#!/usr/bin/env python
# coding: utf-8

# # Friends of convex
# 
# For each of the following statements, determine TRUE or FALSE. If your answer is TRUE, then prove that statement. If your answer is FALSE, then give a counterexample.

# (a) (10 points) Hyperbolic function $f(x_{1},x_{2}) = x_{1}x_{2}$ is quasiconcave on ${\rm{dom}}(f) = \mathbb{R}^{2}_{>0}$.

# #### Solution:
# TRUE. Here $\mathbf{dom}(f)$ is the positive orthant - a convex set (in fact, a convex cone). Since all superlevel sets of the function $f$, i.e., sets of the form $\{(x_{1},x_{2})\in\mathbb{R}^{2} \:\mid\: f(x_{1},x_{2}) \geq \alpha\}$, $\alpha \in \mathbb{R}$, are convex, hence by Lecture 7, p. 16, the function $f$ is quasiconcave.

# (b) (10 points) Logistic function $f(x) = \displaystyle\frac{\exp(x)}{1 + \exp(x)}$ is log-concave on ${\rm{dom}}(f) = \mathbb{R}$.

# #### Solution:
# TRUE. We have 
# $$\log\left(\displaystyle\frac{\exp(x)}{1 + \exp(x)}\right) = x - \log\left(1 + \exp(x)\right).$$
# The first term above is linear, hence concave. By second order condition, the function $\log(1+\exp(x))$ is convex, so the second term being its negative, is concave. Since the sum of concaves is concave, therefore $\log(f(x))$ is concave.

# (c) (10 points) $f(x) = \begin{cases}x &\text{for}\; x<0,\\ x+1 &\text{for}\; x>0,\end{cases}$ is pseudo-convex but not quasiconvex on ${\rm{dom}}(f) = \mathbb{R}\setminus\{0\}$.

# #### Solution:
# TRUE. Since the domain is non-convex, hence $f(x)$ is NOT quasiconvex. However, the convexity of the domain is not needed in the definition of pseudo-convex function given in Section 1 of the paper sent to the class: O.L. Mangasarian, "Psedo-convex functions", J. SIAM Control, 1965. Section 3 of this paper discusses this example.

# (d) (15 points) $f(X) = \log X$ is operator monotone and operator concave on ${\rm{dom}}(X) = \mathbb{S}_{++}^{n}$.

# #### Solution:
# TRUE. Recall the limiting definition of $\log(x)$ over $x>0$, given by $\log(x) = \lim_{p\rightarrow 0}\displaystyle\frac{1}{p}(x^{p}-1)$. Likewise, for $X\in\mathbb{S}^{n}_{++}$, we have
# $$\log(X) = \displaystyle\lim_{p\rightarrow 0}\frac{1}{p}\left(X^{p}-I\right).$$
# Since $X^{p}$ is operator monotone and operator concave for all $0 \leq p \leq 1$ [(see p. 10-11 of these notes sent to class)](http://www.ueltschi.org/AZschool/notes/EricCarlen.pdf), hence the same holds in the small $p$ case appearing in the limit above.

# (e) (15 points) $f(X) = X^{-1}$ is operator monotone and operator convex on ${\rm{dom}}(X) = \mathbb{S}_{++}^{n}$.

# #### Solution:
# TRUE. To prove monotonicity, we will show that if $X \succeq Y$, then $X^{-1} \preceq Y^{-1}$ for all $X,Y\in\mathbb{S}^{n}_{++}$ (this is like decreasing function). Let $Z:= X^{-1/2}YX^{-1/2} \in \mathbb{S}^{n}_{++}$ (congruence transform preserves sign-definiteness). Then
# $$X \succeq Y \quad \Leftrightarrow \quad X - Y = X^{1/2}\left(I - X^{-1/2}YX^{-1/2}\right)X^{1/2} \succeq 0 \quad \Leftrightarrow \quad I - X^{-1/2}YX^{-1/2} \succeq 0 \quad \Leftrightarrow \quad Z - I \preceq 0.$$
# On the other hand, 
# $$X^{-1} -Y^{-1} = X^{-1/2}\left(I - Z^{-1}\right)X^{-1/2} = X^{-1/2}Z^{-1/2}\left(Z-I\right)Z^{-1/2}X^{-1/2} \preceq 0,$$
# which follows from the fact that the LHS of the last inequality is congruence transform by the non-singular matrix $X^{-1/2}Z^{-1/2}$, and that $Z - I \preceq 0$. 
# 
# 
# To prove operator convexity of $f(X)$, we need to show
# $$f(\theta X + (1-\theta)Y) \preceq \theta f(X) \: + \: (1-\theta)f(Y), \quad\text{for all}\;X,Y\in\mathbb{S}^{n}_{++}, \quad 0\leq\theta\leq 1.$$
# 
# Since $f$ is continuous, it suffices to prove mid-point convexity, i.e., to prove the above inequality for $\theta = 1/2$. Thus, we want to prove
# $$\begin{aligned}
# \left(\displaystyle\frac{X+Y}{2}\right)^{-1} \preceq \displaystyle\frac{1}{2}\left(X^{-1} + Y^{-1}\right) \quad \Leftrightarrow \quad \displaystyle\frac{1}{2}\left(X^{-1} + Y^{-1}\right) - \left(\displaystyle\frac{X+Y}{2}\right)^{-1} = X^{-1/2}\left[\frac{1}{2}I + \frac{1}{2}Z^{-1} - \left(\displaystyle\frac{I+Z}{2}\right)^{-1}\right]X^{-1/2} \succeq 0,
# \end{aligned}$$
# where $Z:= X^{-1/2}YX^{-1/2} \in \mathbb{S}^{n}_{++}$ (congruence transform preserves sign-definiteness). Therefore, it is enough to prove that the matrix within the square bracket appearing in the last inequality is positive semidefinite, i.e., $\frac{1}{2}(I + Z^{-1}) \succeq \left(\frac{I+Z}{2}\right)^{-1}$. By arithmetic mean-harmonic mean inequality, we know that $(a+b)/2 \geq ((a^{-1}+b^{-1})/2)^{-1}$ for all $a,b>0$, specializing which for $a=1$ and $b=$ any eigenvalue of $Z^{-1}\in\mathbb{S}^{n}_{++}$, followed by the application of Spectral Theorem, gives us
# $$\frac{1}{2}(I + Z^{-1}) \succeq \left(\frac{I+Z}{2}\right)^{-1} \quad \Leftrightarrow \quad \frac{1}{2}I + \frac{1}{2}Z^{-1} - \left(\displaystyle\frac{I+Z}{2}\right)^{-1} \succeq 0,$$
# as desired.

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