$|f(x)-f(y)|≤(x-y)^2$ for all real $x,y$. What can we say about $f$?

Question

I can see that the absolute value of the difference quotient (for $y≠x$) is always less than $|x-y|$. From this I conclude that the limit of the difference quotient at $y$ i.e., the derivative at any point $y$ is $0$. So the function $f$ is a constant function. Is my math correct?

Answer 1

Your solution is correct if you change the words "less than" to "less than or equal to".

It may be interesting to note that one can give a solution without using an calculus. For every $x$ and every positive integer $n$ the triangle inequality shows that $$|f(x)-f(0)| \le\sum_{j=1}^n|f(jx/n)-f((j-1)x/n)|\le n\left(\frac xn\right)^2 =\frac{x^2}n,$$so $|f(x)-f(0)|=0$.