1
$\begingroup$

Let us suppose that $0 \leq p \leq 1$. All variables are assumed to be non-negative.

The function $x \mapsto x^{p+1}$ is strictly convex upwards, so $$ \left( \frac1n \sum_{i=1}^n x_i^{p+1} \right) \geq \left( \frac1n \sum_{i=1}^n x_i \right)^{p+1} $$ with equality iff the $x_i$ are all equal; while $x \mapsto x^p$ is convex downwards, so $$ \left( \frac1n \sum_{i=1}^n x_i^p \right) \leq \left( \frac1n \sum_{i=1}^n x_i \right)^p $$ with equality if the $x_i$ are all equal.

I am not really sure why this is the case. Can anyone explain this more? I try to follow from the definition of convex/concave functions, but it's not working out.

$\endgroup$
1
$\begingroup$

I try to follow from the definition of convex/concave functions, but it's not working out.

Yes, it is. If the function $u$ is convex, then $$ u\left(\frac1n\sum_{k=1}^nx_k\right)\leqslant\frac1n\sum_{k=1}^nu(x_k). $$ For $u:x\mapsto x^{p+1}$ on $x\geqslant0$ this is your first inequality. For $u:x\mapsto-x^{p}$ on $x\geqslant0$ this is your second inequality.

$\endgroup$
  • $\begingroup$ Quite often the definition given for a convex function is the simpler inequality $f(t x + (1-t) y) \le t f(x) + (1-t) f(y)$. $\endgroup$ – J. J. Oct 15 '13 at 13:19
  • $\begingroup$ @J.J. This might be why the OP is stuck. Yet another example of the "advantage" of putting nothing into a question except its flat statement... $\endgroup$ – Did Oct 15 '13 at 13:21
0
$\begingroup$

Check out Jensen's inequality. For a convex function $f$ we have $$f\left(\frac{x_1 + \dots + x_n}{n}\right) \le \frac{f(x_1) + \dots + f(x_n)}{n}$$ and the inequality sign is reversed for concave functions.

$\endgroup$
0
$\begingroup$

A common definition is that $\phi: [a, b] \to \mathbb{R}$ is convex if $\phi(t x + (1-t)y) \leq t\phi(x) + (1-t)\phi(y)$ whenever $0 \leq t \leq 1$ and $x, y \in [a, b]$. Or if you like, $\phi(sx+ty) \leq s\phi(t) + t\phi(y)$ whenever $s+t = 1$ and $s, t \geq 0$.

Now proceed by induction to show that if $t_1 + t_2 + \ldots + t_n = 1$ with $t_i \geq 0$, and if $x_i \in [a, b]$, then $\phi(t_1 x_1 + \ldots + t_n x_n) \leq t_1\phi(x_1) + \ldots + t_n\phi(x_n)$.

Finally, apply this to the situation where $t_i = 1/n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.