📂Lebesgue Spaces

The Riesz Representation Theorem for Lp Spaces

Theorem[^1]

${L}^{\ p}$ Representation Theorem for Spaces

Let’s say $1<p<\infty$ and $L\in \big( {L}^{\ p} \big)^{\ast}$. Then, $({L}^{\ p})^{\ast}$ is the dual of the ${L}^{\ p}$ space. Therefore, for every $u\in {L}^{\ p}$, there exists a $v \in {L}^{\ p^{\prime}}$ that satisfies the following equation.

$$L(u)=L_{v}(u)=\int_{\Omega} u(x)v(x)dx$$

Explanation

Note that the case where $p=1$ is not included.

$\left\| v \right\|_{p^{\prime}} =\left\| L\ ; ({L}^{\ p})^{\ast}\right\|$ satisfies that $f\ :\ ({L}^{\ p})^{\ast} \rightarrow {L}^{\ p^{\prime}}$ becomes an isometric mapping. Therefore, since $({L}^{\ p})^{\ast}\cong {L}^{\ p^{\prime}}$, ${L}^{\ p^{\prime}}$ can be considered as the dual of the ${L}^{\ p}$ space.

At this time, $p^{\prime}$ is the conjugate exponent of $p$. Originally, the representation theorem by Riesz is a theorem that holds for Hilbert spaces. Since the ${L}^{\ p}$ space is not generally a Hilbert space, the representation theorem by Riesz cannot be directly applied. However, the above theorem states that the ${L}^{\ p}$ space also has the properties mentioned in the Riesz representation theorem. To explain the content more clearly, it is as follows.

Let’s say a certain linear functional $L \in ({L}^{\ p})^{\ast}$ defined on the ${L}^{\ p}$ space is selected. Then, there exists a unique $v \in {L}^{\ p^{\prime}}$ corresponding to it. And it satisfies the following equation.

$$L(u)=\int u(x)v(x) dx$$

In other words, the element $L$ of $({L}^{\ p})^{\ast}$ space and the element $v$ of ${L}^{\ p^{\prime}}$ space are paired so that the value of $u$ substituted into $L$, which is $L(u)$, and the value obtained by multiplying $v(x)$ by $u(x)$ and integrating it, which is $\int u(x)v(x)dx$, are the same. Furthermore, since $({L}^{\ p})^{\ast}$ and ${L}^{\ p^{\prime}}$ are isometric, it can be thought that the ${L}^{\ p^{\prime}}$ space is virtually the dual of the ${L}^{\ p}$ space. Also, since $p$ and $p^{\prime}$ are each other’s conjugate exponent, conversely, the ${L}^{\ p}$ space can also be considered virtually the dual of the ${L}^{\ p^{\prime}}$ space.

Riesz Representation Theorem

Let’s call $X$ a Hilbert Space. Then the following two statements are equivalent to each other.

$(a)$ $x^{\ast}$ is a linear functional defined on $X$.

$(b)$ For all $y\in X$, there exists a unique $x\in X$ that satisfies $x^{\ast}(y)=\langle x,\ y\rangle_{X}$. Also, $| x^{\ast}\ ; X^{\ast}|=|x\ ;X|$. $\langle \cdot,\ \cdot \rangle_{X}$ is the inner product defined in $X$.

The Riesz Representation Theorem explains what relationship exists between $X$, a Hilbert space, and its dual $X^{\ast}$. In simple terms, given a linear functional $x^{\ast} \in X^{\ast}$, there exists a unique $x\in X$ corresponding to it, meaning that the values of $x^{\ast}(y)$ and $\langle x,\ y \rangle_{X}$ are the same for all $y \in X$.

Auxiliary Theorem

Let $1 < p < \infty$.

• $(c)$ If $L\in ({L}^{\ p})^{\ast}$ and $| L\ ; ({L}^{\ p})^{\ast}|=1$, there exists a unique $w \in {L}^{\ p}$ satisfying $\left\| w \right\|_{p}=L(w)=1$.

