📂Lebesgue Spaces

Inverse Holder's Inequality: A Sufficient Condition for Lp Functions

Theorem¹

Let’s consider $\Omega \subset \mathbb{R}^{n}$ as an open set. The necessary and sufficient condition for the measurable function $u$ to be included in the $L^{p}$ space is

$$\sup \left\{ \int_{\Omega} \left| u(x) \right| v(x) dx : v(x) \ge 0 \text{ on } \Omega, \left\| v \right\|_{p^{\prime}} \le 1 \right\} \lt \infty$$

Furthermore, the above supremum is as follows $\left\| u \right\|_{p}$. Here, $p^{\prime} = \dfrac{p}{p-1}$ is the Hölder conjugate.

Explanation

Assuming $1 \lt p \lt \infty$ and $v \in L^{p^{\prime}}$, then the Hölder’s inequality tells us that if $u \in L^{p}$, then $uv \in L^{1}$.

$$u\in L^{p}(\Omega) \implies \int_{\Omega} \left| u(x) v(x) \right| dx \le \left\| u \right\|_{p} \left\| v \right\|_{p^{\prime}}$$

Conversely, the theorem implies that if $uv \in L^{1}$, then $u \in L^{p}$.

$$u\in L^{p}(\Omega) \impliedby \sup \left\{ \int_{\Omega} \left| u(x) \right| v(x) dx : v(x) \ge 0 \text{ on } \Omega, \left\| v \right\|_{p^{\prime}} \le 1 \right\} \lt \infty$$

Proof

$$u \in L^{p}(\Omega) \iff \sup \left\{ \int_{\Omega} \left| u(x) \right| v(x) dx : v(x) \ge 0 \text{ on } \Omega, \left\| v \right\|_{p^{\prime}} \le 1 \right\} \lt \infty$$

• $(\implies)$

  The case of $\left\| u \right\|_{p} = 0$ is trivial. Let’s assume $0 \lt \left\| u \right\|_{p} \lt \infty$. Applying Hölder’s inequality for $v \in L^{p^{\prime}}$ where $0 \le v$ and $\left\| v \right\|_{q} \le 1$ gives,

  $$\int_{\Omega} \left| u(x) \right| v(x) dx \le \left\| u \right\|_{p} \left\| v \right\|_{p^{\prime}} \le \left\| u \right\|_{p} \lt \infty$$

  Also, setting $v = \left( \dfrac{ \left| u \right| }{ \left\| u \right\|_{p} } \right)^{p/p^{\prime}}$, we have $\left\| v \right\|_{p^{\prime}} = 1$ and equality holds. Since $p/p^{\prime} = p \dfrac{p-1}{p} = p-1$,

  $$\begin{align*} \int_{\Omega} \left| u(x) \right| v(x) dx =&\ \int_{\Omega} \left| u(x) \right| \dfrac{ \left| u(x) \right|^{p/p^{\prime}} }{ \left\| u \right\|_{p}^{p/p^{\prime}} } dx \\ =&\ \int_{\Omega} \left| u(x) \right| \dfrac{ \left| u(x) \right|^{p-1} }{ \left\| u \right\|_{p}^{p-1} } dx \\ =&\ \dfrac{1}{ \left\| u \right\|_{p}^{p-1} } \int_{\Omega} \left| u(x) \right|^{p} dx \\ =&\ \dfrac{1}{ \left\| u \right\|_{p}^{p-1} } \left\| u \right\|_{p}^{p} \\ =&\ \left\| u \right\|_{p} \end{align*}$$

  Hence,

  $$\sup \left\{ \int_{\Omega} \left| u(x) \right| v(x) dx : v(x) \ge 0 \text{ on } \Omega, \left\| v \right\|_{p^{\prime}} \le 1 \right\} = \left\| u \right\|_{p} \lt \infty$$

• $(\impliedby)$

  Prove by contraposition. That is, we aim to show the following:

  $$\left\| u \right\|_{p} = \infty \implies \sup \left\{ \int_{\Omega} \left| u(x) \right| v(x) dx : v(x) \ge 0 \text{ on } \Omega, \left\| v \right\|_{p^{\prime}} \le 1 \right\} = \infty$$

  ---

  Assume $\left\| u \right\|_{p} = \infty$. Then, for some simple functions $s_{j}$ forming an increasing sequence $\left\{ s_{j} \right\}$ that satisfies $\Omega$ on top, we can think of $0 \le s_{j}(x) \le \left| u(x) \right|$. This satisfies $\lim \limits_{j \to \infty} \left\| s_{j} \right\|_{p} = \infty$. Now, let’s set $v_{j} = \left( \dfrac{ \left| s_{j} \right| }{ \left\| s_{j} \right\|_{p} } \right)^{p/p^{\prime}}$. Then $v_{j} \ge 0$ and the following $\left\| v_{j} \right\|_{p^{\prime}} = 1$ is satisfied.

  $$\begin{align*} \left\| v_{j} \right\|_{p^{\prime}}^{p^{\prime}} =&\ \int_{\Omega} \left( \dfrac{ \left| s_{j} \right| }{ \left\| s_{j} \right\|_{p} } \right)^{p} dx \\ =&\ \dfrac{1}{ \left\| s_{j} \right\|_{p}^{p} } \int_{\Omega} \left| s_{j} \right|^{p} dx \\ =&\ 1 \end{align*}$$

  Moreover, since $p/p^{\prime} = p \dfrac{p-1}{p} = p-1$, the following equation holds.

  $$\int_{\Omega} s_{j}(x) v_{j}(x) dx = \dfrac{1}{\left\| s_{j} \right\|_{p}^{p-1}}\int_{\Omega} \left| s_{j}(x) \right|^{p} dx = \left\| s_{j} \right\|_{p}$$

  Then, the following also holds.

  $$\int_{\Omega} \left| u(x) \right| v_{j}(x) dx \ge \int_{\Omega} s_{j}(x) v_{j}(x) dx = \left\| s_{j} \right\|_{p}$$

  Therefore, since $\lim \limits_{j \to \infty} \left\| s_{j} \right\|_{p} = \infty$,

  $$\sup \left\{ \int_{\Omega} \left| u(x) \right| v(x) dx : v(x) \ge 0 \text{ on } \Omega, \left\| v \right\|_{p^{\prime}} \le 1 \right\} = \infty$$

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