Let $F:\mathbb{R}^n\longrightarrow \mathbb{R}$ be convex such that $$ |F(x)|\leq A|x|^2+B$$ for $A,B>0$. How can I deduce that, $$ |DF(x)| \leq C(1+|x|)?$$
This fact was used multiple times in Evans book "Weak convergence method for nonlinear PDEs", but I don't know why?
On
Also, using a quite big theorem of convex functions (refer to "Measure Theories and Fine Properties of Functions", L.C.Evans) Page 267, Theorem 6.7
where he proved that a function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is convex then there exists a contant C only depending on n such that $$ \underset { B \left( x , \frac { r } { 2 } \right) }{\mathtt{ess.sup}} | D f | \leq \frac { C } { r } \dfrac{1}{|B_r|}\int _ { B ( x , r ) } | f | d y. $$ This provides us the result rigourously.
Argue by contradiction, suppose that there exists suitable $\{x_k\}_{k=1}^{\infty}$ such that $$|DF(x_k)|>k(1+|x_k|).$$
For $k\gg 2A+100B,$ after a rotation, and note that by the convexity $D^2F\geq 0$ we can assume with out loss of generality that $\partial_1F(x_k+cI)\geq \partial_1F(x_k)=|DF(x_k)|\geq k(1+|x_k|),\ \ \forall c\geq 0$. Hence we " integral on this line" will provide us a contradiction to the growth condition of $|F|\leq A|x|+B$.
This may not be a rigorous proof, but it seems enough to make sure this result holds. Hoping for a more rigorous proof come up. Best Regards!