I want to show a set $$A = \{ ax^2 + bx + c : a_0 \leq a \leq a_1, b_0 \leq b \leq b_1, c_0 \leq c \leq c_1 \}$$ is equicontinius, i.e. there exists a common Lipschitz constant for all f $\epsilon$ A.
So far I have:
$|f(x) - f(y)| = |(ax^2 + bx + c) - (ay^2 + yx + c)| = |(ax^2 + bx - ay^2 - yx)| = |a(x^2-y^2) + b(x-y)| \leq |a(x^2-y^2)| + |b(x-y)|$
Where the inequality comes into play because of the triangle inequality. Now I see that I have $d(x,y)$ here in the form of $|x-y|$, but I can't seem to isolate that $|x-y|$ due to the $|x^2-y^2|$ also there. I'm not sure if I'm missing some algebraic trick or approaching finding this Lipschitz constant incorrectly.
$f(x)=x^{2}$ belongs to $A$ (for suitable values of $a_i,b_i$'s) but it is not a Lipschitz function on $\mathbb R$ so there is hope of proving that the given set is uniformly Lipschitz on $\mathbb R$. It is uniformly Lipschitz on any bounded subset of $\mathbb R$.