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I am working on a homework problem from Avner Friedman's Advanced Calculus (#1 page 68) which asks

Suppose that $f(x)$ is a continuous function on the interval $[0,\infty)$. Prove that if $\lim_{x\to\infty} f(x)$ exists (as a real number), then $f(x)$ is uniformly continuous on this interval.

Intuitively, this argument makes sense to me. Since the limit of $f(x)$ exists on $[0,\infty)$, we will be able to find a $\delta > |x_0 - x_1|$ and this implies that, for any $\epsilon>0$, we have $\epsilon > |f(x_0) - f(x_1)|$ (independent of the points chosen). I am aware that the condition of uniform continuity requires that $\delta$ can only be a function of $\epsilon$, not $x$.

What information does the existence of a real-valued limit provide that implies $f(x)$ is uniformly continuous on this interval?

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    $\begingroup$ You can cut off the function on some $[0, N]$ for a large $N$ and then use the limit condition. $\endgroup$
    – JT_NL
    Oct 24, 2011 at 19:21
  • $\begingroup$ I think this result is more deeper and true in a general spaces such as locally compact metric spaces. See the proof below $\endgroup$
    – Guy Fsone
    Oct 8, 2017 at 17:29

5 Answers 5

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Remember the definition of "uniformly continuous":

$f(x)$ is uniformly continuous on $[0,\infty)$ if and only if for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that for all $x,y\in [0,\infty)$, if $|x-y|\lt \delta$, then $|f(x)-f(y)|\lt \epsilon$.

We also know that the limit exists. Call $$\lim_{x\to\infty}f(x) = L.$$ That means that:

For every $\varepsilon\gt 0$ there exists $N\gt 0$ (which depends on $\varepsilon$) such that if $x\gt N$, then $|f(x)-L|\lt \varepsilon$.

Finally, you probably know that if $f(x)$ is continuous on a finite closed interval, then it is uniformly continuous on that interval.

So: let $\epsilon\gt 0$. We need to show that there exists $\delta\gt0$ such that for all $x,y\in [0,\infty)$, if $|x-y|\lt \delta$, then $|f(x)-f(y)|\lt\epsilon$.

We first use a common trick: if you know that any value of $f(x)$ in some interval is within $k$ of $L$, then you know that any two values of $f(x)$ in that interval are within $2k$ of each other: because if $|f(x)-L|\lt k$ and $|f(y)-L|\lt k$, then $$|f(x)-f(y)| = |f(x)-L + L-f(y)| \leq |f(x)-L| + |L-f(y)| \lt k+k = 2k.$$

So: pick $N\gt 0$ such that for all $x\gt N$, $|f(x)-L|\lt \epsilon/2$. That means that if $x,y\gt N$, then $|f(x)-f(y)|\lt \epsilon$, by the argument above. So we are "fine" if both $x$ and $y$ are greater than $N$.

Now, we just need to worry about what happens if both $x$ and $y$ are in $[0,N]$, or if one of $x$ and $y$ is in $[0,N]$ and the other one is in $(N,\infty)$.

For both in $[0,N]$, we are in luck: since $f$ is continuous on $[0,\infty)$, then it is continuous on the finite closed interval $[0,N]$, hence is uniformly continuous there. So we know there exists $\delta_1\gt 0$ such that for all $x,y\in [0,N]$, if $|x-y|\lt\delta_1$, then we have $|f(x)-f(y)|\lt \epsilon$. So we just need to ensure that $x$ and $y$ are within $\delta_1$ of each other; that will ensure the inequality we want if $x$ and $y$ are both in $[0,N]$, or if they are both in $(N,\infty)$.

Now we run into a slight problem: what if, say, $x\in [0,N]$ and $y\in (N,\infty)$? Well, since $f$ is continuous at $N$, we know that we can ensure that $f(x)$ and $f(y)$ are both as close as we want to $f(N)$ provided that $x$ and $y$ are both very close to $N$. But if $x$ and $y$ are within some $\ell$ of $N$, then they are within $2\ell$ of each other (same argument as before); and if $f(x)$ and $f(y)$ are both within some $k$ of $f(N)$, then they will be within $2k$ of each other.

So: let $\delta_2$ be such that if $|a-N|\lt\delta_2$, then $|f(a)-f(N)|\lt \epsilon/2$. Then, if $x$ and $y$ are both within $\delta_2$ of $N$, then $|f(x)-f(y)|\lt \epsilon$, and we'll be fine.

In summary: we want to select a $\delta\gt 0$ that will ensure that if $|x-y|\lt\delta$, then:

  • If $x$ and $y$ are both less than $N$, then $|x-y|\lt \delta_1$;
  • If $x$ and $y$ are both greater than $N$, then it doesn't matter how close to one another they are; and
  • If one of $x$ and $y$ is less than $N$ and the other is larger than $N$, then they are each within $\delta_2$ of $N$.

To make sure the first condition happens, we just need to make sure that $\delta\leq\delta_1$. The second condition is easy. What should we require of $\delta$ in order for the second condition to hold? If we can find a $\delta$ that makes all three things happens simultaneously, we'll be done.

