Inequality of Hölder Means

Theorem
Let $p, q \in \overline \R$ be extended real numbers such that $p < q$.

Let $x_1, x_2, \ldots, x_n \ge 0$ be real numbers.

If $p < 0$, then we require that $x_1, x_2, \ldots, x_n \ge 0$.

Then the Hölder mean satisfies the inequality:
 * $\map {M_p} {x_1, x_2, \ldots, x_n} \le \map {M_q} {x_1, x_2, \ldots, x_n}$

Equality holds :
 * $x_1 = x_2 = \cdots = x_n$

or:
 * $q < 0$ and $x_k = 0$ for at least one $k \in \set {1, 2, \ldots, n}$.

Note that in particular:
 * $\forall p \in \R: \map {M_{-\infty} } {x_1, x_2, \ldots, x_n} \le \map {M_p} {x_1, x_2, \ldots, x_n}$

and:
 * $\forall p \in \R: \map {M_\infty} {x_1, x_2, \ldots, x_n} \ge \map {M_p} {x_1, x_2, \ldots, x_n}$

Proof
For real $p \ne 0$, the Hölder mean is defined as:
 * $\ds \map {M_p} {x_1, x_2, \ldots, x_n} = \paren {\frac 1 n \sum_{k \mathop = 1}^n {x_k}^p}^{1 / p}$

whenever the above expression is defined.

First we note that by definition of Hölder mean with $p = \infty$:
 * $\map {M_\infty} {x_1, x_2, \ldots, x_n} = \max \set {x_1, x_2, \ldots, x_n}$

This is justified by Limit of Hölder Mean as Exponent tends to Infinity:
 * $\ds \lim_{p \mathop \to +\infty} \map {M_p} {x_1, x_2, \ldots, x_n} = \max \set {x_1, x_2, \ldots, x_n}$

From Maximum is Greater than or Equal to Hölder Mean:
 * $\max \set {x_1, x_2, \ldots, x_n} \ge \map {M_p} {x_1, x_2, \ldots, x_n}$

and so:
 * $\forall p \in \R: \map {M_\infty} {x_1, x_2, \ldots, x_n} \ge \map {M_p} {x_1, x_2, \ldots, x_n}$

Similarly, by definition of Hölder mean with $p = -\infty$:
 * $\map {M_{-\infty} } {x_1, x_2, \ldots, x_n} = \min \set {x_1, x_2, \ldots, x_n}$

This is justified by Limit of Hölder Mean as Exponent tends to Negative Infinity:
 * $\ds \lim_{p \mathop \to -\infty} \map {M_p} {x_1, x_2, \ldots, x_n} = \min \set {x_1, x_2, \ldots, x_n}$

From Minimum is Less than or Equal to Hölder Mean:
 * $\min \set {x_1, x_2, \ldots, x_n} \le \map {M_p} {x_1, x_2, \ldots, x_n}$

and so:
 * $\forall p \in \R: \map {M_{-\infty} } {x_1, x_2, \ldots, x_n} \le \map {M_p} {x_1, x_2, \ldots, x_n}$

Let either $p = 0$ or $q = 0$.

By definition of Hölder mean with $p = 0$:
 * $\map {M_0} {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$

where $G$ denotes the geometric mean.

This is justified by Limit of Hölder Mean as Exponent tends to Zero is Geometric Mean:
 * $\ds \lim_{p \mathop \to 0} \map {M_p} {x_1, x_2, \ldots, x_n} = \map G {x_1, x_2, \ldots, x_n}$

It remains to resolve the inequality for $p, q \in \R_{\ge 0}$.

Consider the function $\phi: \R_{\ge 0} \to \R_{\ge 0}$ defined as:
 * $\forall x \in \R_{\ge 0}: \map \phi x = x^{q/p}$

By the Power Rule for Derivatives:
 * $\map {D_x} {x^{q / p} } = \dfrac q p x^{q / p - 1}$

From Real Function is Strictly Convex iff Derivative is Strictly Increasing:
 * $\phi$ is strictly convex if $q > 0$.

From Real Function is Strictly Concave iff Derivative is Strictly Decreasing:
 * $\phi$ is strictly concave if $q < 0$.

Now apply Jensen's inequality to ${x_1}^p, {x_2}^p, \ldots, {x_n}^p$.

For $q > 0$, this gives:
 * $\ds \paren {\frac 1 n \sum_{k \mathop = 1}^n {x_k}^p}^{q/p} \le \frac 1 n \sum_{k \mathop = 1}^n {x_k}^q$

For $q < 0$, the reverse inequality holds.

Also by Jensen's inequality, equality holds $x_1 = x_2 = \cdots = x_n$.

In either case, the result follows.