\(L^p\) Space

Modern Real analysis focuses on digging into the structure of Space combined by these functions. Lebesgue Integral Theory is the most successful one.

Definitions

(i) Assume \(E\) is a measurable set, \(1\leq p<\infty\), denote

\[ L^p(E)=\{f: f\text{ is measurable on } E \text{ and } \int_E |f(x)|^pdx<\infty \}. \]

for \(f\in L^p(E)\), denote

\[ \|f\|_p=\left(\int_E |f(x)|^pdx\right)^{\frac{1}{p}} \]

as norm of \(f\). Notice that \(L^1(E)\) equals \(L(E)\).

(ii) Assume \(m(E)>0\). If there exists \(M>0\), such that \(|f(x)|\leq M,\quad a.e.x\in E\), then we call \(f\) is essentially bounded on \(E\), \(M\) is called the essential upper bound of \(f(x)\). A set composed by all essentially bounded function is denoted as \(L^\infty(E)\). If \(f\in L^{\infty}(E)\), define

\[ \|f\|_{\infty}:=\inf\{M: |f(x)|\leq M,\quad a.e.x\in E\} \]

as the norm or essential bound of \(f\).

Example. Prove: when \(0< m(E)<\infty\), we have

\[ \lim_{p\rightarrow \infty}\|f\|_p=\|f\|_\infty. \]

Proof

Let \(M=\|f\|_\infty\), then choose \(M'<M\), and define set

\[ A=\{x\in E: |f(x)|>M'\} \]

which has positive measure. Then by

\[ \|f\|_p=\left(\int_E |f(x)|^pdx\right)^{\frac{1}{p}}\geq M' [m(A)]^{\frac{1}{p}} \]

let \(p\rightarrow \infty\), we have

\[ \underline{\lim}\limits_{p\rightarrow \infty}\|f\|_p\geq M' \]

let \(M'\rightarrow M\), we have \(\underline{\lim}\limits_{p\rightarrow \infty}\|f\|_p\geq M\).

On the other hand, we have

\[ \|f\|_p\leq \left(\int_E M^pdx\right)^{\frac{1}{p}}=M[m(E)]^{\frac{1}{p}} \]

let \(p\rightarrow \infty\), we have

\[ \overline{\lim}\limits_{p\rightarrow \infty}\|f\|_p\leq M. \]

So we are done.

Theorem

If \(m(E)<\infty\), then \(L^p(E)\subset L^1(E)\).

Proof

Let \(A=\{f\geq 1\}\cap E\), \(B=\{f<1\}\cap E\), if \(f\in L^p(E)\), then

\[ \begin{align*} \int_E |f|^p dx<\infty. \end{align*} \]

So

\[ \begin{align*} \int_E |f| dx&=\int_A |f|dx+\int_B|f|dx\\ &<\int_A |f|^pdx+m(B)\\ &<\int_E |f|^pdx+m(B)<\infty. \end{align*} \]

Example. \(E=(0,\infty)\), \(f(x)=\frac{1}{1+x}\), then \(f\in L^p(E)\) but \(f\notin L^1(E)\).

\(L^p(E)\) is a linear space

Assume \(f,g\in L^p(E)\), \(0< p\leq \infty\), \(\alpha,\beta\in \mathbb{R}\), then

\[ \alpha f + \beta g\in L^p(E). \]

Proof

• \(0<p<\infty\),

\[ |\alpha f(x)+\beta g(x)|^p\leq 2^p(|\alpha|^p|f(x)|^p+|\beta|^p|g(x)|^p). \]

• \(p=\infty\).

\[ |\alpha f(x)+\beta g(x)|\leq |\alpha|\|f\|_\infty+|\beta|\|g\|_\infty, \quad a.e.x\in E. \]

so

\[ \|\alpha f+\beta g\|_\infty\leq |\alpha|\|f\|_\infty+|\beta|\|g\|_\infty. \]

Properties of \(L^p\) Space

Hölder Inequation

If \(1<p,q<\infty\), and

\[ \frac{1}{p}+\frac{1}{q}=1, \]

then we call \(p\) and \(q\) are conjugate indexes.

