📂Lebesgue Spaces

Embedding Theorems in Lp Spaces

Theorem¹

If $\Omega \subset \mathbb{R}^{n}$ is an open set and let’s assume $\text{vol}(\Omega) = \int_{\Omega} 1 dx \lt \infty$.

(a) For $1 \le p \le q \le \infty$, if $u \in L^{q}(\Omega)$ then, $u \in L^{p}(\Omega)$ and

$$ \begin{equation} \left\| u \right\|_{p} \le \left( \text{vol}(\Omega) \right)^{\frac{1}{p} - \frac{1}{q}} \left\| u \right\|_{q} \end{equation} $$

And $L^{q}$ is embedded into $L^{p}$.

$$ \begin{equation} L^{q}(\Omega) \to L^{p}(\Omega) \end{equation} $$

(b) For $1 \le p \le q \le \infty$, if $u \in L^{\infty}(\Omega)$ then,

$$ \begin{equation} \lim \limits_{p \to \infty} \left\| u \right\|_{p} = \left\| u \right\|_{\infty} \end{equation} $$

(c) For all $1 \le p \lt \infty$, if $u \in L^{p}(\Omega)$ and there exists $K$ that is $\left\| u \right\|_{p} \le K$,

$$ \begin{equation} u \in L^{\infty}(\Omega) \quad \text{and} \quad \left\| u \right\|_{\infty} \le K \end{equation} $$

Explanation

(a) When $1 \le p \le q$, generally, there is no inclusion relation between the space of $L^{p}$ and $L^{q}$. However, if the domain’s volume is finite, then $L^{q} \subset L^{p}$ holds.

(b) Since the above holds when $q = \infty$, if $u \in L^{\infty}$, for all $1 \le p \lt \infty$, $u \in L^{p}$ and the limit of $\left\| \cdot \right\| _{p}$ converges to $\left\| \cdot \right\|_{\infty}$.

(c) The assumption about constant $K$ does not mean there exists one for all $p$ respectively, but means one $K$ satisfies $\left\| u \right\|_{p} \le K$ for all $p$. As (a) and (b) suggest, the larger $p$ becomes, it turns into a smaller space, which can be considered as ${L^{\infty}(\Omega) = \bigcap\limits_{1 \le p \lt \infty} L^{p}(\Omega)}$.

Proof

(a)

If either $p = q$ or $q = \infty$ holds, it is trivial that $(1)$ and $(2)$ are true. So, let’s assume $1 \le p \lt q \lt \infty$ and $u \in L^{q}(\Omega)$.

Lemma: Generalized Hölder’s Inequality

If three constants $\alpha \gt 0, \beta \gt 0, \gamma \gt 0$ satisfy $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{1}{\gamma}$ and $f \in {L}^{\alpha}(\Omega), g \in {L}^{\beta}(\Omega)$, then $fg \in L^{\gamma}(\Omega)$ and the following inequality holds.

$$ \| fg \|_{\gamma} = \left( \int_{\Omega} |f(x)g(x)|^{\gamma} dx \right)^{1 / \gamma} \le \| f \|_{\alpha} \| g \|_{\beta} $$

Substituting $\alpha = q, \beta = \dfrac{1}{p}-\dfrac{1}{q}, \gamma = p$, $f = u$, and $g = 1$ into the lemma above gives $u \in L^{q}(\Omega), 1 \in L^{\frac{1}{p}-\frac{1}{q}}(\Omega)$, therefore,

$$ \begin{align*} && \left( \int_{\Omega} \left| u(x) \cdot 1 \right|^{p} dx \right)^{1 / p} \le& \left\| u \right\|_{q} \left\| 1 \right\|_{\frac{1}{p} - \frac{1}{q}} \\ \implies && \left\| u \right\|_{p} \le& \left( \int_{\Omega} 1 dx \right)^{\frac{1}{p} - \frac{1}{q}} \left\| u \right\|_{q} \\ && =&\ \left( \text{vol}(\Omega) \right)^{\frac{1}{p} - \frac{1}{q}} \left\| u \right\|_{q} \end{align*} $$

Thus, $u \in L^{p}(\Omega)$ holds.

