Measure Theory/L^p spaces

Recall that an ${ℒ}^p$ space is defined as ${ℒ}^p(X)=\{f:X\to ℂ:f\text{ is measurable,}\underset{X}{\int }|f{|}^{p}d\mu <\mathrm{\infty }\}$

Jensen's inequality

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

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

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

$\varphi \left(\underset{X}{\int }fd\mu \right)\le \underset{X}{\int }\varphi \circ fd\mu$

Proof

Let $t=\underset{X}{\int }fd\mu$. As $\mu$ is a probability measure, $a<t<b$

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

Let $t<u<b$; then $\beta \le \frac{\varphi (u)-\varphi (t)}{u-t}$

Thus, $\frac{\varphi (t)-\varphi (s)}{t-s}\le \frac{\varphi (u)-\varphi (t)}{u-t}$, that is $\varphi (t)-\varphi (s)\le \beta (t-s)$

Put $s=f(x)$

$\varphi \left(\underset{X}{\int }fd\mu \right)-\varphi f(x)\le \beta (f(x)-\underset{X}{\int }fd\mu )$, which completes the proof.

Corollary

1. Putting $\varphi (x)=e^x$,
   $e^{(\underset{X}{\int }fd\mu )}\le \underset{X}{\int }{e}^{f}d\mu$

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

For every $f\in {ℒ}^p$, define $\Vert f{\Vert }_p={(\underset{X}{\int }|f{|}^{p}d\mu )}^{\frac{1}{p}}$

Holder's inequality

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

Then, $fg\in {ℒ}^1$ and

$\Vert fg\Vert \le \Vert f{\Vert }_p\Vert g{\Vert }_q$

Proof

We know that $\log$ is a concave function

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

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

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

$\frac{|f|}{\Vert f{\Vert }_p}\frac{|g|}{\Vert g{\Vert }_q}\le \frac{1}{p}\frac{|f{|}^p}{\Vert f{\Vert }_p^p}+\frac{1}{q}\frac{|g{|}^q}{\Vert g{\Vert }_q^q}$

Then, $\frac{1}{\Vert f{\Vert }_p\Vert g{\Vert }_q}\underset{X}{\int }|f||g|d\mu \le \frac{1}{p\Vert f{\Vert }_p^p}\underset{X}{\int }|f{|}^{p}d\mu +\frac{1}{p\Vert g{\Vert }_q^q}\underset{X}{\int }|g{|}^{q}d\mu =1$,

which proves the result

Corollary

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

Proof

Let $\varphi \in {ℒ}^s$, $p=\frac{r}{s}\ge 1$, $g\equiv 1$

Then, $f=|\varphi {|}^s\in {ℒ}^1$, and hence $\underset{X}{\int }|\varphi {|}^{s}d\mu \le {\left(\underset{X}{\int }{(|\varphi {|}^s)}^{\frac{r}{s}}d\mu \right)}^{\frac{s}{r}}\mu (X{)}^{1-\frac{s}{r}}$

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

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

Theorem

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

1. $\Vert f{\Vert }_p\ge 0$ for every $f\in L^p$, further, $\Vert f{\Vert }_p=0⟺f=0$
2. $\Vert \lambda {\Vert }_p=|\lambda |\Vert f{\Vert }_p$
3. $\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$ . . . (Minkowski's inequality)

Proof

1. and 2. are clear, so we prove only 3. The cases $p=1$ and $p=\mathrm{\infty }$ (see below) are obvious, so assume that $0<p<\mathrm{\infty }$ 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 $p/q=p-1$:

$\underset{X}{\int }|f+g{|}^{p}d\mu =\underset{X}{\int }|f+g{|}^{p-1}|f+g|d\mu \le \underset{X}{\int }|f+g{|}^{p-1}(|f|+|g|)d\mu$

$\le {(\underset{X}{\int }|f+g{|}^{(p-1)q}d\mu )}^{\frac{1}{q}}\Vert f{\Vert }_p+{(\underset{X}{\int }|f+g{|}^{(p-1)q}d\mu )}^{\frac{1}{q}}\Vert g{\Vert }_p=\Vert f+g{\Vert }_p^{\frac{p}{q}}\Vert f{\Vert }_p+\Vert f+g{\Vert }_p^{\frac{p}{q}}\Vert g{\Vert }_p.$

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

$\frac{|f+g{|}^p}{2^p}={|\frac{f}{2}+\frac{g}{2}|}^p\le {(\frac{|f|}{2}+\frac{|g|}{2})}^p\le \frac{1}{2}|f{|}^p+\frac{1}{2}|g{|}^p.$

This shows that $\Vert f+g{\Vert }_p<\mathrm{\infty }$ so that we may divide by it in the previous calculation to obtain $\Vert f+g{\Vert }_p\le \Vert f{\Vert }_p+\Vert g{\Vert }_p$.

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

Template:BookCat