Find all $f:\mathbb{R}\to\mathbb{R}$ so that $|f(x+y)-f(x-y) – y|\leq y^2$ for all $x,y\in\mathbb{R}$.
A standard approach when solving functional equations or functional inequalities is to guess which function might work, and then show that the guessed function is unique. One can show uniqueness by showing that if $g$ is the guessed function and $f$ is any function satisfying the given conditions, then $f-g$ is identically zero. The function $f(x)=x/2$ clearly works. It's definitely the only linear function with zero constant term that works. Note that we may assume $f(0) = 0$ because if we let $h(x)=f(x)+C$ for an arbitrary constant $C$, then $h$ satisfies the same inequality as $f$. In general, any function satisfying the given inequality is the sum of a constant and a function $f$ satisfying the inequality and $f(0) = 0$. If $f(x)=cx$ for all x then the inequality reads $|c(x+y) – c(x-y) – y|\leq y^2\,\forall x,y\Rightarrow \,\forall y, |y(2c-1)|\leq y^2.$ But as $y\to 0,|\dfrac{y^2}{y}|\to 0$ so eventually $y^2$ will be less than $|y(2c-1)|$ for sufficiently small y, and hence $2c-1$ must be equal to $0$. Thus we've shown $c=1/2$ if $f(x)=cx$ for all $x$. Let $g(x)=f(x)-x/2.$ It could be useful to show $g(x)=0$ for all x. We clearly have $|g(x+y)-g(x-y)| = |f(x+y)-(x+y)/2 – f(x-y) +(x-y)/2| = |f(x+y)-f(x-y) – y|\leq y^2$ for all $x,y\in\mathbb{R}$. Plugging in $x=y$ gives that $|g(2y)-g(0)| \leq y^2$ for all $y$ and by our assumption, $g(0)$ can be assumed to be zero, so $|g(2y)|\leq y^2$ for all $y$. But I'm not sure how to show that $g$ must be identically zero, assuming $f(0) = 0$.
Best Answer
You have made great progress.
Here are two further ideas.
Let $g(x)=f(x)-(x/2+f(0))$. Then $g(0)=0$. $$\begin{aligned}&g(x+y)-g(x-y)\\ =&f(x+y)-((x+y)/2+f(0))-(f(x-y)-((x-y)/2+f(0)))\\ =&f(x+y)-f(x-y) - y \end{aligned}$$
Hence, we have $$|g(x+y)-g(x-y)|\le y^2$$ for all $x, y\in\Bbb R$. Replacing $(x,y)$ by $((x+y)/2, (x-y)/2)$, we have $$ |g(x)-g(y)|\le((x-y)/2)^2$$
For any $x\in\Bbb R$, $$\begin{aligned}&|g(x)-g(0)|\\ =&\left|\sum_{i=1}^n |g(ix/n)-g((i-1)x/n)\right|\\ \le&\sum_{i=1}^n |g(ix/n)-g((i-1)x/n)|\\ \le&\sum_{i=1}^n ((ix/n-(i-1)x/n)/2)^2\\ =&\sum_{i=1}^n (\frac x{2n})^2\\ =&\frac {x^2}{4n} \end{aligned}$$ Since $n$ can be arbitrarily large, the inequality above implies $|g(x)-g(0)|$ must be smaller than any positive number. That means it must be $0$. Hence $g(x)=0$ for all $x$. $f(x)=x/2+f(0)$ for all $x$.
This answer is prepared especially for students who have not learned calculus. For people who have, the given condition implies immediately the derivative of f(x) is $\frac12$ everywhere.
Here is a minor variation as an easy exercise.
Find all $f:\mathbb{R}\to\mathbb{R}$ so that $|f(x+2y)-f(x-3y) - y|\leq |y|^{\frac32}$ for all $x,y\in\mathbb{R}$.