\|\leq\|f\|_p\|g\|_q$

Proof

We know that $\log$ is a concave function

Let $0\leq t\leq 1$ , $0<a<b$ . Then $t\log a+(1-t)\log b\leq \log(at+b(1-t))$

That is, $a^tb^{1-t}\leq ta+(1-t)b$

Let $t=\frac{1}{p}$ , $a=\left(\frac{|f|}{\|f\|_p}\right)^p$ , $b=\left(\frac{|f|}{\|f\|_q}\right)^q$

$\displaystyle\frac{|f|}{\|f\|_p}\frac{|g|}{\|g\|_q}\leq\frac{1}{p}\frac{|f|^p}{\|f\|^p_p}+\frac{1}{q}\frac{|g|^q}{\|g\|^q_q}$

Then, $\displaystyle\frac{1}{\|f\|_p\|g\|_q}\int_X|f||g|d\mu\leq\frac{1}{p\|f\|_p^p}\int_X|f|^pd\mu+\frac{1}{p\|g\|_q^q}\int_X|g|^qd\mu=1$ ,

which proves the result

Corollary

If $\mu(X)<\infty$ , $1<s<r<\infty$ then $\mathcal{L}^r\subset \mathcal{L}^s$

Proof

Let $\phi\in\mathcal{L}^s$ , $p=\frac{r}{s}\geq 1$ , $g\equiv 1$

Then, $f=|\phi|^s\in\mathcal{L}^1$ , and hence $\displaystyle\int_X|\phi|^sd\mu\leq\left(\int_X\left(|\phi|^s\right)^{\frac{r}{s}}d\mu\right)^{\frac{s}{r}}\mu(X)^{1-\frac{s}{r}}$

We say that if $f,g:X\to\mathbb{C}$ , $f=g$ almost everywhere on $X$ if $\mu(\{x|f(x)\neq g(x)\})=0$ . Observe that this is an equivalence relation on $\mathcal{L}^p$

If $(X,\Sigma,\mu)$ is a measure space, define the space $L^p$ to be the set of all equivalence classes of functions in $\mathcal{L}^p$

Theorem

The $L^p$ space with the $\|\cdot\|_p$ norm is a normed linear space, that is,

$\|f\|_p\geq0$ for every $f\in L^p$ , further, $\|f\|_p=0\iff f=0$
$\|\lambda\|_p=|\lambda|\|f\|_p$
$\|f+g\|_p\leq\|f\|_p+\|g\|_p$ . . . (Minkowski's inequality)

Proof

1. and 2. are clear, so we prove only 3. The cases $p=1$ and $p=\infty$ (see below) are obvious, so assume that $0<p<\infty$ and let $f,g\in L^p$ be given. Hölder's inequality yields the following, where $q$ is chosen such that $1/q+1/p = 1$ so that $p/q = p-1$ :

$\displaystyle\int_X|f+g|^pd\mu=\int_X|f+g|^{p-1}|f+g|d\mu\leq\int_X|f+g|^{p-1}(|f|+|g|)d\mu$

$\leq\displaystyle\left(\int_X|f+g|^{(p-1)q}d\mu\right)^\frac{1}{q}\|f\|_p+\left(\int_X|f+g|^{(p-1)q}d\mu\right)^{\frac{1}{q}}\|g\|_p=\|f+g\|_p^{\frac{p}{q}}\|f\|_p+\|f+g\|_p^{\frac{p}{q}}\|g\|_p.$

Moreover, as $t\mapsto t^p$ is convex for $p>1$ ,

$\displaystyle \frac{|f+g|^p}{2^p} = \left|\frac{f}{2}+\frac{g}{2}\right|^p\leq \left(\frac{|f|}{2}+\frac{|g|}{2}\right)^p \leq \frac{1}{2}|f|^p + \frac{1}{2}|g|^p.$

This shows that $\|f+g\|_p<\infty$ so that we may divide by it in the previous calculation to obtain $\|f+g\|_p\leq\|f\|_p+\|g\|_p$ .

Define the space $L^{\infty}=\{f|X\to\mathbb{C},f\text{ is bounded almost everywhere}\}$ . Further, for $f\in L^{\infty}$ define $\|f\|_{\infty}=\sup\{|f(x)|:x\notin E\}$

This article is issued from Wikibooks. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.