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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!

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3 Answers 3

up vote 2 down vote accepted

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 $f(x_1)\geq f(x) \;\;\;\forall x\in R$.

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A function is uniformly continuous if for every small "range error" $\Delta y$, there is some small \emph{non-zero} "domain error" $\Delta x$ such that a change of $\pm \Delta x$ around \emph{any} element $x$ in the domain will result in a change of only $\pm \Delta y$ in the range.

Now the derivative $dy/dx$ is the ratio between "change in range" and "change in domain". Going back to the definition of uniform continuity, we fixed $\Delta y$ and wanted $\Delta x$ to be bounded \emph{from below} (think why we want a bound from below rather than from above). So the derivative needs to be bounded from above.


For the second question, suppose that the function is not constant. Take some point which has a value different from $L$, say $M > L$. Now $M$ is not necessarily the maximum, but perhaps there is a way to identify the maximum, as some kind of supremum?

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Q1: Try proving by contradiction. If $f$ is uniformly continuous, there exists some $\delta$ such that $|f(x) - f(y)| \le 1$ whenever $|x-y| \le \delta$. Now find an interval $[a,b]$ of length longer than $\delta$ such that $f' > 1/\delta$ on $[a,b]$. If we pick $x=a$, $y=a+\delta$, what does the mean value theorem tell you about $|f(x)-f(y)|$?

Q2: If $f$ is identically equal to $L$ we are done. Otherwise there exists $x_0$ with, say, $f(x_0) > L$. Now find a closed interval $[a,b]$ with $f(x) < f(x_0)$ outside $[a,b]$, and consider the maximum of $f$ on $[a,b]$...

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Regarding Q1: How exactly I find the interval [a,b]? and if I understand correctly, then $|f(x) - f(y)| > 1$. –  Ma.H Jan 30 '11 at 22:36
    
@Ma.H: Actually, the interval can be of the form $[a,\infty)$. Use the fact that $f'(x) \to +\infty$ as $x \to +\infty$... –  Nate Eldredge Jan 31 '11 at 1:10
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