Strict convexity of the convex conjugate of the sum of infinity norm and Euclidean norm

Question

Define a function

$$f(x)=\min_{y \in \mathbb{R}_+^n} \Big\{ \|y\|_\infty + y^\top Q y : x^\top y = 1 \Big\}$$

where $Q$ is a positive definite matrix. Is $f$ strictly convex on $x\geq 0$?

Answer 1

The function is not even convex on $\mathbb{R}^n_+$. Take $n=2$, $Q=I$, $x_1=(1, 0)$, $x_2=(0,1)$ and $x_3 = 0.5(x_1+x_2)$, then: $$\begin{align} f(x_1) &= \min_{y \geq 0} \{ \max\{y_1, y_2\} + y_1^2+y_2^2 : y_1=1 \} = 2, \\ f(x_2) &= \min_{y \geq 0} \{ \max\{y_1, y_2\} + y_1^2+y_2^2 : y_2=1 \} = 2, \text{ and} \\ f(x_3) &= \min_{y \geq 0} \{ \max\{y_1, y_2\} + y_1^2+y_2^2 : y_1 + y_2 = 2 \} = 3. \end{align}$$

The initial question with $x \in \mathbb{R}^n$: for $n=1$ and $Q=I$ you get $f(x) = \frac{1}{|x|} + \frac{1}{x^2}$, which is not convex on $\mathbb{R}$.