Suppose $f(x)$ is differentiable everywhere and satisfy first-order Lipschitz condition, i.e., $$ |f'(x)-f'(y)| \leq \beta \cdot |x-y|, \ \forall x,y $$ Show that $$ f(x)-f(y)-f'(y) \cdot (x-y) \leq \frac{\beta}{2} |x-y|^2, \ \forall x,y $$
By applying the mean value theorem, we can easily prove that $$ f(x)-f(y)-f'(y) \cdot (x-y) \leq \beta |x-y|^2, \ \forall x,y $$ Since we do not know if the second-order derivative exists, I was wondering where this $\frac{1}{2}$ comes from ...
Let $y \lt x$ be fixed. For any $u \in [y,x]$, we have
$$-\beta(u-y) \le f^\prime(u) - f^\prime(y) \le \beta (u-y).$$ Hence integrating those inequalities in $u$ from $y$ to $x$ and using the continuous of$ f^\prime$ which is Lipschitz
$$- \frac{\beta}{2}(x-y)^2 \le f(x) - f(y) - f^\prime(y)(x-y) \le \frac{\beta}{2}(x-y)^2$$ and we're done.