Sergeev I.n.

The function $f$ assigns to each vector $v$ (with a common origin at point $O$) a number $f(\boldsymbol{v})$, and for any vectors $\boldsymbol{u}, \boldsymbol{v}$ and any numbers $\alpha, \beta$, the value $f(\alpha \boldsymbol{u}+\beta \boldsymbol{v})$ does not exceed at least one of the numbers $f(\boldsymbol{u})$ or $f(\boldsymbol{v})$. What is the maximum number of values that such a function can take?
Evaluation. Sequentially, we will prove that the function $f$ described in the condition takes:

a) no more than two different values on any line (passing through point $O$): indeed, if it takes different values on three vectors of the same line, and on vector $\boldsymbol{v}$ - the greatest value, and on some non-zero (such will exist) vector $\boldsymbol{u}$ - a smaller value, then for some number $\alpha$ we will get a contradiction: $\boldsymbol{v} = \alpha \boldsymbol{u} \Rightarrow f(\boldsymbol{v}) = f(\alpha \boldsymbol{u} + 0 \cdot \boldsymbol{u}) \leq f(\boldsymbol{u})$

b) no more than three different values on any plane (passing through point $O$): indeed, if it takes different values on four vectors of the same plane, and on vector $\boldsymbol{v}$ - the greatest value, and on some non-collinear (such will exist due to the previous point) vectors $\boldsymbol{u}_{1}, \boldsymbol{u}_{2}$ - smaller values, then for some numbers $\alpha_{1}, \alpha_{2}$ we will get a contradiction: $\boldsymbol{v} = \alpha_{1} \boldsymbol{u}_{1} + \alpha_{2} \boldsymbol{u}_{2} \Rightarrow f(\boldsymbol{v}) = f\left(\alpha_{1} \boldsymbol{u}_{1} + \alpha_{2} \boldsymbol{u}_{2}\right) \leq \max \left\{f\left(\boldsymbol{u}_{1}\right), f\left(\boldsymbol{u}_{2}\right)\right\}$

c) no more than four different values in the entire space: indeed, if it takes different values on five vectors, and on vector $\boldsymbol{v}$ - the greatest value, and on some non-coplanar (such will exist due to the previous point) vectors $\boldsymbol{u}_{1}, \boldsymbol{u}_{2}, \boldsymbol{u}_{3}$ - smaller values, then for some numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}$ we will get a contradiction:

$\boldsymbol{v} = \alpha_{1} \boldsymbol{u}_{1} + \alpha_{2} \boldsymbol{u}_{2} + \alpha_{3} \boldsymbol{u}_{3} \Rightarrow f(\boldsymbol{v}) = f\left(\alpha_{1} \boldsymbol{u}_{1} + \left(\alpha_{2} \boldsymbol{u}_{2} + \alpha_{3} \boldsymbol{u}_{3}\right)\right) \leq \max \left\{f\left(\boldsymbol{u}_{1}\right), f\left(\alpha_{2} \boldsymbol{u}_{2} + \alpha_{3} \boldsymbol{u}_{3}\right)\right\} \leq \max \left\{f\left(\boldsymbol{u}_{1}\right), \max \left\{f\left(\boldsymbol{u}_{2}\right), f\left(\boldsymbol{u}_{3}\right)\right\}\right\} = \max \left\{f\left(\boldsymbol{u}_{1}\right), f\left(\boldsymbol{u}_{2}\right), f\left(\boldsymbol{u}_{3}\right)\right\}$.

Example of a function $f$ that satisfies the condition of the problem and takes exactly four different values: introducing Cartesian coordinates in the space with the origin at point $O$, we define $\boldsymbol{v} = (x, y, z)$,

$$
f(\boldsymbol{v}) = \begin{cases}
0, & x = y = z = 0 \\
1, & x = y = 0 \neq z \\
2, & x = 0 \neq y \\
3, & x \neq 0
\end{cases}
$$