Let $f : \mathbb R \rightarrow \mathbb R $ be a twice differentiable function. Suppose that for all $x, y \in \mathbb R$ the function $f$ satisfies $f'(x) − f'(y) ≤ 3|x − y|$ then show that for all $x$ and $y$, we must have $$|f(x)−f(y)-f'(y)(x-y)|≤\frac{3}{2}(x-y)^2$$
Setting $y=0$ gives $c-3x≤f'(x)≤c+3x$, where $c=f'(0)$ $\Rightarrow \int_0^t c-3x≤\int_0^t f'(x)≤\int_0^t c+3x$ $\Rightarrow ct-\frac{3}{2}t^2-c≤f(t)-c_1≤ct+\frac{3}{2}t^2-c$ Where $c_1=f(0)$. Now let $c_1-c=c_0$
$\Rightarrow ct-\frac{3}{2}t^2+c_0≤f(t)≤ct+\frac{3}{2}t^2+c_0$
For the given expression to be maximum, $f(x)$ must be maximum, $f(y)$ must be minimum and for $x≥y$ $f'(y)$ must also be minimum (we will look at the case $x<y$ separately). Substituting all these values, we get
$(cx+\frac{3}{2}x^2+c_0)-(cy-\frac{3}{2}y^2+c_0)-(c-3y)(x-y)$
$\implies \frac{3}{2}(x^2+y^2)+6xy-3y^2$ $\implies \frac{3}{2}(x-y)^2+3y(2x-y)$
Under assumption $x>y$, for all $y>0$ this expression is clearly $>\frac{3}{2}(x-y)^2$. So where did I go wrong, and what will be the correct solution?