---
tag: 1V11QM5C07
kind: statement
role: theorem
name: Clarkson's inequalities
msc: ["46E30", 46B20]
uses: [SASP14MA1Z, P15VQS5DJ9, 026HPGYGS4, BEAMC5011M]
about: [SASP14MA1Z]
sources:
  - url: https://en.wikipedia.org/wiki/Clarkson%27s_inequalities
same_as:
  wikidata: Q5127621
  wikipedia: Clarkson's inequalities
---

## Statement

Let $(\Omega, \Sigma, \mu)$ be a [[measure space|P15VQS5DJ9]] and let $f, g \in L^p(\mu)$, the [[Lp space|SASP14MA1Z]].
Write $\norm{\cdot}$ for the norm of $L^p(\mu)$.

1. If $2 \le p < \infty$, then
$$\Big\| \frac{f + g}{2} \Big\|^p + \Big\| \frac{f - g}{2} \Big\|^p \le \frac{1}{2} \big( \norm{f}^p + \norm{g}^p \big).$$
2. If $1 < p \le 2$ and $q = p/(p-1)$, then
$$\Big\| \frac{f + g}{2} \Big\|^q + \Big\| \frac{f - g}{2} \Big\|^q \le \Big( \frac{1}{2} \norm{f}^p + \frac{1}{2} \norm{g}^p \Big)^{q/p}.$$

## Remarks

The first inequality follows from the same inequality for numbers $a, b$, integrated over $\Omega$.
For numbers, with $\alpha = (a+b)/2$ and $\beta = (a-b)/2$, one has $\abs{\alpha}^p + \abs{\beta}^p \le (\abs{\alpha}^2 + \abs{\beta}^2)^{p/2}$ since $p \ge 2$.
The [[parallelogram law|026HPGYGS4]] gives $\abs{\alpha}^2 + \abs{\beta}^2 = (\abs{a}^2 + \abs{b}^2)/2$.
Convexity of $t \mapsto t^{p/2}$ then bounds $((\abs{a}^2 + \abs{b}^2)/2)^{p/2}$ by $(\abs{a}^p + \abs{b}^p)/2$.
The second inequality needs a finer argument.

For $p = 2$ both inequalities are the parallelogram law, with equality.
Clarkson proved them in 1936 to show that $L^p(\mu)$ is [[uniformly convex|BEAMC5011M]] for $1 < p < \infty$.
If $\norm{f} = \norm{g} = 1$ and $\norm{f - g} \ge \varepsilon$, then for $p \ge 2$ the first inequality gives $\norm{(f+g)/2}^p \le 1 - (\varepsilon/2)^p$.

## Explanations

### L1 human

In a space of functions with the $L^p$ norm, two unit vectors that are far apart have a midpoint that is shorter than $1$ by a definite amount.
Clarkson's inequalities make this precise with explicit numbers.
They generalise the parallelogram law of Euclidean geometry.

### L2 human

The inequalities are the standard route to uniform convexity of $L^p$ for $1 < p < \infty$.
By the Milman–Pettis theorem, uniform convexity then gives reflexivity of $L^p$.
Through uniform convexity, they can also be used to prove the duality of $L^p$ and $L^q$ without the Radon–Nikodym theorem.
The spaces $L^1$ and $L^\infty$ are not uniformly convex, and the inequalities have no analogue there.