Minimum Volume Enclosing Ellipsoid and convex function

Question

I need your help in solving the following.

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Let $f : \mathbb{R}^n \to [0,\infty]$ be a convex function and $L = \left\lbrace x \in \mathbb{R}^n \mid f(x) \leq 1\right\rbrace$ be a sub-level set of $f$ (We assume that $L$ is not empty).

Let $E = (G,v)$ denote the minimum volume ellipsoid that encloses $L$, which is centered at $v \in \mathbb{R}^n$ and $G \in \mathbb{R}^{n \times n}$ be a positive definite matrix which can represent the ellipsoid.

Then, does $f$ attains it's minimal value (minimum value) at $v$ (the center of the enclosing ellipsoid of $L$)?

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Please advise and thanks in advance.

Answer 1

No. To illustrate, let’s take the simplest possible dimension for this, n=1. An ellipsoid in one dimension is of the form $$(x-a)^2=b^2$$

Let $f$ be defined by $$f(x)=\left\{\begin{array}{lr} |x|&, x <0\\ \frac{x}{3}&, x \ge 0\end{array}\right.$$

Then $L=[-1,3]$ and $E$ is given by $(x-1)^2=2^2$. In other words, $v$ is the point $x=1$ but $f$ attains its minimum at $x=0$.

Example in 2D

Let $f$ be defined by $$f(x, y)=\left\{\begin{array}{lr} |x|+|y|&, x<0\\ \frac{x}{3}+|y|&, x \ge 0\end{array}\right.$$

Then the minimum of $f$ is at the origin, but $$E: \frac{(x-1)^2}{2^2}+\frac{y^2}{\frac{4}{3}}=1$$

has center at $(1,0)$.

It is just as easy to construct an example in higher dimensions as well.

Answer 2

No, certainly not. Write $f = f_1 + f_2$, where $f_1 = \delta_L : \mathbb R^n \to \{0,\infty\}$ is the indicator function of $L$ and $f_2 : \mathbb R^n \to \mathbb R$ is an arbitrary convex function. Then, the minimum value of $f$ is the minimal value of $f_2$ on $L$ and this can be any point in $L$.