calculus - Let $f:[a,infty] rightarrow mathbb R$ and $lim_{xrightarrowinfty} f'(x) = infty$, show that $f$ is not uniformly continuous

I am stuck on this question from a problem set. The question itself has two parts so I post them under the same topic (they're pretty much related to each other too - at least in the book). I hope it's ok.

Q1: Let $f:[a,\infty] \rightarrow \mathbb R$ derivative function. $\lim_{x\rightarrow\infty} f'(x) = \infty$. Show that $f$ is not uniformly continuous on $[a,\infty]$

Attempts:

• I tried to show that if $f'(x)$ isn't blocked, the function cannot be uniformly continuous, but it only works in the opposite direction.


• Showing that $\lim_{x\rightarrow\infty} f(x)$ doesn't exist doesn't help me either.

Q2: Let $f:\mathbb R \rightarrow \mathbb R$ continuous function. $\lim_{x\rightarrow\infty} f(x) = \lim_{x\rightarrow-\infty} f(x) = L$. Show that $f$ has either minimum or maximum in $\mathbb R$

Attempts:

• Exists $M_1$ (for $\lim_{x\rightarrow\infty} f(x)$) and $M_2$ (for $\lim_{x\rightarrow-\infty} f(x)$) and if function gets max or min in $[M_1, M_2]$ which is equal to L, then we're done.

Thanks in advance!

Answer

1.Fix $\epsilon,\delta>0$.

Since lim$_{x\rightarrow\infty}f'(x)=\infty,\exists N$ such that $f'(x)>\frac{\epsilon}{\delta}$, if $x>N$.

Then, by MVT, we have that$\;\;\;\forall x>N\;\;\;\exists c\in(x,x+\delta)$ such that $f(x+\delta)-f(x)=f'(c)\delta>\epsilon$.

Since $\epsilon$ and $\delta$ were arbitrary, this shows f can not be uniformly continuous.

2.Suppose f is not constant. Then $\exists x_0$ such that, say, $f(x_0)>L$.

Since $lim_{x\rightarrow\infty}f(x)=lim_{x\rightarrow-\infty}f(x)=L, \exists N$ such that $f(x_0)>f(x)$ if $|x|>N$.

Now since $f$ is continuous it obtains a maximum on $[-N,N]$, say $f(x_1)$.

Clearly, $f(x_1)\geq f(x_0)$. It follows that $f(x_1)\geq f(x) \;\;\;\forall x\in R$.