If \(f\in L^p(E)\), \(g\in L^q(E)\), then \(fg\in L(E)\) and

\[ \|fg\|_1\leq \|f\|_p\cdot \|g\|_q. \]

Proof

Use inequation

\[ ab\leq \frac{a^p}{p}+\frac{b^q}{q}. \]

When \(p=2\) in the above inequation, it is called Schwartz Inequation.

Example. Assume \(m(E)<\infty\), and \(1\leq p< q <\infty\), then \(L^q(E)\subset L^p(E)\), and

\[ \|f\|_p\leq [m(E)]^{\frac{1}{p}-\frac{1}{q}}\|f\|_q. \]

Proof

Use Hölder inequation. Let \(r=\frac{q}{p}\) and \(r'=\frac{r}{r-1}=\frac{q}{q-p}\).

\[ \begin{align*} \int_E |f|^p\cdot 1dx&\leq \| |f|^p\|_r\cdot \|1\|_{r'}\\ \|f\|_p^p&=\left(\int_E |f|^{p\cdot r}dx\right)^\frac{1}{r}\cdot \left(\int_E 1dx\right)^\frac{1}{r'}\\ \|f\|_p^p&=\left(\int_E |f|^{q}dx\right)^\frac{p}{q}\cdot \left(m(E)\right)^\frac{q-p}{q}\\ \|f\|_p^p&=\|f\|_q^p\cdot \left(m(E)\right)^\frac{q-p}{q}\\ \Rightarrow \|f\|_p&=\|f\|_q\cdot\left(m(E)\right)^{\frac{1}{p}-\frac{1}{q}} \end{align*} \]

which means \(L^q(E)\subset L^p(E)\).

Example. Assume \(f\in L^r(E)\cap L^s(E)\), where \(1\leq r< q < s \leq \infty\), and

\[ \frac{1}{p}=\frac{\lambda}{r}+\frac{1-\lambda}{s},\quad 0<\lambda<1. \]

then

\[ \|f\|_p\leq \|f\|_r^\lambda \|f\|_s^{1-\lambda}. \]

Proof

Minkowski Inequation

Assume \(f,g\in L^p(E)\), \(1\leq p\leq\infty\), then \(f+g\in L^p(E)\), and

\[ \|f+g\|_p\leq \|f\|_p+\|g\|_p. \]

Proof

Example. Assume \(1\leq p\leq \infty\). If \(f_k(x)\in L^p(E), k=1,2,\cdots\), and \(\sum\limits_{k=1}^\infty f_k(x)\) converges \(a.e.x\in E\), then

\[ \left\|\sum_{k=1}^\infty f_k\right\|_p\leq \sum_{k=1}^\infty \|f_k\|_p. \]

Proof

• \(p=\infty\).

Notice that

\[ \left|\sum\limits_{k=1}^\infty f_k(x)\right|\leq \sum\limits_{k=1}^\infty \left|f_k(x)\right|\leq \sum_{k=1}^\infty \left\|f_k(x)\right\|_\infty \]

then by definition of \(\left\|\sum\limits_{k=1}^\infty f_k\right\|_\infty\), we have

\[ \begin{align*} \left\|\sum_{k=1}^\infty f_k\right\|_\infty&\leq \sum_{k=1}^\infty \left\|f_k\right\|_\infty. \end{align*} \]

• \(1\leq p<\infty\).

\[ \begin{align*} \left\|\sum_{k=1}^\infty f_k\right\|_p^p&=\int_E \left|\sum_{k=1}^\infty f_k\right|^pdx\\ &\leq \int_E \left(\sum_{k=1}^\infty \left|f_k\right|\right)^p dx\\ &=\int_E \lim_{n\rightarrow \infty} \left(\sum_{k=1}^n \left|f_k\right|\right)^p dx\\ &=\lim_{n\rightarrow \infty} \int_E \left(\sum_{k=1}^n \left|f_k\right|\right)^p dx\quad \text{Levi Theorem}\\ &=\lim_{n\rightarrow \infty} \left\|\sum_{k=1}^n \left|f_k\right|\right\|_p^p\\ &\leq \lim_{n\rightarrow \infty}\left(\sum_{k=1}^n \left\|f_k\right\|_p\right)^p \quad \text{Minkowski} \end{align*} \]

To introduce distance in \(L^p(E)\), we assume that if \(f,g\in L^p(E)\), we have \(f(x)=g(x),a.e.x\in E\), then we denote this relationship as \(f=g\).

