Is function $f(x) = \left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^k$ convex for $k \ge 1/2$

Question

Let ${\bf v}$ and ${\bf w}$ be column vector of dimension $n$. Is function $f(x) = \left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^k$ convex for $k \ge 1/2$ ?

I want to show this via second derivative argument.

My attempt : First, compute the first derivative

\begin{align} \frac{d}{dx}\left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^k&= \frac{d}{dx}\left(\sum_{i=1}^n (w_i-x v_i)^2 \right)^k \\ &=k \left(\sum_{i=1}^n (w_i-x v_i)^2 \right)^{k-1} \left(-2 \sum_{i=1}^n (w_i-x v_i) v_i \right) \\ &= -2k \left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^{k-1} ( {\bf w} -x {\bf v})^T {\bf v} \end{align}

\begin{align} &\frac{d^2}{dx^2} \left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^k \\ &= -2k \frac{d}{dx} \left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^{k-1} ( {\bf w} -x {\bf v})^T {\bf v} \\ &= -2k \left( \frac{d}{dx} \left(\sum_{i=1}^n (w_i-x v_i)^2 \right)^{k-1} \right) \left(-2 \sum_{i=1}^n (w_i-x v_i) v_i \right)-2k \left( \left(\sum_{i=1}^n (w_i-x v_i)^2 \right)^{k-1} \right) \left(-2 \frac{d}{dx}\sum_{i=1}^n (w_i-x v_i) v_i \right) \\ &= -2k \left( (k-1) \left(\sum_{i=1}^n (w_i-x v_i)^2 \right)^{k-2} \left(-2 \sum_{i=1}^n (w_i-x v_i) v_i \right)\right) \left(-2 \sum_{i=1}^n (w_i-x v_i) v_i \right)-2k \left( \left(\sum_{i=1}^n (w_i-x v_i)^2 \right)^{k-1} \right) \left(2 \sum_{i=1}^n v_i^2 \right) \\ &=4k (k-1) \left( ( {\bf w} -x {\bf v})^T ( {\bf w} -x {\bf v}) \right)^{k-2} ( {\bf w} -x {\bf v})^T {\bf v} ( {\bf w} -x {\bf v})^T {\bf v}-4k {\bf v}^T {\bf v} \end{align}

This approach seem very cumbersom is there a better approach?

Answer 1

Let $g(x)=(W-xV)^T(W-xV)\geq 0$. Then $g'(x)=-2(W-xV)^TV$ and $g''(x)=2V^TV$. Now $f(x)=g^{k}(x)$ implies that $f''=kg^{k-2}((k-1)g'^2+gg'')$; if $k>0$, then $f''$ has same signum as $(k-1)g'^2+gg''=4(k-1)((W-xV).V)^2+2||W-xV||^2||V||^2$. By Cauchy-Schwarz, $((W-xV).V)^2\leq ||W-xV||^2||V||^2$; thus, if $k\geq 1/2$, then $f''(x)$ is always $\geq 0$ and $f$ is convex.

EDIT. Necessarily, for every $x$, $W-xV$ must be non-zeo. Then we must assume that $W,V$ are linearly independent.