In Oksendal's book on SDEs, theorem 5.2.1 states
Let $T>0$ and $b(\cdot,\cdot):[0,T]\times\mathbb{R}^n\to\mathbb{R}^n,\sigma(\cdot,\cdot):[0,T]\times\mathbb{R}^n\to\mathbb{R}^{n\times m}$ be measurable functions satisfying $$|b(t,x)|+|\sigma(t,x)|\le C(1+|x|), x\in\mathbb{R}^n, t\in[0,T]$$ for some constant $C$, (where $|\sigma^2|=\sum|\sigma_{ij}|^2$), and such that $$|b(t,x)-b(t,y)| + |\sigma(t,x)-\sigma(t,y)|\le D |x-y|, x,y\in\mathbb{R}^n, t\in[0,T]$$ for some constant $D$.
Does the second condition imply the first condition? By the triangle inequality $$|b(t,x)-b(t,0)|+|\sigma(t,x)+\sigma(t,0)| \ge |b(t,x)|-|b(t,0)|+|\sigma(t,x)|-|\sigma(t,0)|$$ Suppose the second condition holds. Then $$|b(t,x)| + |\sigma(t,x)| \le |b(t,0)|+|\sigma(t,0)| + D|x|$$ Let $C = |b(t,0)|+|\sigma(t,0)|$. Then we have $$|b(t,x)| + |\sigma(t,x)|\le \max(C,D)(1+|x|)$$
Did I miss something here?
Edit: After typing all this, I think what I missed is that the constants in the theorem cannot depend on $t$. And that would mean that the 2nd condition does imply the first if $b,\sigma$ are continuous or bounded in $t$. Does that sound right?