Problem 10.2. Consider the strictly monotonic function $f: R \rightarrow R$. Determine all functions $g: R \rightarrow R$ such that

$$
f(g(x+2017)) \geq f(x) \geq f(g(x)+2017),(\forall) x \in R
$$
Solution. Consider the strictly increasing function $f$. From the conditions of the problem, we have

$$
\left\{\begin{array} { l } 
{ g ( x + 2 0 1 7 ) \geq x , } \\
{ x \geq g ( x ) + 2 0 1 7 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
x \leq g(x+2017) \\
x \geq g(x)+2017
\end{array}\right.\right.
$$

In the first inequality, substitute $x$ with $x-2017$ and we obtain:

$$
\left\{\begin{array} { l } 
{ x - 2 0 1 7 \leq g ( x ) , } \\
{ x \geq g ( x ) + 2 0 1 7 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
g(x) \geq x-2017 \\
g(x) \leq x-2017
\end{array}\right.\right.
$$

Thus, $g(x)=x-2017$ for any $x \in \mathbb{R}$.

We verify the solution:

$$
f(g(x+2017))=f(x+2017-2017)=f(x) ; f(g(x)+2017)=f(x-2017+2017)=f(x)
$$

The inequalities from the problem statement are satisfied, they become equalities.

The case of the strictly decreasing function is absolutely identical.