Prove that $f(x) = \frac{x}{2} + c$ if $|f(x+y)-f(x-y)-y|\le y^2\ \forall \ x,y \in \mathbb{R}$

Question

Problem Statement: Let $f:\mathbb{R} \to \mathbb{R}$ be a function such that $\forall \ x,y \in \mathbb{R}$, $$|f(x+y)-f(x-y)-y|\le y^2$$ Prove that $f(x) = \frac{x}{2} + c$ for some $c\in \mathbb{R}$

My try: $$|f(x+y)-f(x-y)-y|\le y^2 \\ \implies \frac{y-y^2}{2y}\le \frac{f(x+y)-f(x-y)}{2y} \le \frac{y+y^2}{2y}\ \ \text{when}\ y \ne 0$$ Now, taking the limit as $y \to 0$ in the above inequality and using the Squeeze theorem we have $$\lim_{y\to 0}{\frac{f(x+y)-f(x-y)}{2y}} = \frac{1}{2}\ \ \forall x \in \mathbb{R}\ \ \ (*)$$ Now, if it is given that $f$ is differentiable on $\mathbb{R}$, then we have $f'(x)=\frac{1}{2}\ \forall \ x \in \mathbb{R} $ and hence $f(x) = \frac{x}{2} + c$ for some $c \in \mathbb{R}$
But since the problem does not explicitly state that $f$ is differentiable on $\mathbb{R}$, I am not sure on how to proceed after step $(*)$.

Thanks for any answers!!

Answer 1

Put $x = a+h/2$, $y = h/2$ to have $|f(a+h) - f(a) - h/2|< h^2/4$, or $$\left| \frac{f(a+h)-f(a)}{h} - \frac12\right| < \frac{h}{4}$$. What can you say about the diffentiablity?