Use p-norm as Metric

Now in the following parts, we assume \(1\leq p\leq \infty\).

Definitions

Assume \(f,g\in L^p(E)\), we introduce distance (or to be more specific, metric)

\[ \rho(f,g)=\|f-g\|_p,\quad 1\leq p\leq \infty. \]

then \(\rho\) satisfies three conditions of metric, and \((L^p(E),\rho)\) is a metric space.

Convergence

With metric \(\rho\), we could introduce the limit in a sense of \(\rho\). This is a special case of convergence.

Definition of Convergence in \(L^p\) Space

Assume \(f_n\in L^p(E)\), if there exists \(f\in L^P(E)\), such that

\[ \lim_{n\rightarrow \infty} \rho(f_n,f)=\lim_{n\rightarrow \infty}\|f_n-f\|_p=0, \]

then we call \(\{f_n\}\) converges to \(f\) in a sense of \(L^p(E)\), which is denoted by \(f_n\rightarrow f(L^p)\).

Properties of Convergence in \(L^p\) Space

(i) Uniqueness. If \(\lim\limits_{n\rightarrow \infty} \rho(f_n,f)=0=\lim\limits_{n\rightarrow \infty} \rho(f_n,g)\), then

\[ f(x)=g(x),\quad a.e.x\in E. \]

(ii) If \(\lim\limits_{n\rightarrow \infty} \rho(f_n,f)=0\), then

\[ \lim\limits_{n\rightarrow \infty} \|f_n\|_p=\|f\|_p. \]

Proof

(i) easy to see.

(ii) We have

\[ |\|f_k\|_p-\|f\|_p|\leq \|f_k-f\|_p. \]

Example. Still the same function of Example. We could know that

\[ f_n(x)\nrightarrow 0,\quad a.e.x\in E, \]

but \(f_n\overset{L^p}{\rightarrow}0\), since

\[ \int_0^1 |f_n-f|^pdx=\int_{\left[\frac{i}{2^k},\frac{i+1}{2^k}\right]}1^pdx=\frac{1}{2^k}\rightarrow 0(n\rightarrow \infty). \]

Example. \(f_k(x)=k\chi_{(0,\frac{1}{k})}\), \(E=(0,\infty)\), then

\[ f_k(x)\rightarrow 0,\quad \forall x\in E. \]

but \(f_k\overset{L^p}{\nrightarrow} f\), since

\[ \int_E |f_n-f|^pdx=\int_{(0,\frac{1}{k})}|k|^pdx=|k|^{p-1}\rightarrow \infty. \]

Similar to Cauchy sequence in \(\mathbb{R}\), we have the following Cauchy sequence in a sense of \(L^p(E)\).

Definition of Cauchy Sequence in \(L^p\) Space

Assume \(f_k\in L^p(E)\), if

\[ \lim_{n,m\rightarrow \infty}\rho(f_m,f_n)=\lim_{n,m\rightarrow \infty}\|f_m-f_n\|_p=0, \]

then we call \(\{f_k\}\) is Cauchy Sequence in \(L^p(E)\).

Property of Cauchy Sequence

Assume \(f_n\) is a Cauchy Sequence in \(L^p(E)\). If there exists a subsequence \(\{f_{n_j}\}_{j\geq 1}\) which is convergent in a sense of \(L^p\), then \(f_n\) itself also converges in a sense of \(L^p\).

Proof

\[ \|f_n-f\|_p\leq \|f_n-f_{n_j}\|_p+\|f_{n_j}-f\|_p \]

Completeness

Similar to proof in \(\mathbb{R}\), we could prove that a convergent sequence must be Cauchy Sequence. Readers could use Minkowski inequation

\[ \|f_{n+l}-f_{n}\|_p\leq \|f_{n+1}-f\|_p+\|f_n-f\|_p\rightarrow 0 (n\rightarrow \infty) \]

To show its completeness, we have the following lemma.