• $(d)$ Conversely, if $w \in {L}^{\ p}$ and $\left\| w \right\|_{p}=1$, there exists a unique $L \in ({L}^{\ p})^{\ast}$ satisfying $|L\ ;({L}^{\ p})^{\ast}|=L(w)=1$.

Proof

To explain the flow of the proof simply: $|L|=1$ let’s assume $\implies$ By the auxiliary theorem $(c)$, a unique $w$ exists $\implies$ By the auxiliary theorem $(d)$, a unique $L$ exists $\implies$ The defined $v$ satisfies the same properties as $L$ $\implies$ Since $L$ is unique, $L=L_{v}$ applies, and so does the rest of the theorem

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• Part 1 $L=0$

  If we set $v=0$, the theorem is satisfied.

• Part 2 $L\ne 0$

  Let’s assume that any $L$ satisfies $|L\ ; ({L}^{\ p})^{\ast}|=\alpha$. Then, by multiplying the constant $\frac{1}{\alpha}$, the norm can be made to be 1, and let’s call $\frac{1}{\alpha}L$ again $L$. Based on this process, we can assume $|L\ ; ({L}^{\ p})^{\ast}|=1$ without loss of generality. Then, the auxiliary theorem $(c)$ exists a $w \in {L}^{\ p}$ that satisfies the following equation.

  $$\left\| w \right\|_{p}=1,\quad L(w)=1$$

  And let’s define $v$ as follows.

  $$v(x) = \begin{cases} |w(x)|^{p-2}\overline{w(x)}, & w(x)\ne0 \\ 0, & \mathrm{otherwise} \end{cases}$$

  Then, it can be seen through the following calculation that it is $v \in {L}^{\ p^{\prime}}$.

  $$\begin{align*} \left\| v \right\|_{p^{\prime}}^{p^{\prime}} =&\ \int \left| |w(x)|^{p-2}\overline{w(x)} \right| ^{p^{\prime}} dx \\ =&\ \int |w(x)|^{(p-1)p^{\prime}} dx \\ =&\ \int |w(x)|^p dx \\ =&\ \left\| w \right\|_{p}^p =1< \infty \quad \cdots (1) \end{align*}$$

  And let’s define $L_{v}$ as follows. Demonstrating that it satisfies $|L_{v}|=L_{v}(w)=1$, we aim to show $L=L_{v}$.

  $$L_{v}(u)=\int u(x)v(x)dx,\quad u\in {L}^{\ p}$$

  According to the definition of the norm of Dual Norm,

  $$\begin{align*} |L_{v}(u)| \le& \int |u(x)v(x)| dx \\ \le& \| u \|_{p}\ \left\| v \right\|_{p^{\prime}} \\ =&\ \left\| v \right\|_{p^{\prime}}=1 \end{align*}$$

  The second line is due to the Hölder’s inequality, and the last line is valid due to the condition $\left\| u \right\| \le 1$ and $(1)$. Therefore, $|L_{v}\ ; ({L}^{\ p})^{\ast}|=1$. Also, it can be confirmed that $L_{v}(w)=1$ is satisfied.

  $$\begin{align*} L_{v}(w)=&\ \int |w(x)|^{p-2}\overline{w(x)}w(x) dx \\ =&\ \int |w(x)|^p dx \\ =&\ \left\| w \right\|_{p}^p=1 \end{align*}$$

  Therefore, since the auxiliary theorem $(d)$ asserts that the $L$ satisfying $|L|=L(w)=1$ is unique, $L=L_{v}$. Thus, for $L\in ({L}^{\ p})^{\ast}$, a unique $v \in {L}^{\ p^{\prime}}$ exists

  $$L(u)=L_{v}(u)=\int u(x)v(x)dx$$

  satisfying it, and $\left\| v \right\|_{p^{\prime}}=|L; ({L}^{\ p})^{\ast}|$ is true.

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