📂Lebesgue Spaces

Linear Functionals on Lp Spaces

Definition¹

Let $\Omega \subset \mathbb{R}^{n}$ be an open set. Suppose that $1 \le p \le \infty$ and $p^{\prime}=\frac{p}{p-1}$. For each $v \in L^{p^{\prime}}(\Omega)$, define the linear functional $L_{v}\ :\ L^p(\Omega) \rightarrow \mathbb{C}$ on the space $L^p(\Omega)$ as follows.

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

Theorem

Let the norm on the space $L^p$ be denoted by $\| \cdot \|_{p}$. Then, by the Hölder’s inequality, the following inequality holds.

$$ \left\| L_{v}(u) \right\| \le \left\| u \right\|_{p} \left\| v \right\|_{p^{\prime}} $$

Then the norm of $L_{v}$ satisfies the following inequality.

$$ \left\|L_{v}; (L^p)^{\ast} \right\| :=\sup \left\{ |L_{v}(u)| \ :\ \left\| u \right\|\le1 \right\} \le \left\| v \right\|_{p^{\prime}} $$

Here, $(L^p)^{\ast}$ is the dual of $L^p$. In fact, equality can be shown to hold.

Proof

• Case 1. $1<p\le \infty$

  Let $u(x)$ be as follows.

  $$ u(x) =\begin{cases} |v(x)|^{p^{\prime}-2}\overline{v(x)} & \mathrm{if}\ v(x)\ne 0 \\ 0 & \mathrm{otherwise} \end{cases} $$

  Then, it can be shown that $u(x)\in L^p$ as follows.

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

  Since $\frac{1}{p}+\frac{1}{p^{\prime}}=1$, by organizing properly, one can obtain $pp^{\prime}-p=p^{\prime}$. Using this, the fourth equation holds. Also, since $v\in L^{p^{\prime}}$, the last inequality holds. Since $u \in L^p$, substituting into $L_{v}$ gives

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

  Since $\frac{1}{p}+\frac{1}{p^{\prime}}=1$, by organizing properly, one can obtain $\frac{p^{\prime}}{p}=p^{\prime}-1$. Using this, the sixth equation holds. Also, the last equation holds by using the result obtained above, $\left\| u \right\|_{p}=\left\| v \right\|_{p^{\prime}}^{\frac{p^{\prime}}{p}}$. Therefore

  $$ | L_{v}(u)| = \left\| u \right\|_{p} \left\| v \right\|_{p^{\prime}}, \quad | L_{v}; (L^p)^{\ast}|=\left\| v \right\|_{p^{\prime}} $$

• Case 2. $p=1$

  If $p=1$, then $p^{\prime}=\infty$. Initially, if $\left\| v \right\|_{p^{\prime}}=\left\| v \right\|_{\infty}=0$, let’s say $u(x)=0$. Then the equation holds. Otherwise, suppose $0 < \epsilon < \left\| v \right\|_{\infty}$. Also, let $A$ be a measurable subset of $\Omega$ for which $0< \mu (A) < \infty$ holds, and suppose $|v(x)| \ge \left\| v \right\|_{\infty} -\epsilon$ holds on $A$. Now, let’s say $u(x)$ as follows.

  $$ u(x)= \begin{cases} \overline{v(x)}/|v(x)| & \mathrm{on}\ A \\ 0 & \mathrm{otherwise} \end{cases} $$ Then, one can see that $u\in L^p=L^1$.

  $$ \begin{align*} \left\| u \right\|_{1} =&\ \int_{A} \left| \overline{v(x)}/|v(x)| \right| dx \\ =&\ \int_{A}dx= \mu (A) < \infty \end{align*} $$

  Since $u \in L^1$, substituting into $L_{v}$ gives

  $$ \begin{align*} L_{v}(u) =&\ \int u(x)v(x)dx \\ =&\ \int |v| dx \\ \ge& \int (\left\| v \right\|_{\infty}-\epsilon )dx \\ =&\ (\left\| v \right\|_{\infty}-\epsilon)\int dx \\ =&\ (\left\| v \right\|_{\infty}-\epsilon) \mu (A) \\ =&\ \left\| u \right\|_{1}\ ( \left\| v \right\|_{\infty} -\epsilon ) \end{align*} $$

  Since $| L_{v}; (L^p)^{\ast}|=\sup \left\{ |L_{v}(u) |\ :\ \left\| u \right\|_{1} \le 1\right\}$, by applying $\sup\limits_{\left\| u \right\|\le 1}$ to both sides of the obtained $|L_{v}(u)| \ge \left\| u \right\|_{1}\ ( \left\| v \right\|_{\infty}-\epsilon)$,

  $$ | L_{v};\ (L^p)^{\ast} | \le \left\| u \right\|_{1}\ ( \left\| v \right\|_\infty -\epsilon ) \le \left\| v \right\|_{\infty}-\epsilon $$

  Therefore

  $$ \left\| v \right\|_{\infty}-\epsilon \le | L_{v} | \le \left\| v \right\|_{\infty} $$

  And since this holds for any $\epsilon$,

  $$ | L_{v}; (L^P)^{\ast}| =\left\| v \right\|_{\infty} $$

■