Lemma: find convergent function

Assume \(\{f_k\}\subset L^p(E)\), \(1\leq p<\infty\). If

\[ \begin{equation} \sum_{k=1}^\infty\|f_k\|_p<\infty,\label{lemma-condition} \end{equation} \]

then there exists \(f\in L^p(E)\), such that

\[ \sum_{k=1}^\infty f_k\overset{L^p}{=}f. \]

Proof

Let \(g_k(x)=\sum\limits_{j=1}^k |f_j|\), and \(g=\lim\limits_{k\rightarrow \infty} g_k(x)=\sum\limits_{k=1}^\infty |f_k|\). so

\[ g_k^p(x)=\left(\sum\limits_{j=1}^k |f_j|\right)^p\rightarrow g^p=\left(\sum\limits_{k=1}^\infty |f_k|\right)^p. \]

Since \(g^p_k\) are non-negative and monotonically increasing, by Levi Theorem, we have

\[ \begin{align*} \int_E g^pdx&=\lim_{k\rightarrow \infty}\int_E g_k^pdx\\ &=\lim_{k\rightarrow \infty}\|g_k\|^p_p\\ &\leq \lim_{k\rightarrow \infty} \left(\sum_{j=1}^k \|f_j\|_p\right)^p\quad \text{Minkowski}\\ &=\left(\sum_{j=1}^\infty \|f_j\|_p\right)^p \end{align*} \]

by condition \(\ref{lemma-condition}\) of this lemma, we have \(g(x)<\infty\), i.e. \(\sum\limits_{k=1}^\infty f_k\) converges absolutely \(a.e.x\in E\). So there exists measurable function \(f\), such that

\[ f=\sum\limits_{k=1}^\infty f_k, \quad a.e.x\in E. \]

Notice this limit function is just measurable, we have to prove that \(f\in L^p(E)\). Notice that

\[ \left|\sum\limits_{j=1}^k f_j\right|^p\leq \left(\sum\limits_{j=1}^k |f_j|\right)^p\leq \left(\sum\limits_{j=1}^\infty |f_j|\right)^p=g^p \]

where the last item \(g^p\) is L integral, so \(F_k=\left(\sum\limits_{j=1}^k f_j\right)^p\) is controlled by \(g^p\). Since \(|f|^p=\lim\limits_{k\rightarrow \infty}|F_k|^p\), by LDCT, we have

\[ \begin{align*} \int_E |f|^pdx=\lim_{k\rightarrow \infty}\int_E \left|\sum\limits_{j=1}^k f_j\right|^pdx\leq \int_E g^pdx<\infty. \end{align*} \]

which means \(f\in L^p(E)\). Finally, we have to show that \(F_k\) converges to \(f\) in a sense of \(L^p(E)\). That is, Notice that

\[ |f-F_k|^p=\left|\sum_{j=k+1}^\infty f_j\right|\leq \left(\sum_{j=k+1}^\infty|f_j|\right)^p\leq g^p \in L^1(E), \]

by LDCT, we have

\[ \lim_{k\rightarrow \infty}\int_E |f-F_k|^pdx=\int_E \lim_{k\rightarrow \infty} |f-F_k|^pdx=0. \]

Now we have to prove Cauchy Sequence must be a convergent sequence, and its limit must be in \(L^p(E)\).

Riesz-Ficher: Completeness

\(L^p(E)\) is a complete metric space. That is, Cauchy sequence in \(L^p(E)\) must converges.

Proof

• \(1\leq p<\infty\).

