Some Main Theorems on NLS - 3. The space $l_p^n(\R)$, $1 \leq p \leq \infty$

Underlying space $\R^n$, let $x = (x_1,\, x_2,\, \ldots,\, x_n) \in \R^n$. Define

\[ \norm{x}_p = \begin{cases} \sum_{i=1}^{n} \abs{x_i}, & \quad p=1 \\ \left( \sum_{i=1}^{n} \abs{x_i}^p \right)^{1/p}, & \quad 1<p<\infty \\ \max_{1\leq i \leq n} \abs{x_i}, & \quad p=\infty. \end{cases}\]

We want to show that $\norm{\cdot}_p$ is a norm on $\R^n$. This is so-called the Minkowski's inequality:

\[ \norm{x+y}_p \leq \norm{x}_p + \norm{y}_p, \]

for $x,\,y \in \R^n$, $1 \leq p \leq \infty$.

This inequality comes from the Hölder's inequality:

\[ \abs{\sum_{i=1}^{n} x_iy_i} \leq \left( \sum_{i=1}^{n} \abs{x_i}^p \right)^{\frac{1}{p}} \left( \sum_{i=1}^{n} \abs{y_i}^q \right)^{\frac{1}{q}}, \]

for $x,\,y \in \R^n$, $1 \leq p \leq \infty$, and $\frac{1}{p} + \frac{1}{q} = 1$. Note that the H{\"o}ler's inequality is the same as: $\abs{\ip{x}{y}} \leq \norm{x}_p \norm{x}_q$.

When $\frac{1}{p} + \frac{1}{q} = 1$, we say that $p$ and $q$ are conjugate of each other.

(1) $p=1 \implies q=\infty$, $q=q \implies p=\infty$.

(2) When $p=q=2$, we get

\[ \abs{\sum_{i=1}^{n} x_iy_i}^2 \leq \left( \sum_{i=1}^{n} \abs{x_i}^2 \right) \left( \sum_{i=1}^{n} \abs{y_i}^2 \right). \]

This is called the Cauchy-Schwarz inequality.

Lemma 2.3.1

Let $1< p < \infty$. Then we know $1 < q < \infty$.

From the above figure, for all $a,\,b \geq 0$, we have

$\begin{aligned} ab &\leq \operatorname{area}(I) + \operatorname{area}(II) \\ &= \int_{0}^{a} x^{p-1} \text{d}{x} + \int_{0}^{b} y^{q-1} \text{d}{y} \\ &= \frac{a^p}{p} + \frac{b^q}{b}. \end{aligned}$

Theorem 2.3.2 (Hölder's inequality)

For $x,\,y \in \R^n$, $1 \leq p \leq \infty$, and $\frac{1}{p} + \frac{1}{q} = 1$,

\[ \abs{\ip{x}{y}} \leq \norm{x}_p \norm{x}_q. \]

Proof. It is easy to show for $p=1$ and $p=\infty$: When $p=1$ and $q=\infty$,

\[ \abs{\ip{x}{y}} = \abs{\sum_{i=1}^{n} x_i y_i} \leq \left( \max_i \abs{y_i} \right) \sum_{i=1}^{n} \abs{x_i} = \norm{x}_1 \norm{y}_\infty. \]

Consider the case $1 < p,\,q < \infty$. If $\norm{x}_p = 0$ or $\norm{y}_q = 0$, the inequality trivially holds. Thus, assume $\norm{x}_p \neq 0 \neq \norm{y}_q$. Put

\[ a_i := \frac{\abs{x_i}}{\norm{x}_p},\ b_i \frac{\abs{y_i}}{\norm{y}_q} \qquad \text{for $i=1;n$}. \]

Use the lemma to get

\[ \frac{\abs{x_iy_i}}{\norm{x}_p \norm{y}_q} \leq \frac{1}{p} \left( \frac{\abs{x_i}^p}{\norm{x}_p^p} \right) + \frac{1}{q} \left( \frac{\abs{y_i}^q}{\norm{y}_q^q} \right) \]

for each $i=1;n$. Add all of these,

\[ \begin{aligned} \frac{\sum_{i=1}^{n} \abs{x_iy_i}}{\norm{x}_p \norm{y}_q} &\leq \frac{1}{p} \underbrace{\left( \frac{\sum_{i=1}^{n} \abs{x_i}^p}{\norm{x}_p^p} \right)}_{=1} + \frac{1}{q} \underbrace{\left( \frac{\sum_{i=1}^{n} \abs{y_i}^q}{\norm{y}_q^q} \right)}_{=1} \\ &= \frac{1}{p} + \frac{1}{q} \\ &= 1 \end{aligned} \]

This implies $\sum_{i=1}^{n} \abs{x_iy_i} \leq \norm{x}^p \norm{y}^q$. Thus,

\[ \abs{\ip{x}{y}} \abs{\sum_{i=1}^{n} x_iy_i} \leq \sum_{i=1}^{n} \abs{x_iy_i} \leq \norm{x}^p \norm{y}^q. \]

Hence, the proof is completed. ■

Theorem 2.3.3 (Minkowski's inequality)

For $x,\,y \in \R^n$, $1 \leq p \leq \infty$,

\[ \norm{x+y}_p \leq \norm{x}_p + \norm{y}_p. \]

