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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 $x'<x''$ and $\lambda\in(0,1)$.
If $x'<x''\le0$ or $0 \le x'<x''$ 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 $x'<0<x''$ 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.