By definition of Cauchy sequence, \(\forall m\geq 1\), \(\exists j_m\), such that

\[ \|f_{j_m}-f_k\|_p<\frac{1}{2^m},\quad \forall k>j_m. \]

Choose \(k=j_m+1\), so we get a subsequence \(\{f_{j_m}\}\) such that

\[ \|f_{j_m}-f_{j_m+1}\|_p<\frac{1}{2^m} \]

Now let \(g_1=f_{j_1}\), \(g_k=f_{j_k}-f_{j_{k-1}}, k\geq 2\). Then we have

\[ \begin{align*} \sum_{k=1}^\infty\|g_k\|_p&=\|f_{j_1}\|_p+\sum_{k=2}^\infty \|f_{j_k}-f_{j_{k-1}}\|_p\\ &<\|f_{j_1}\|_p+\sum_{k=2}^\infty\frac{1}{2^k}<\infty \end{align*} \]

which satisfies the condition \(\ref{lemma-condition}\) of lemma. So there exsits \(f\in L^p(E)\) such that

\[ G_k:=\sum_{j=1}^k g_j=f_{j_k}\rightarrow f(L^p). \]

Then by Property of Cauchy Sequence, convergence of subseuquence could deduce original sequence \(f_k\) converges to \(f\) in a sense of \(L^p(E)\).

• \(p=\infty\).

Assume \(\{f_k\}\in L^\infty(E)\) is a Cauchy sequence, then by its definition, \(\forall m\in \mathbb{N}\), \(\exists k_m\), such that

\[ \|f_j-f_k\|_\infty<\frac{1}{2^m},\quad \forall j,k>k_m. \]

By definition of \(L^\infty\), there exsits a zero-measure set \(Z_{j,k,m}\), such that

\[ |f_j(x)-f_k(x)|\leq \|f_j-f_k\|_\infty<\frac{1}{2^m},\quad \forall x\in E-Z_{j,k,m},j,k>k_m. \]

Let \(Z=\bigcup_{j,k,m\in \mathbb{N}}Z_{j,k,m}\) which is also zero-measure, and

\[ \begin{align} |f_j(x)-f_k(x)|<\frac{1}{2^m},\quad x\in E-Z, j,k>k_m.\label{completeness} \end{align} \]

The above inequation implies \(\{f_n\}\) is Cauchy number sequence (\(\mathbb{R}\)), so

\[ \lim_{n\rightarrow \infty}f_n(x)=f(x),\quad x\in E-Z. \]

let \(j\rightarrow \infty\) in inequation \(\ref{completeness}\), we have

\[ |f(x)-f_k(x)|\leq\frac{1}{2^m},\quad x\in E-Z, k>k_m. \]

which still by definition of \(L^\infty\), we have

\[ \|f(x)-f_k(x)\|_\infty \leq\frac{1}{2^m},\quad x\in E-Z, k>k_m. \]

which means \(f\in L^\infty(E)\) and \(f_k\overset{L^\infty}{\rightarrow} f\).

Lebesgue's dominated control theorem in \(L^p\) Space

Assume \(f_k\), \(f\) are measurable function on \(E\). If \(f_k\rightarrow f,a.e.x\in E\), and there exists a function \(F(x)\in L^p(E)\) such that

\[ |f_k(x)|\leq F(x),\quad a.e.x\in E, \]

then \(f_k\overset{L^p}{\rightarrow}f\).

Proof

Let

\[ g_n(x)=|f_k(x)-f(x)|^p\rightarrow 0(k\rightarrow \infty). \]

Notice that

\[ |g_k(x)|\leq (|f_k|+|f|)^p\leq 2^p|f|^p\in L^1(E), \]

So by LDCT in \(L^1(E)\), we have

\[ \lim_{k\rightarrow \infty}\int_E |f_k-f|^p=\int_E 0dx=0. \]

Separability

Definitions of Separations

(i) Assume \(\varGamma\subset L^p(E)\). If \(\forall f\in L^p(E)\), \(\forall\varepsilon>0\), \(\exists g\in \varGamma\), such that

\[ \|f-g\|_p<\varepsilon, \]

then we call \(\varGamma\) is a dense subset of \(L^p(E)\). Apparently, its equivalent definition is \(\forall f\in L^p(E)\), \(\exists \{g_n\}\subset \varGamma\), such that \(g_n\rightarrow f(L^p)\), i.e.

\[ \lim_{n\rightarrow \infty}\|f-g_n\|_p=0. \]

(ii) If \(L^p(E)\) has dense subsets whose elements are denumerable, then we call \(L^p(E)\) is separable.

Separability of \(L^p(E)\)

If \(1\leq g<\infty\), then \(L^p(E)\) is separable.

Proof