Sum of real powers: $\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$

Question

Let $\{x_i\}_{i=1}^{N}$ be positive real numbers and $\beta \in \mathbb{R}$. Can we say that: $$ \sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$$

I know that this holds if $\beta \in \mathbb{N}$. Does the above inequality have a name in case it's true?

@kagami Try taking $N=100$, all $x_i=0.01$, and $\beta=\frac{1}{2}$. The LHS is 10. The RHS is 1. As a result of that (+ umpteen similar examples) and the fact that you admitted to making zero effort to see if your inequality was true, you now (0516Z) have 4 out of the necessary 5 close votes. — almagest, Jun 27 '16 at 05:17
@almagest some of the answers correct that. The inequality is reversed for $\beta<1$. — AD - Stop Putin -, Jun 27 '16 at 07:50
@AD. Sure, but that will not be enough to stop closure. The question needs editing. — almagest, Jun 27 '16 at 07:51

score 6 · Answer 1 · answered Jun 26 '16 at 21:38

6

Note that this amounts to showing that $$ \left(\sum_{i=1}^{N}{x_i}^{\beta}\right)^{1/\beta} \leq \sum_{i=1}^N x_i $$ This fact holds for all $\beta \geq 1$, and it is an instance of the Minkowski inequality.

answered Jun 26 '16 at 21:38

Ben Grossmann

225,327

How can we see that this is an instance of Minkowski inequality.? – kagami Jun 26 '16 at 21:44
Let $$ v_i = (\overbrace{0,\dots,0}^{i-1},x_i,0,\dots,0) $$ and note that $$ \left| \sum_i v_i \right|{\beta} \leq \sum_i |v_i|{\beta} $$ – Ben Grossmann Jun 27 '16 at 00:48

score 3 · Answer 2 · answered Jun 26 '16 at 21:36

3

If $\beta \ge 1$, this is a consequence of Jensen's inequality applied to the convex function $t \mapsto t^{\beta}$.

If $\beta < 1$, this is false with $x_i = 1$.

answered Jun 26 '16 at 21:36

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

The case $0<\beta<1$ we have re reversed inequality $$\left(\sum_{i=1}^N x_i\right)^\beta\leq \sum_{i=1}^N x_i^\beta$$ Also, this can actually be used to prove the inequality of the present post. Here are some details (for the case $\beta>1$ choose $p=1$ and $q=\beta$).

edited Apr 13 '17 at 12:21

Community

1

answered Jun 26 '16 at 21:52

AD - Stop Putin -

10,970

do you happen to know the name of the inequality in the general case with $p \leq q$? – kagami Jun 26 '16 at 22:01
I've followed the proof but I would be grateful if you refer me to the reference from which you have inequality (1) – kagami Jun 26 '16 at 22:23
1

@kagami No, I have never seen a name for it. We could call it the embedding inequality of $\ell^p$ into $\ell^q$ for $p<q$. – AD - Stop Putin - Jun 26 '16 at 22:26
1

@kagami If I remember correctly it is an exercise in Zelazko's book Banach Algebras. – AD - Stop Putin - Jun 26 '16 at 22:31

score 2 · Answer 4 · edited Apr 13 '17 at 12:19

If $0<\beta<1$, then the opposite inequality is true, since the function $x\mapsto x^\beta$ is subadditive in this case. See Prove variant of triangle inequality containing p-th power for 0 < p < 1 and An inequality concerning concave functions. $$\left(\sum_{i=1}^{N}{x_i}\right)^{\beta} \leq \sum_{i=1}^{N}{x_i^{\beta}}.$$

For $\beta=1$ then there is equality $$\sum_{i=1}^{N}{x_i^{\beta}} = \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}.$$

If $\beta>1$, then $x\mapsto x^\beta$ is superadditive and thus. $$\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}.$$

In general, concave function $f(x)$ with $f(0)=0$ is subadditive and concave function $f(x)$ with $f(0)=0$ is superadditive. The proof can be found, for example, in:

The Wikipedia article on subadditivity
This answer: Prove variant of triangle inequality containing p-th power for 0 < p < 1

See also Exercise 16.6.4, p.480, in M. Kuczma: An introduction to the theory of functional equations and inequalities.

score 0 · Answer 5 · answered Jun 28 '16 at 10:05

The best part about the inequality $$\sum_{n = 1}^{N}x_{i}^{\beta} \leq \left(\sum_{i = 1}^{N}x_{i}\right)^{\beta}\tag{1}$$ is that it is homogeneous in variables $x_{i}$ and hence we can assume without any loss of generality that $\sum x_{i} = 1$. Since each each $x_{i}$ is positive and they sum to $1$ it follows that each $x_{i}$ lies between $0$ and $1$ and hence $x_{i}^{\beta} < x_{i}$ if $\beta > 1$ and therefore $$\sum x_{i}^{\beta} < \sum x_{i} = 1 = 1^{\beta} = \left(\sum x_{i}\right)^{\beta}$$ If $0 < \beta < 1$ or $\beta < 0$ then we have $x_{i}^{\beta} > x_{i}$ and hence the inequality is reversed.

Sum of real powers: $\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$

5 Answers5

Linked