Lp spaces

Let $\mu$ be a positive measure on the set $E$, and $u:E\to \mathbb{C}$ a measurable function, then we can define

$$\Vert u{\Vert }_{L^p}:={\left({\int }_E|u{|}^{p}d\mu \right)}^{\frac{1}{p}},p\in (0,+\infty ),$$ $$\Vert u{\Vert }_{L^{\infty }}:=\inf \{C:|u|\le Ca.e.\}$$

For $p\in [1,+\infty ]$ these functions are a norm. The only non-trivial condition to check is the triangle inequality: Minkowsky’s inequality.

Also $\Vert u\Vert =0$ if $u=0$ a.e., so it’s a semi-norm. Just work with the equivalence classes, where $u\sim v$ iff $u=v$ a.e.

In the case of finite measure space, there is a nice inclusion property: Proposition Suppose $(X,\mathcal{A},\mu )$ is a finite measure space $(\mu (X)<\infty )$ and $0<p<q<\infty$. Then

$$\Vert u{\Vert }_p\le \mu (X{)}^{\frac{q-p}{pq}}\Vert u{\Vert }_q$$

in particular $L^q\subseteq L^p$.

Oss In a sense $L^{\infty }$ is the limit of $L^p$ when $p\to \infty$, for a finite measure space, since:

$$\Vert f{\Vert }_p\le \mu (X{)}^{\frac{1}{p}}\Vert f{\Vert }_{\infty }\to \Vert f{\Vert }_{\infty }\text{ when }p\to \infty$$

on the other hand, for any $ϵ>0$ the set

$$\mu (x\text{ s.t. }\Vert f{\Vert }_{\infty }-f(x)<ϵ)=\delta >0$$

has positive measure. This implies

$$\int |f{|}^{p}d\mu \ge \delta (\Vert f{\Vert }_{\infty }-ϵ{)}^p$$

taking the $\mathrm{liminf}$ on gets that ${\lim }_p\Vert f{\Vert }_p=\Vert f{\Vert }_{\infty }$.

Proof Let $u\in L^q$, Apply Holder inequality to $u^p$ and $1$. Since we want on the right the $q$ norm, use as indices $\frac{q}{p}$ and its conjugated:

$$\Vert u^p\cdot 1{\Vert }_1\le \Vert u{\Vert }_q^p\cdot \Vert 1{\Vert }_{\frac{q}{q-p}}$$ $$\Vert u{\Vert }_p^p\le \Vert u{\Vert }_q^p\mu (X{)}^{\frac{q-p}{q}}\square$$

Remark the result is true only for finite measure spaces. A cool counterexample, in which also the opposite inclusion $L^p\subseteq L^q$ when $p\le q$ is also true is $\mathbb{Z}$ equipped with the counting measure. Indeed

$$\Vert f{\Vert }_q=\sum _{n\in \mathbb{Z}}|f(n){|}^p$$

Proposition Let $p\in [1,+\infty ]$. The space $L^p(\mu )$ is complete: suppose that $\{f_n\}$ is a sequence such that $\forall ϵ>0$, there exists $N\ge 0$ such that

$$\Vert f_j-f_k{\Vert }_p<ϵ,\forall i,k\ge N$$

(The sequence is Cauchy). Then there exists an element $f\in L^p$ such that

$$\underset{n\to \infty }{\lim }\Vert f-f_n{\Vert }_p=0$$

Proof Suppose now that $p\in [1,+\infty )$. First we build a limit candidate, then we show that this limit belongs to $L^p$. Since the sequence is Cauchy, we can fine a subsequence $\{f_{n_k}\}$ such that

$$\Vert f_{n_{k+1}-f_{n_k}}{\Vert }_p<\frac{1}{2^k}$$

to simplify the notation we identify the subsequence as the original sequence.

