# Prove variant of triangle inequality containing p-th power for 0 < p < 1

Sorry if this is a trivial question, but I am kind of stuck with proving the following inequality and have been searching for a while:

$\rho \left( \sum\limits_i^n d_i \right) \leq \sum\limits_i^n \;\rho(d_i)\;$ with $\rho(d_k) := |d_k|^p, \quad 0 < p < 1$ and $d_i \in \mathbb{R}$.

I encountered this situation while dealing with an energy-based regularization approach. While I do not expect this inequality to hold for $p > 1$, I basically believe that it does so for $0 < p < 1$, implying that the energy $\rho$ for the sum of all $d_i$ is lower than the sum of energies of the individual $d_i$, under such a quasi-norm. I attempted to show this for the less general case where $p := \frac{q}{r}$ is a rational number, by raising both sides to the $r$-th power:

$\mid\sum\limits_i^n d_i\mid^p \leq \sum\limits_i^n \mid d_i \mid^p \; \Leftrightarrow \; \mid\sum\limits_i^n d_i\mid^q \leq \left( \sum\limits_i^n \mid d_i \mid^p \right)^r = \sum\limits_i^n \mid d_i \mid^{pr} + u = \sum\limits_i^n \mid d_i \mid^{q} + u$,

where $u$ contains a sum of mixed terms. Next, I tried to get rid of the $q$ exponent in order to somehow make use of the triangle inequality. However, it seems like I am going in circles here.

Could somebody give me a hint on how to better approach this? Is there an existing inequality that I could use in proving this? I guess these are basics that have been studied before, I am just not sure what keywords to search for. Thank you very much!

A function $f: \mathbb R \to \mathbb R$ is called subadditive if $f(x+y)\le f(x)+f(y)$ for each $x,y\in\mathbb R$. You are asking whether the function $f(x)=|x|^p$ is subadditive for $0<p<1$.

According to wikipedia article every concave function with $f(0)=0$ is subadditive. The proof from wikipedia:
$f(x) \ge \frac{y}{x+y} f(0) + \frac{x}{x+y} f(x+y)$
$f(y) \ge \frac{x}{x+y} f(0) + \frac{y}{x+y} f(x+y)$.
By adding these inequalites together you get $$f(x)+f(y)\ge f(x+y).$$ If I understood the proof given at wikipedia correctly, this works only for $x,y\ge 0$.

But since your function has property $f(x)=f(|x|)$ and it is increasing on $[0,\infty)$, I believe this should suffice in your case. Namely if we already have subadditivity for positive $x$ and $y$, we get: $$f(x+y) = f(|x+y|) \le f(|x|+|y|) \le f(|x|) + f(|y|) = f(x)+f(y).$$

So you only need to verify that your function is concave.

Here's another reference for a similar result:

Functional analysis and semi-groups By Einar Hille, Ralph S. Phillips, Theorem 7.2.5:

A necessary and sufficient condition that a measurable concave function $f(t)$ be subadditive on $(0,\infty)$ is that $f(0+)\ge 0$.

• Great, thank you very much! Subadditivity is the term I was looking for. Nov 8 '11 at 13:23
• The same result is also in Eric Schechter's Handbook of analysis and its foundations, 12.15c, p.314. Nov 25 '11 at 8:41