the minimum of the sum of two convex functions

Question

Context: Here is a question jump into my mind, when I met a problem in 1-dim case, I'm wondering if there is a generalization to higher dimensional $\mathbb R^n$ space:

Is the following statement true or false? If it's false, please give counter-example, if it's true, please prove that:

Suppose $f, g$ are two convex function $\mathbb R^n \rightarrow \mathbb R$ with unique minimum in $\mathbb R^n$. The minimum of function $f$ is denoted as $\vec{x^{(1)}}$; the minimum of function $g$ is denoted as $\vec{x^{(2)}}$. Suppose $\vec{x^*}$ is the minimum of $f + g$, then we have \begin{align} \min \{\vec{x^{(1)}_j}, \vec{x^{(2)}_j} \} \leq \vec{x^*_j} \leq \max \{\vec{x^{(1)}_j}, \vec{x^{(2)}_j} \} ~~~~~~~~~~\forall 1 \leq j \leq n. \end{align}

For $1$ dim case, the above statement is of course true; but what if in the high dimensions case $n \geq 2$?

Answer 1

Let $f(x,y)=2(x+1)^2+((x+y)+1)^2$ and $g(x,y)=(x-1)^2+((x+y)-1)^2$. Since the second partial derivatives of $f$ and $g$ are positive and both $H_f=\begin{bmatrix} 6 & 2 \\ 2 & 2 \end{bmatrix}$ and $H_g=\begin{bmatrix} 4 & 2 \\ 2 & 2 \end{bmatrix}$ are positive definite, $f$ and $g$ are strictly convex. The minimum for $f$ is at $(-1,0)$ and for $g$ at $(1,0)$. Since $0=\partial(f+g)/\partial y$ implies $x+y=0$ and $\partial(f+g)/\partial x=6x+2+4(x+y)$, $f+g$ achieves its minimum at $(-1/3,1/3)$. Thus the statement is false.