Show that two functions have the same minimizer Define the following two functions:
\begin{align}
g(x)&=1_{\{  \langle v_1,x \rangle  \ge 0 \} }+1_{\{  \langle v_2,x \rangle \ge 0  \} },\\
f(x)&=1_{\{  \langle v_1,x \rangle  \ge 0 \} } \langle v_1,x \rangle +1_{\{  \langle v_2,x \rangle \ge 0  \} } \langle v_2,x \rangle,
\end{align}
where the vectors $v_1$ and $v_2$ are given.  Here, as usual,  $1_{ \{\cdot \}}$, denotes the indicator function. We further restric $x$ to be $\|x\|=1$.
I would like to show that if $x^\star$ is a minimizer of $f(x)$, then it is also a minimizer of $g(x)$.
Intuitively a minimizer of $f$ forces the inner products to be either negative or very close to zero and this, in turn, should minimize $g$.  However, I am not very sure what arguments to use to show this.
Edit: Based on comments this appears to be true up to a set of measure zero for $(v_1,v_2)$.  However, not all details are clear to me yet.
 A: The set of minimisers is almost the same, basically they can differ by a set of measure $0$ when $v_1$ and $v_2$ are not colinear with different sign.
Assume first that $v_1$ and $v_2$ are not colinear with different sign. In essence, we assume that $v_1 \ne -c v_2$ for some $c > 0$. We observe that $g(x)$ and $f(x)$ have image in $[0,\infty)$ so that if there exists $x$ such that $f(x) = 0$, then the minimising set for $f$ is its zero set, and similarly for $g$. Set
$$\hat x = \left(\frac{v_1}{\|v_1\|} +  \frac{v_2}{\|v_2\|}\right)$$
and
$$x^\star = \frac{\hat x}{\|\hat x\|}$$
It is not too hard to see that $f(x^\star) = g(x^\star) = 0$ and that $\|x^\star\| = 1$, the latter condition holding because $v_1 \ne -c v_2$.  Therefore, since both $f$ and $g$ are nonnegative functions, the set which Now, it is clear that $g(x) = 0$ implies $f(x) = 0$. However, there are always vectors such that $f(x) = 0$ while $g(x) \ne 0$. To have $g(x) = 0$, we need two conditions: $\langle x,v_1 \rangle < 0$ and $\langle x,v_2\rangle < 0$, whereas to have $f(x) = 0$ we need $\langle x,v_1\rangle \le 0$ and $\langle x, v_2 \rangle \le 0$. We therefore have that
$$
\{x : f(x) = 0 \text{ and } g(x) \ne 0\} = (\operatorname{span}(v_1)^\perp \cap \{x : \langle x,v_2\rangle \le 0\}) \cup (\operatorname{span}(v_2)^\perp \cap \{x : \langle x,v_1\rangle \le 0\}).
$$
This is the union of two intersections of a half-space with a codimension 1 subspace, so it has codimension 1 in the unit sphere.
If $v_1 = -c v_2$ for some $c > 0$, the picture is completely different. Then,
$$
f(x) = \begin{cases}
0 & \text{if } x \in \operatorname{span}(v_1)^\perp = \operatorname{span}(v_2)^\perp \\
> 0 & \text{otherwise },
\end{cases}
$$
whereas
$$
g(x) = \begin{cases}
2 & \text{if } x \in \operatorname{span}(v_1)^\perp = \operatorname{span}(v_2)^\perp \\
1 & \text{otherwise }.
\end{cases}
$$
This means that in this situation their minimising sets are completely disjoint.
