# Measure Theory/L^p spaces

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

## Jensen's inequality

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

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

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

${\displaystyle \displaystyle \phi \left(\int _{X}fd\mu \right)\leq \int _{X}\phi \circ fd\mu }$

Proof

Let ${\displaystyle t=\displaystyle \int _{X}fd\mu }$. As ${\displaystyle \mu }$ is a probability measure, ${\displaystyle a

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

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

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

Put ${\displaystyle s=f(x)}$

${\displaystyle \phi \left(\int _{X}fd\mu \right)-\phi f(x)\leq \beta \left(f(x)-\int _{X}fd\mu \right)}$, which completes the proof.

### Corollary

1. Putting ${\displaystyle \phi (x)=e^{x}}$,
${\displaystyle \displaystyle e^{(\int _{X}fd\mu )}\leq \int _{X}e^{f}d\mu }$
1. If ${\displaystyle X}$ is finite, ${\displaystyle \mu }$ is a counting measure, and if ${\displaystyle f(x_{i})=p_{i}}$, then
${\displaystyle \displaystyle e^{\left({\frac {p_{1}+\ldots +p_{n}}{n}}\right)}\leq {\frac {1}{n}}\left(e^{p_{1}}+e^{p_{2}}+\ldots +e^{p_{n}}\right)}$

For every ${\displaystyle f\in {\mathcal {L}}^{p}}$, define ${\displaystyle \|f\|_{p}=\left(\int _{X}|f|^{p}d\mu \right)^{\frac {1}{p}}}$

## Holder's inequality

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

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

${\displaystyle \|fg\|\leq \|f\|_{p}\|g\|_{q}}$

Proof

We know that ${\displaystyle \log }$ is a concave function

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

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

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

${\displaystyle \displaystyle {\frac {|f|}{\|f\|_{p}}}{\frac {|g|}{\|g\|_{q}}}\leq {\frac {1}{p}}{\frac {|f|^{p}}{\|f\|_{p}^{p}}}+{\frac {1}{q}}{\frac {|g|^{q}}{\|g\|_{q}^{q}}}}$

Then, ${\displaystyle \displaystyle {\frac {1}{\|f\|_{p}\|g\|_{q}}}\int _{X}|f||g|d\mu \leq {\frac {1}{p\|f\|_{p}^{p}}}\int _{X}|f|^{p}d\mu +{\frac {1}{p\|g\|_{q}^{q}}}\int _{X}|g|^{q}d\mu =1}$,

which proves the result

### Corollary

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

Proof

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

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

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

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

## Theorem

The ${\displaystyle L^{p}}$ space with the ${\displaystyle \|\cdot \|_{p}}$ norm is a normed linear space, that is,

1. ${\displaystyle \|f\|_{p}\geq 0}$ for every ${\displaystyle f\in L^{p}}$, further, ${\displaystyle \|f\|_{p}=0\iff f=0}$
2. ${\displaystyle \|\lambda \|_{p}=|\lambda |\|f\|_{p}}$
3. ${\displaystyle \|f+g\|_{p}\leq \|f\|_{p}+\|g\|_{p}}$ . . . (Minkowski's inequality)

Proof

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

${\displaystyle \displaystyle \int _{X}|f+g|^{p}d\mu =\int _{X}|f+g|^{p-1}|f+g|d\mu \leq \int _{X}|f+g|^{p-1}(|f|+|g|)d\mu }$

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

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

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

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

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