# Prove constant exists such that a function is uniformly continuous

Let $$p(x)$$ be a given polynomial. Show that there is a constant $$K$$ such that the function $$g$$ defined by

$$g(x)=\ln(p(x)+K)$$

is uniformly continuous on the interval $$[-10,10]$$

• If you make it continuous, by Cantor's theorem it will be uniformly continuous. To make it continuous just make sure that $p(x)+K>0$ on $[-10,10]$. Since $p$ is continuous it attains a minimum $m$ on $[-10,10]$. You could take $K=\epsilon+\min(0,m)$ for some $\epsilon>0$. – conditionalMethod Nov 13 '19 at 20:26

Every continuous function on a compact set is uniformly continuous. However, not all values of $$K$$ will give you a function that is defined on the whole interval, which is the obstruction here. The uniform condition is a red herring.
If $$p(x)$$ is always positive you can set $$K=0$$. Otherwise, let $$M$$ be the minimum value of $$p$$ and set $$K=|M|+1$$. Then $$p(x)+K$$ is always positive, so we have a continuous function on $$[-10,10]$$.
$$p$$ as polynomial is continuous at the compact $$[-10,10]$$, thus $$p$$ is bounded. This means that there exists $$K>0$$ such that $$\forall x\in[-10,10] \;\; -K hence the function $$x\mapsto \ln(p(x)+K)$$ is well defined and continuous at $$[-10,10]$$ and by Heine's theorem, it is uniformly continuous at $$[-10,10]$$.
Observe that $$p$$ is continuous, and hence there exist $$m\le M$$ such that $$m\le p(x)\le M, \quad \text{for all x\in [-10,10].}$$ Hence, clearly $$p(x)-m+1\ge 1, \quad\text{for all x\in [-10,10].}$$ Thus $$f(x)=\ln\big(p(x)-m+1\big)$$ is well defined and continuous in $$[-10,10]$$, and hence uniformly continuous, since $$[-10,10]$$ is compact, i.e., closed and bounded subset of $$\mathbb R$$.