# Strict convexity of a power function of an absolute value

I would like to prove that that function $$f(x) = |a-bx|^k$$ is strictly convex in $$x$$ on $$\mathbb{R}$$ for $$b\ne0$$ and $$k>1$$.

I believe the function is differentiable everywhere (including at $$x=a/b$$) but am not sure if it is twice differentiable everywhere, so have not been able to establish that the second derivative is everywhere strictly positive.

I also wonder if there is a more direct route to a proof.

Any help would be appreciated.

Convexity is preserved under affine maps, as shown here. That is, if $$h(x)$$ is convex in $$\mathbb{R}$$, then $$g(x)=h(cx+d)$$ is also convex in $$\mathbb{R}$$ where $$c,d\in\mathbb{R}$$. The same statement about strict convexity only requires, I believe, $$c\ne 0$$ (which we have since $$b\ne 0$$). So it suffices to show that $$f(x)=|x|^k$$ is strictly convex. $$f(x)$$ is differentiable everywhere with $$f'(x)=\begin{cases} kx^{k-1} & x>0 \\ 0 & x=0 \\ -k(-x)^{k-1} & x<0 \end{cases}$$ and we need to show $$f'(x)$$ is strictly increasing. You can use the second derivative for that: $$f''(x)= \begin{cases} k(k-1)x^{k-2} & x>0\\ k(k-1)(-x)^{k-2} & x<0 \end{cases}$$ Because $$f'(x)$$ is continuous at $$0$$ and $$f''(x)>0$$ for all $$x\ne 0$$, it follows that $$f'(x)$$ is strictly increasing in $$(-\infty,0]$$ and $$[0,\infty]$$, hence in $$(-\infty,\infty)$$ as well.

$$f(x)=|a-bx|^k\ge0$$

Using $$\dfrac{d}{dx}|U|=\dfrac{|U|}{U}\cdot\dfrac{dU}{dx}$$ we have $$\begin{eqnarray} f^\prime(x)&=&-bk|a-bx|^{k-1}\cdot\frac{|a-bx|}{a-bx}\\ &=&-\frac{bx}{a-bx}\cdot|a-bx|^k\\ &=&-\frac{bk}{a-bx}\cdot f(x) \end{eqnarray}$$

$$\begin{eqnarray} f^{\prime\prime}(x)&=&\frac{b^2k}{(a-bx)^2}\cdot f(x)-\frac{bk}{a-bx}\cdot\frac{-bk}{a-bx}\cdot f(x)\\ &=&\frac{b^2k}{(a-bx)^2}\cdot f(x)+\frac{b^2k^2}{(a-bx)^2}\cdot f(x)\\ &=&\frac{b^2k(k+1)}{(a-bx)^2}\cdot f(x)>0\text{ for }k>0 \end{eqnarray}$$ So $$f$$ is concave on any interval not containing $$x=\frac{a}{b}$$.

As here, note that strict convexity of $$g(\cdot)$$ implies strict convexity of $$g(ax+b)$$ for $$a\ne0$$, $$b\in\mathbb{R}$$.

Therefore it is sufficient to prove strict convexity of $$|x|^k$$ for $$k>1$$.

Pick $$x' and $$\lambda\in(0,1)$$.

If $$x' or $$0 \le x' then the fact that $$\begin{eqnarray} |\lambda x'+(1-\lambda)x''|^k<\lambda |x'|^k + (1-\lambda) |x''|^k \end{eqnarray}$$ follows from the strict convexity of $$x^k$$ for $$k>1$$ on $$\mathbb{R}$$.

If $$x'<0 then \begin{align} |\lambda x'+(1-\lambda)x''|^k &<\left(\max\left\{ \left|\lambda x'\right|,\left|\left(1-\lambda\right)x''\right|\right\} \right)^{k} \\ & =\left(\max\left\{ \lambda^{k}\left|x'\right|^{k},\left(1-\lambda\right)^{k}\left|x''\right|^{k}\right\} \right) \\ & < \lambda\left|x'\right|^{k}+\left(1-\lambda\right)\left|x''\right|^{k} \end{align} where the first inequality follows from the signs of $$x',x''$$ and the second from the fact that $$\lambda^{k}<\lambda$$, $$\left(1-\lambda\right)^{k}<\left(1-\lambda\right)$$, and $$x',x''\ne0$$.

Since these are the only three cases, the proof is complete.