# Proving that a polynomial function with complex roots is uniformly continuous

If E is a bounded set and $p(x)$ is a polynomial that has no real roots, explain why $\frac{1}{p(x)}$ is uniformly continuous on E

Let $f(x)=\frac{1}{p(x)}$, a function $f(x)$ is said to be uniformly continuous on E if

$\forall \epsilon>0$, $\exists \delta>0$, such that $\forall x_0 \in E$, $\lvert x-x_0 \rvert < \delta \implies \lvert f(x)-f(x_0) \rvert < \epsilon$

Let $p(x)=ax^2+bx+c$ for some constants a,b,c

We know that $p(x)$ has no real roots if $b^2-4ac<0$

So $\forall \epsilon>0$, we need to find $\delta>0$ such that $\forall x_0 \in E$, $\lvert x-x_0 \rvert <\delta \implies \lvert \frac{1}{ax^2+bx+c}-\frac{1}{ax_0^2+bx_0+c} \rvert < \epsilon$

$=\lvert \frac{(ax_0^2+bx_0+c)-(ax^2+bx+c)}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert= \lvert \frac{a(x^2-x_0^2)+b(x-x_0)}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert=\lvert \frac{[a(x-x_0)(x+x_0)]+b(x-x_0)}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert=\lvert \frac{(x-x_0)[a(x+x_0)+b]}{(ax^2+bx+c)(ax_0^2+bx_0+c)}\rvert=\lvert x-x_0 \rvert \lvert \frac{a(x+x_0)+b}{(ax^2+bx+c)(ax_0^2+bx_0+c)} \rvert$

I'm not sure if I'm going in the right direction but eventually I want to set $b=$min{$\sqrt{4ac}$}

• degre of p can be other than 2 : $(x^2+1)^2$ has no real root Nov 12, 2015 at 2:15

case 1 : $p(x)=C$ then $f$ is constant therefore uniformely continuous
case 2 : $deg(p)>0$
p has no real root so f is $C^\infty$ $$f'(x) = -p'(x)/p^2(x)$$ $$deg(p^2)=2*(deg(p')+1)$$ so $\lim\limits_{x\to\infty}f'(x)=0$ and $\lim\limits_{x\to-\infty}f'(x)=0$
so $f'$ is bounded and $f$ is uniformly continuous
• actually, as you are resticted to a bounded it is even simplier : $f'$ is continuous on a bounded set therefore bounded Nov 12, 2015 at 2:29