Is the "$p$-norm" with $0<p<1$ a concave function

Question

We know that for $p>1$ by the triangle inequality

$$\|tx + (1-t)y\|_p \leq t\|x\|_p + (1-t)\|y\|_p,$$

when $t\in[0,1]$. Now the "$p$-norm" for $0<p<1$ is not actually a norm but we can still define it to be

$$\|x\|_p=\left(\sum_{i=1}^n |x_i|^p\right)^{1/p},$$

we do not have the triangle inequality but can we at least say that this function is concave?

Answer 1

It is not concave, as $$ \frac12\|x\|_p + \frac12 \|-x\|_p \ge \left\| \frac{x-x}2\right\|_p =\|0\|_p=0. $$

Answer 2

This is neither concave nor convex function.

For example- take $p=0.5, \: \theta=0.5, \: n=2 $ and check for the convex combination of the function for the two points $\bar{x}= (x_1,x_2)=(1,0)$ and $\bar{y}= (y_1,y_2)= (0,1)$ $$ f(x)= \sum_{i=1}^{2} (\left | x_i \right |^{0.5})^{^2}=(\left | x_1 \right |^{0.5}+\left | x_2 \right |^{0.5})^{^{2}} $$

Let's take $\theta= 0.5,$

$$ f(\theta \bar{x}+ (1-\theta) \bar{y})= f(\frac{1}{2}(\bar{x}+\bar{y}))=f(\frac{1}{2},\frac{1}{2})= 2\\ \theta f(\bar{x})+(1-\theta)f(\bar{y})= \frac{1}{2}(1)+\frac{1}{2}(1)= 1 \\$$ Here, $$f(\theta \bar{x}+ (1-\theta) \bar{y})>\theta f(\bar{x})+ (1-\theta) f(\bar{y}) $$ Hence, not convex.

Similarly, if we check for $\bar{x}= (x_1,x_2)=(-1,0)$ and $\bar{y}= (y_1,y_2)= (1,0)$,

we get $$f(\theta \bar{x}+ (1-\theta) \bar{y})=f(0,0)=0\\ \theta f(\bar{x})+ (1-\theta) f(\bar{y})= \frac{1}{2}(1+1)=1 $$ Here, $$f(\theta \bar{x}+ (1-\theta) \bar{y})<\theta f(\bar{x})+ (1-\theta) f(\bar{y}) $$ Hence, not concave.

Answer 3

It is true for nonnegative $x$ and $y$. We want to prove for $0 < p < 1$ that $$\|x\|_p + \|y\|_p \leq \|x + y\|_p,$$ when $x,y$ are nonnegative $n$-dimensional vectors. Let $a_i = (x_i)^{p}$ and $b_i = (y_i)^{p}$. Take both sides of the inequality above to the $p$-th power, we have $$\bigg(\Big(\sum_{i=1}^n a_i\Big)^{1/p} + \Big(\sum_{i=1}^n b_i\Big)^{1/p}\bigg)^p \leq \sum_{i=1}^n \big|a_i^{1/p} + b_i^{1/p}\big|^{p} ,$$ which is equivalent to $$ \Bigg\|\begin{pmatrix}\sum_{i=1}^n a_i \\ \sum_{i=1}^n b_i\end{pmatrix}\Bigg\|_{1/p} \leq \sum_{i=1}^n \Bigg\|\begin{pmatrix}a_i \\ b_i\end{pmatrix}\Bigg\|_{1/p}, $$ which is clearly true using the triangular inequality for the $1/p$-norm.