Embedding

• $X$ is a subspace of $Y$.
• $\exists M \gt 0 \text{ such that } \left\| Ix \right\|_{{Y}} \le M \left\| x \right\|_{X},\quad x \in X$

Also, if we let $M = \left( \text{vol}(\Omega) \right)^{\frac{1}{p} - \frac{1}{q}}$

$$ \left\| u \right\|_{p} \le M \left\| u \right\|_{q} $$

Therefore, by the definition of embedding, we can see that $L^{q}(\Omega) \to L^{p}(\Omega)$ is an embedding.

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(b)

From the result of (a), the following holds.

$$ \limsup _{p \to \infty} \left\| u \right\|_{p} \le \left\| u \right\|_{\infty} $$

Meanwhile, for any $\epsilon \gt 0$, there exists a set $A \subset \Omega$ with a positive measure $\mu (A)$ satisfying the following condition.

$$ \left| u(x) \right| \ge \left\| u \right\|_{\infty} - \epsilon,\quad \text{if } x \in A $$

If such $A$ does not exist, $\left\| u \right\|_{\infty}$ would not satisfy the definition of $\left\| \cdot \right\|_{\infty}$, hence the existence of $A$ is guaranteed. Therefore, the following formula holds.

$$ \int_{\Omega} \left| u(x) \right|^{p} dx \ge \int_{A} \left| u(x) \right|^{p} dx \ge \int_{A} \left( \left\| u \right\|_{\infty} - \epsilon \right)^{p} dx \ge \mu (A) \left( \left\| u \right\|_{\infty} - \epsilon \right)^{p} $$

Thus, we obtain the following.

$$ \left\| u \right\|_{p} \ge \left( \mu (A) \right)^{1/p} \left( \left\| u \right\|_{\infty} - \epsilon \right) $$

This must hold for all $\epsilon$ and any corresponding $A$, thus

$$ \liminf \limits_{p \to \infty} \left\| u \right\|_{p} \ge \left\| u \right\|_{\infty} $$

Therefore,

$$ \liminf \limits_{p \to \infty} \left\| u \right\|_{p} \le \limsup _{p \to \infty} \left\| u \right\|_{p} \le \left\| u \right\|_{\infty} \le \liminf \limits_{p \to \infty} \left\| u \right\|_{p} \le \limsup _{p \to \infty} \left\| u \right\|_{p} $$

$$ \implies \liminf _{p \to \infty} \left\| u \right\|_{p} = \left\| u \right\|_{\infty} = \limsup \limits_{p \to \infty} \left\| u \right\|_{p} $$

$$ \implies \lim \limits_{p \to \infty} \left\| u \right\|_{p} = \left\| u \right\|_{\infty} $$

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(c)

Let’s assume for all $1 \le p \lt \infty$, there exists $K$ that is $\left\| u \right\|_{p} \le K$. And suppose either $u \in L^{\infty}(\Omega)$ or $\left\| u \right\|_{\infty} \le K$ does not hold. Then, according to the definition of $\left\| \cdot \right\|_{\infty}$, we can find a constant $K_{1}$ and a set $A \subset \Omega$ that satisfies the following.

$$ K_{1} \gt K \quad \text{and} \quad \mu (A) \gt 0 \quad \text{and} \quad \left| u(x) \right| \gt K_{1} \text{ for } x \in A $$

Then, as shown in the proof of (b), the following holds.

$$ \liminf \limits_{p \to \infty} \left\| u \right\|_{p} \ge K_{1} \gt K $$

This contradicts the assumption that for all $p$, $\left\| u \right\|_{p} \le K$ holds. Therefore,

$$ u \in L^{\infty}(\Omega) \quad \text{and} \quad \left\| u \right\|_{\infty} \le K $$

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