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Suppose that $f: [0,\infty) \to \mathbb{R}$ is a continuous function and $\displaystyle \lim_{x \to \infty}f(x)$ exists. Prove that $f$ is uniformly continuous on $[0,\infty)$.

Here's my solution, but because the problem has been given in a math contest preparation program I'm afraid that there's some point that I might be missing.

Since $\displaystyle \lim_{x \to \infty}f(x)$ exists, there exists $L \in \mathbb{R}$ such that:

$\forall \epsilon>0, \exists M>0 \text{ such that } x > M \implies |f(x)-L|<\epsilon/2$

$\forall \epsilon>0, \exists M>0 \text{ such that } y > M \implies |f(y)-L|<\epsilon/2$

Therefore, $\forall x,y \in (M,\infty)$ we have shown that: $|f(x)-f(y)|<\epsilon$.

Now, let's consider $f$ on $[0,M]$. Since $[0,M]$ is a compact interval and $f$ is continuous, $f$ is uniformly continuous on $[0,M]$:

$\forall \epsilon>0, \exists \delta_1>0, \forall x,y \in [0,M]: |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon$.

Now let us focus our attention at $x=M$.

Since $f: [0,\infty) \to \mathbb{R}$ is continuous, it is continuous at $x=M$ as well. Therefore:

$\epsilon>0, \delta_2>0, \forall x: |x-M|<\delta_2 \implies |f(x)-f(M)|<\epsilon/2$

So, for any $x,y \in (M-2\delta_2,M+2\delta_2)$ we have $|f(x)-f(y)|<\epsilon$

Now if we set $\delta=\min\{\delta_1,\delta_2\}$ we see that all the $3$ cases above become true and we conclude that $\forall x,y \in [0,\infty): |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$. This proves that $f$ is uniformly continuous on $[0,\infty)$.

Is my nonsense considered a proof?

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    $\begingroup$ Yes, it seems correct. $\endgroup$ – Berci May 10 '14 at 19:22
  • $\begingroup$ @Berci: Thank you, Berci. Do I actually need to consider continuity around $x=M$ in my proof or it's redundant? $\endgroup$ – math.n00b May 10 '14 at 19:26
  • $\begingroup$ math.stackexchange.com/questions/768984/… $\endgroup$ – Clin May 10 '14 at 19:42
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Did you make a mistake in ".. for any $x, y \in (x-2\delta_2, x+2\delta_2)$..."? Because M'd disappeared. But I think that you are almost there.

It worth trying this with sequence. If the result is false, then there is $\epsilon>0$ and sequences $x_n$ and $y_n$ such that

$$|x_n - y_n| < 1/n$$ and $$|f(x_n)-f(y_n)|\geq \epsilon.$$

The hypothesis shows that $y_n$ and $x_n$ are bounded, and Bolzano Weierstrass gives a contradiction.

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  • $\begingroup$ Yes, it should be $\forall x,y \in (M-2\delta,M+2\delta)$. Thanks for pointing it out. Would you please state what "Bolzano Weirestrauss" states and explain your proposed solution a little bit more? $\endgroup$ – math.n00b May 10 '14 at 19:33
  • $\begingroup$ You can find more information here: en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem . The $\epsilon$, $x_n$ and $y_n$ appears when you suppose that $f$ is not unif continuously. If one of the sequences converges, the other also coverges to the same limit (including infinity). So, there is no subsequence 'going' to infinity, because of the existence of L. So, both sequences are bounded. The BW theorem and the same argument give the contradiction $\endgroup$ – user85353 May 10 '14 at 19:35
  • $\begingroup$ Ah, I see. So, $x_n$ and $y_n$ will have converging subsequences but the image of those subsequences always stay at an $\epsilon$ distance away, hence contradiction (because $f$ is continuous). Is this what you have in mind? Does it help us avoid dividing $[0,\infty)$ to $[0,M]$ and $[M,\infty)$? And finally, is my proof correct now? $\endgroup$ – math.n00b May 10 '14 at 19:41
  • $\begingroup$ Yes, you got it. I think you just have to use $\delta_2/2$ instead of $2\delta_2$. :) $\endgroup$ – user85353 May 10 '14 at 19:45
  • $\begingroup$ Why $\delta_2/2$? Both $x$ and $y$ lie in an interval centered at $M$ with radius $\delta_2$, but they can be at the opposite sides of the interval, that makes their maximum distance equal to $2\delta_2$. Right? $\endgroup$ – math.n00b May 10 '14 at 19:54

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