# Convexity of $x \mapsto \frac{\|Ax-b\|^2_2}{1-\|x\|^2_2}$

Given matrix $$A \in \Bbb R^{n \times n}$$ and vector $$b \in \Bbb R^n$$, determine whether the following function is convex.

$$f: \{x \in \Bbb R^n : \|x\|_2 < 1\} \to \Bbb R, \qquad x \mapsto \frac{\|Ax-b\|^2_2}{1-\|x\|^2_2}$$

The numerator appears to be similar to the Least square function which I know is convex. Since: for convex function f over a convex set for any α ≥ 0, the function α*f is a convex function. Is setting:

$$a=\frac{1}{1-\|x\|^2_2} \geq 0$$

a valid proof for convexity? I am very thankful for any tips!

The function is convex. Indeed, clearly $$S = \{x\in \mathbb{R}^n: \|x\|_2 < 1\}$$ is a convex set, $$\|Ax-b\|_2$$ is convex and nonnegative on $$S$$, $$1-\|x\|_2^2$$ is concave and positive on $$S$$ (since $$\|x\|_2^2$$ is convex on $$\mathbb{R}^n$$), from Fact 1 below, the desired result follows.
Fact 1: Let $$S\subseteq \mathbb{R}^n$$ be a convex set. Let $$f: S \to \mathbb{R}$$ be convex and $$f(x) \ge 0, \forall x \in S$$. Let $$g: x \to \mathbb{R}$$ be concave and $$g(x) > 0, \forall x \in S$$. Then $$\frac{f^2}{g}$$ is convex on $$S$$.
Proof of Fact 1: For any $$x_1, x_2 \in S$$ and $$t\in [0,1]$$, we have $$f(t x_1 + (1-t)x_2) \le tf(x_1) + (1-t)f(x_2)$$ and $$g(tx_1 + (1-t)x_2) \ge tg(x_1) + (1-t)g(x_2).$$ Thus, we have \begin{align} &t\frac{(f(x_1))^2}{g(x_1)} + (1-t)\frac{(f(x_2))^2}{g(x_2)} - \frac{(f(tx_1 + (1-t)x_2))^2}{g(tx_1 + (1-t)x_2)}\\ \ge\ & t\frac{(f(x_1))^2}{g(x_1)} + (1-t)\frac{(f(x_2))^2}{g(x_2)} - \frac{(tf(x_1) + (1-t)f(x_2))^2}{tg(x_1) + (1-t)g(x_2)}\\ =\ & \frac{t(1-t)[f(x_1)g(x_2) - f(x_2)g(x_1)]^2}{g(x_1)g(x_2)[tg(x_1) + (1-t)g(x_2)]}\\ \ge\ & 0. \end{align} Q.E.D.
• To avoid the definition mess: define $L(x,t)=-g(x)t^2+2tf(x)$. The function is convex in $x$ for all $t\ge 0$, hence, $\max_{t\ge 0}L(x,t)=\frac{f^2}{g}$ is convex. – A.Γ. Mar 30 at 1:10