Proof. It is simple to verify when $p=1$ and $p=\infty$. So, let $1<p<\infty$. Assume $\norm{x+y}_p \neq 0$, otherwise the inequality is obviously true. Then,

\[ \begin{aligned} & \sum_{i=1}^{n} \abs{x_i + y_i}^p \\ & \qquad = \sum_{i=1}^{n} \abs{x_i + y_i}^{p-1}\abs{x_i + y_i} \\ & \qquad \leq \sum_{i=1}^{n} \abs{x_i + y_i}^{p-1}\abs{x_i} + \sum_{i=1}^{n} \abs{x_i + y_i}^{p-1}\abs{y_i}. \end{aligned} \]

By the Hölder;s inequality,

\[ \begin{aligned} & \qquad \leq \left( \sum_{i=1}^{n} \abs{x_i}^p \right)^{\frac{1}{p}} \left( \sum_{i=1}^{n} \abs{x_i + y_i}^{(p-1)q} \right)^{\frac{1}{q}} \\ & \qquad\qquad + \left( \sum_{i=1}^{n} \abs{y_i}^p \right)^{\frac{1}{p}} \left( \sum_{i=1}^{n} \abs{x_i + y_i}^{(p-1)q} \right)^{\frac{1}{q}} \\ & \qquad = \left( \norm{x}_p \norm{y}_p \right) \left( \sum_{i=1}^{n} \abs{x_i + y_i}^{p} \right)^{\frac{1}{q}} \end{aligned} \]

Here, we used the fact that $\frac{1}{p} + \frac{1}{q} = 1 \Leftrightarrow (p-1)q = p$. Hence,

\[ \begin{aligned} \norm{x}_p \norm{y}_p &\geq \frac{\sum_{i=1}^{n} \abs{x_i + y_i}^p}{\left( \sum_{i=1}^{n} \abs{x_i + y_i}^{p} \right)^{\frac{1}{q}}} \\ &= \left( \sum_{i=1}^{n} \abs{x_i + y_i}^{p} \right)^{1-\frac{1}{q}} \\ &= \left( \sum_{i=1}^{n} \abs{x_i + y_i}^{p} \right)^{\frac{1}{p}} \\ &= \norm{x+y}_p. \end{aligned} \]

This completes the proof. ■

Theorem 2.3.4

For $1<p<\infty$, $\left( l_p^n(\R) \right)^* = l_q^n(\R)$.

This theorem means:

(1) For each $f \in \left( l_p^n(\R) \right)^*$, there exists $a \in l_q^n(\R)$ such that $f(x) = \ip{a}{x}$ for any $x \in l_p^n(\R)$ and $\norm{f} = \norm{a}_q$.

(2) For each $a \in l_q^n(\R)$, the functional $f(x) = \ip{a}{x}$ belongs to $\left( l_p^n(\R) \right)^*$ and $\norm{f} = \norm{a}_q$.

Remark. When $p=1$ or $p=\infty$, we have

(1) $\left( l_1^n(\R) \right)^* = l_\infty^n(\R)$.

(2) $\left( l_\infty^n(\R) \right)^* = l_1^n(\R)$.

Now, we construct the $L_p$-spaces. Let $(X,\, \mathcal{A},\,\mu)$ be a measure space, where $X$ is a nonempty set, $\mathcal{A}$ is a $\sigma$-algebra on $X$, and $\mu$ is a measure on $(X,\,\mathcal{A})$. 

Let a function $f : X \to [-\infty,\, \infty]$ be measurable, that is, $f^{-1}((a,\,b)) \in \mathcal{A}$ for any open interval $(a,\,b) \subseteq \R$.

Then, for $1 \leq p < \infty$, define

\[ \norm{f}_p := \left[ \int_X \abs{f}^p \text{d}{\mu} \right]^{\frac{1}{p}}. \]

A number $a$ is called an essential upper bound of $f$ if the measurable set $f^{-1}(a,\, \infty)$ is a set of measure zero, i.e.,  i.e., if $f(x) \leq a$ for almost all $x$ in $X$. Let

\[ U_f^{\mathrm{ess}} := \set{a \in \R}{\mu(f^{-1}(a,\, \infty)) = 0}, \]

which is the set of all essential upper bounds. Then the essential supremum of $f$ is defined as

$\operatorname{ess\,sup} f := \inf U_f^{\mathrm{ess}}.$

If $U_f^{\mathrm{ess}} = \emptyset$, then $\operatorname{ess\,sup} f = \infty$.

Now, let $norm{f}_\infty = \operatorname{ess\,sup} f$. Combining all the above concepts, we now define

\[ \begin{aligned} L_p(X,\, \mathcal{A},\, \mu) := \{f : X \to [-\infty,\, \infty] \,|\, & \text{$f$ is measurale} \\ & \text{and $\norm{f}_p < \infty$} \}. \end{aligned} \]

Example 2.3.5

(1) Let $\begin{aligned}[t] X &= \{1,\,2,\, \ldots,\, n\} \\ \mathcal{A} &= \text{power set of $X$} \quad (\text{i.e., $\mathcal{A} = 2^X$}) \\ \mu &= \text{counting measure} \end{aligned}$
Then $L_p(X,\, \mathcal{A},\, \mu) = l_p^n(\R)$.

(2) Let $ \begin{aligned}[t] X &= \text{$[a,\,b]$ in $\R$} \\ \mathcal{A} &= \text{Lebesgue measurable sets in $[a,\,b]$} \\ \mu &= \text{Lebesgue measure} \end{aligned} $
Then, we can define  $L_p[a,\,b]$ with
\[ \norm{f}_p := \left[ \int_a^b \abs{f}^p \text{d}{\mu} \right]^{1/p}. \]