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The function $f(x) = \mathbf{max}( 1/2, x, x^2 )$ is convex.

The function $f(x) = \mathbf{min}( 1/2, x, x^2)$ is concave.

The function $f(x) = \mathbf{min}( 1/2, x, x^2)$ is quasilinear.

The square of a convex nonnegative function is convex.

The reciprocal of a positive concave function is convex.

$f(x) = (x^2 + 2)/(x+2)$, with $\mathbf{dom}f = (-\infty, -2)$.

$f$ is convex.

$f$ is concave.

$f(x) = 1/(1-x^2)$, with $\mathbf{dom} f = (-1, 1)$.

$f$ is convex.

$f$ is log-convex.

$f(x) = \max_i x_i - \min_i x_i$ is convex.

$f(x) = \cosh x = (e^x+e^{-x})/2$.

$f$ is convex.

• Correct! $f''(x) = \cosh x >0$.
• Incorrect.

$f$ is log-concave.

• Incorrect.
• Correct! In fact, $f$ is log-convex.

For $x \in \mathbb{R}^n$, we define $f(x) = \min\{ k \mid \sum_{i=1}^k |x_i| > 1 \}$, with $f(x) = \infty$ if $\sum_{i=1}^n |x_i| \leq 1$.

Conjugate function.

$f(x) = \mathbf{1}^T(x)_+$ where $(x)_+ = \max\{0,x\}$. What is $f^*$?

We define $(x)_-$ to be $\max\{0,-x\}$, such that $x = (x)_+ - (x)_-$.

The constraint $\mathbf{1}^T(x)_- \leq (1/2) \mathbf{1}^T(x)_+$ defines a convex set.

• Correct! To see this, we re-write it as $(1/2) \mathbf{1}^T(x)_- - (1/2) \mathbf{1}^Tx \leq 0$; here the lefthand side is a convex function.
• Incorrect.