求证L^p范数相关比值在p→∞时取得最大值

Let's break this problem down step by step. We're working with an essentially bounded function $u$ on a compact real interval $X$, so by definition $|u|\infty = \text{ess sup}{x \in X} |u(x)| < \infty$. Our goal is to show the ratio $$R(p) = \frac{\int_X |u|^{p+1} , dx}{\int_X |u|^{p} , dx}$$ reaches its maximum value as $p$ tends to infinity.

Step 1: Establish an upper bound for $R(p)$ #

First, since $|u(x)| \leq |u|\infty$ for almost every $x \in X$, multiply both sides by the non-negative term $|u(x)|^p$ and integrate over $X$:
$$\int_X |u|^{p+1} , dx \leq |u|\infty \int_X |u|^p , dx$$
Dividing both sides by $\int_X |u|^p , dx$ (which is positive, as $u$ isn't essentially zero—if it were, the ratio is trivially zero), we get:
$$R(p) \leq |u|_\infty \quad \text{for all finite } p > 0$$

Step 2: Show the limit of $R(p)$ as $p \to \infty$ equals $|u|_\infty$ #

Let $M = |u|\infty$. By the definition of essential supremum, for any $\epsilon > 0$, there exists a measurable subset $E\epsilon \subseteq X$ with positive measure $m(E_\epsilon) > 0$ such that $|u(x)| \geq M - \epsilon$ for almost every $x \in E_\epsilon$.

Estimate the numerator and denominator of $R(p)$ using this set:

• Denominator: $\int_X |u|^p , dx \geq \int_{E_\epsilon} (M - \epsilon)^p , dx = (M - \epsilon)^p m(E_\epsilon)$
• Numerator: $\int_X |u|^{p+1} , dx \geq \int_{E_\epsilon} (M - \epsilon)^{p+1} , dx = (M - \epsilon)^{p+1} m(E_\epsilon)$

Dividing these inequalities gives:
$$R(p) \geq \frac{(M - \epsilon)^{p+1} m(E_\epsilon)}{(M - \epsilon)^p m(E_\epsilon)} = M - \epsilon$$

Since $\epsilon > 0$ was arbitrary, combining this with our upper bound from Step 1, we conclude:
$$\lim_{p \to \infty} R(p) = M = |u|_\infty$$

Step 3: Conclude the maximum is attained as $p \to \infty$ #

We've shown two critical points:

• For every finite $p$, $R(p)$ can never exceed $|u|_\infty$
• As $p$ grows without bound, $R(p)$ approaches $|u|_\infty$ exactly

This means the supremum of $R(p)$ over all $p > 0$ is $|u|_\infty$, and this value is the limit as $p \to \infty$. In other words, the ratio reaches its maximum value in the limit as $p$ tends to infinity.

Note: If $u$ is essentially constant (i.e., $|u(x)| = M$ almost everywhere on $X$), then $R(p) = M$ for all $p$, so the maximum is attained for every $p$—this is just a special case of our general result.

内容的提问来源于stack exchange，提问作者Jon Warneke