Clearly one can write with a telescopic series

$$f_n=f_1+\sum _{k=1}^{n-1}(f_{k+1}-f_k)$$

we now define the candidate limit function $f$ as a series

$$f:=f_1+\sum _{k\ge 1}(f_{k+1}-f_k)$$

and similalry

$$g:=\underset{n\to \infty }{\lim }{g}_n=|f_1|+\sum _{k\ge 1}|f_{k+1}-f_k|$$

Note that $g_n\uparrow g$. Applying the Minkowsky’s inequality and taking the limit we get

$$\underset{n}{\lim }\Vert g_n{\Vert }_p\le \Vert f_1{\Vert }_p+\sum _{k\ge 0}\Vert f_{k+1}-f_k{\Vert }_p<\infty$$

using Monotone Convergence Theorem

$$\Vert g{\Vert }_p=\underset{n}{\lim }\Vert g_n{\Vert }_p<\infty$$

so that $g\in L^p$. This implies that $g$ is finite a.e., so that $f_n$ is absolutely convergent a.e., meaning that it’s also convergent a.e. and the definition of $f$ makes sense. Also $f_n\le g_n$ and we can conclude that also $f\in L^p$.

By construction

$$f_{n_k}\to fa.e.$$

we need to prove that this converges in $L^p$ also

$$|f(x)-f_n(x){|}^p\le [2\max (|f(x)|,|f_n(x)|){]}^p$$ $$\le 2^p|f(x){|}^p+2^p|f_n(x){|}^p$$ $$\le 2^p|g(x){|}^p+2^p|g_n(x){|}^p\le 2^{p+1}|g(x){|}^p$$

taking the limit and using Dominated convergence theorem we get

$$\underset{n\to \infty }{\lim }\Vert f-f_n{\Vert }_p^p=\int \underset{n}{\lim }|f-f_n{|}^{p}d\mu =0$$

The last step is to show that also the original sequence converges to $f$.

$$\Vert f-f_n{\Vert }_p\le \Vert f-f_{n_k}{\Vert }_p+\Vert f_{n_k}-f_n{\Vert }_p$$

since $f_{n_k}\to f$ there exists an $N\ge 0$ such that $\forall k\ge N$ the first term on the left is less than $\frac{ϵ}{2}$. Since the original sequence was Cauchy there also exists an $M\ge 0$ such that the second term on the right is less than $\frac{ϵ}{2}$. We can conclude that

$$\Vert f-f_n{\Vert }_p\to 0$$

The completeness of $L^{\infty }$ is straight forward, since the sequence is Cauchy $\forall ϵ>0$ there exists $N\ge 0$ such that $\forall m>n\ge N$ it holds almost everywhere that

$$|f_m(x)-f_n(x)|<ϵa.e.$$

For a fixed $ϵ$ these sets of measure zero where the inequality does not holds can depend on $m$ and $n$, consider the (contable) union

$$Z=\bigcup _{m>n\ge N}{Z}_{m,n}$$

this set has also measure zero. Clearly for each $x\in X\setminus Z$ we have pointwise convergence (since $\mathbb{R}$ is complete), hence it makes sense to define $f$ as the pointwise limit of $f_n$ on this set. The convergence is also uniform on $X\setminus Z$, so that

$$\Vert f-f_n{\Vert }_{\infty }\to 0\square$$

Dense subsets §

Note

Let $1\le p<+\infty$. Simple functions are dense in $L^p(\Omega )$: for any $f\in L^p(\Omega )$ there exists a sequence $\{h_k\}$ of simple functions such that:

$$\underset{n\to \infty }{\lim }\Vert f-h_k{\Vert }_p=0$$

Note f = f_+ - f_-, then we can assume w.l.o.g. that f \geq 0. By the definition of Lebesgue's integral, we can find a sequence of simple functions such that

Write

$$\int h_{n}d\mu \stackrel{n\to \infty }{\to }\sup \left\{\int g_{k}d\mu \mid 0\le g_k\le |f{|}^p,g_k\text{ simple}\right\}$$

So the case $p=1$ is proven. To prove the general case observe that:

$$|f-h_n{|}^p\le |f{|}^p$$

Since $|f-h_n{|}^p\to 0$ pointwise, using Dominated convergence theorem:

$$\Vert f-h_n{\Vert }_p^p=\int |f-h_n{|}^{p}d\mu \stackrel{n\to \infty }{\to }0\square$$

Note

Let $1\le p<+\infty$. Test functions $C_c^{\infty }(\Omega )$ are dense in $L^p(\Omega )$: for any $f\in L^p(\Omega )$ there exists a sequence $\{{\psi }_k\}\subset C_c^{\infty }(\Omega )$ such that:

$$\underset{n\to \infty }{\lim }\Vert f-{\psi }_k{\Vert }_p=0$$

Note

By definition