Clarkson's inequalities

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 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 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 $\norm{(f+g)/2}^p \le 1 - (\varepsilon/2)^p$.

Going deeper

For a student

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.