# Measure Theory/L^p spaces

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

## 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,
$\displaystyle \phi \left(\int _{X}fd\mu \right)\leq \int _{X}\phi \circ fd\mu$
Proof
Let $t=\displaystyle \int _{X}fd\mu$. 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 $s=f(x)$
$\phi \left(\int _{X}fd\mu \right)-\phi f(x)\leq \beta \left(f(x)-\int _{X}fd\mu \right)$, which completes the proof.

### Corollary

* Putting $\phi (x)=e^{x}$, $\displaystyle e^{(\int _{X}fd\mu )}\leq \int _{X}e^{f}d\mu$
* 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 $\|f\|_{p}=\left(\int _{X}|f|^{p}d\mu \right)^{\frac {1}{p}}$

## 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, $a^{t}b^{1-t}\leq ta+(1-t)b$
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, $\displaystyle {\frac {1}{\|f\|_{p}\|g\|_{q}}}\int _{X}|f||g|d\mu \leq {\frac {1}{p\|f\|_{p}^{p}}}\int _{X}|f|^{p}d\mu +{\frac {1}{p\|g\|_{q}^{q}}}\int _{X}|g|^{q}d\mu =1$,
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 $\displaystyle \int _{X}|\phi |^{s}d\mu \leq \left(\int _{X}\left(|\phi |^{s}\right)^{\frac {r}{s}}d\mu \right)^{\frac {s}{r}}\mu (X)^{1-{\frac {s}{r}}}$
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,
* $\|f\|_{p}\geq 0$ 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 $p/q=p-1$:
$\displaystyle \int _{X}|f+g|^{p}d\mu =\int _{X}|f+g|^{p-1}|f+g|d\mu \leq \int _{X}|f+g|^{p-1}(|f|+|g|)d\mu$
$\leq \displaystyle \left(\int _{X}|f+g|^{(p-1)q}d\mu \right)^{\frac {1}{q}}\|f\|_{p}+\left(\int _{X}|f+g|^{(p-1)q}d\mu \right)^{\frac {1}{q}}\|g\|_{p}=\|f+g\|_{p}^{\frac {p}{q}}\|f\|_{p}+\|f+g\|_{p}^{\frac {p}{q}}\|g\|_{p}.$
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 $\|f\|_{\infty }=\sup\{|f(x)|:x\notin E\}$