GeorgeRP wrote: Find all real numbers $a$ for which there exist functions $f,g: \mathbb{R} \to \mathbb{R}$, where $g$ is strictly increasing, such that $f(1) = 1$, $f(2) = a$, and \[ f(x) - f(y) \leq (x-y)(g(x) - g(y)) \]for all real numbers $x$ and $y$. Swapping $x,y$, we get $f(y)-f(x)\le (x-y)(g(x)-g(y)$ and so $|f(x)-f(y)|\le (x-y)(g(x)-g(y))$ And so $\left|\frac{f(x)-f(y)}{x-y}\right|\le g(x)-g(y)$ $\forall x>y$ Then, for $x>y$ : $g(x)-g(y)=\sum_{k=0}^{n-1}\left(g(y+(k+1)\frac{x-y}n)-g(y+k\frac{x-y}n)\right)$ $\ge \sum_{k=0}^{n-1}\left| \frac{f(y+(k+1)\frac{x-y}n)-f(y+k\frac{x-y}n)}{\frac{x-y}n} \right|$ $\ge \left|\sum_{k=0}^{n-1}\frac{f(y+(k+1)\frac{x-y}n)-f(y+k\frac{x-y}n)}{\frac{x-y}n} \right|$ $=n\left|\frac{f(x)-f(y)}{x-y}\right|$ And setting $n\to +\infty$ in $g(x)-g(y)\ge n\left|\frac{f(x)-f(y)}{x-y}\right|$, we get impossibility if $f(x)\ne f(y)$ And so only solution if $f(x)$ is constant and so $\boxed{a=1}$ (which indeed fits for example with $f\equiv 1$ and $g\equiv Id$)