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    $\begingroup$ To answer your questions, I believe that we should require $\delta \le \delta_2$ in order for the second condition to hold. The $\delta$ that is sufficient for all conditions to hold is $\delta = min(\delta_1\ \text{,} \delta_2)$. $\endgroup$
    – Jubbles
    Oct 25, 2011 at 0:23
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    $\begingroup$ I appreciate the prompt and instructive answer. This is my first posted question on math.statexchange.com and I'm impressed by how helpful people have been so far (people are helpful at stackoverflow, but not quite this helpful). $\endgroup$
    – Jubbles
    Oct 25, 2011 at 3:07
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    $\begingroup$ @ArturoMagidin the last conclusion where u begin with "So:" is because the function is continuous and u can always find a finite closed interval around N ? $\endgroup$ Feb 19, 2015 at 2:13
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    $\begingroup$ @ArturoMagidin Excellent answer! $\endgroup$ Mar 1, 2016 at 19:09
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    $\begingroup$ It isn´t clear to me that those $\delta$s are independent from the points $x,y$, could you please clarify this? $\endgroup$
    – user561334
    Aug 17, 2018 at 16:30
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We know that for all $\varepsilon > 0$ there exists $X \in \mathbf R$ such that for all $x \geqslant X$ we have $|f(x) - \ell| < \varepsilon$ where $\displaystyle \ell = \lim_{x \to \infty} f(x)$.

So pick $\epsilon > 0$. Then we get from the previous condition a real number $X_\varepsilon > 0$. $f$ is uniformly continuous on $[0, X_\varepsilon]$ because that interval is compact.

Now, on $(X_\varepsilon, \infty)$ we have $|f(x) - \ell| < \varepsilon$. So we will always have $|f(x) - f(y)| \leq 2\varepsilon$ for $x, y$ in $(X_\varepsilon, \infty)$. Can you finish this?

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    $\begingroup$ Replace $|f(x)|$ by $|f(x)-\ell|$ twice, with $\ell=\lim\limits_{x\to+\infty}f(x)$. $\endgroup$
    – Did
    Oct 24, 2011 at 19:31
  • $\begingroup$ Oh, I had assumed that the limit was $0$. I'll modify. $\endgroup$
    – JT_NL
    Oct 24, 2011 at 19:32
  • $\begingroup$ @JonasTeuwen: You can assume WLOG that the limit is $0$, eventually by considering the function $g(x)=f(x)-\ell$. Then if $g$ is uniformly continuous so is $f$. $\endgroup$ Oct 25, 2011 at 6:28
  • $\begingroup$ @BeniBogosel Yes, that's true. I will leave it this way I suppose, people can read your comment. $\endgroup$
    – JT_NL
    Oct 25, 2011 at 9:11
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Consider for example the function $\tan : [0,\pi/2]\to [0,\infty]$ with the convention $\tan(\pi/2)=\infty$. This function is increasing and $C^\infty$ on $(0,\pi/2)$.

Then you may consider $g:[0,\pi/2] \to \Bbb{R}$ defined by

$$ g(x)= \begin{cases} f(\tan x), & x \in [0,\pi/2) \\ \lim_{x \to \infty}f(x)=f(\infty) & x=\pi/2\end{cases}$$

Then $g$ is continuous on a compact set, therefore it is uniformly continuous. We can obtain $f$ by using the composition $f(x)=g(\arctan x)$. We know that $(\arctan x)'=\frac{1}{1+x^2}\leq 1$, which means, by the intermediate value theorem, that $|\arctan x-\arctan y| \leq |x-y|$ for every $x,y \in [0,\infty)$. Now pick $\varepsilon >0$ in the uniform continuity of $g$. Then there exists $\delta >0$ such that $|x-y|<\delta \Rightarrow |g(x)-g(y)|<\varepsilon$. But then $|\arctan x-\arctan y|\leq |x-y|<\delta$, therefore

$$ |f(x)-f(y)|=|g(\arctan x)-g(\arctan y)|<\varepsilon $$

This means that for every $\varepsilon >0$ there exists $\delta$ (the same as in the uniform continuity of $g$) such that every $x,y \in [0,\infty)$ with $|x-y|<\delta$ it follows that $|f(x)-f(y)|<\varepsilon$. Therefore $f$ is uniformly continuous.


What I did above was just translating the structure of the space $[0,\infty]$ which is compact, to a usual compact interval. The condition that $f$ has a limit at $\infty$ means that $f$ is continuous on the space $[0,\infty]$, which is the compactification of $[0,\infty)$ by adding another point, namely $\infty$. Why is $[0,\infty]$ compact?

  • if $(y_n) \subset [0,\infty]$ then either $(y_n)$ has a bounded subsequence which by the Weierstrass theorem implies that there is a convergent subsequence, either $(y_n)$ is unbounded, which means that there is a subsequence converging to $\infty$.

    Then the theorem that says that any continuous function on a compact set is uniformly continuous can be applied. The arguments above are a workaround this.

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    $\begingroup$ this proof is elegant and cool $\endgroup$
    – Guy Fsone
    Oct 8, 2017 at 17:34
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If you've learned that continuous functions on compact sets are uniformly continuous, then this turns out to be a simple exercise with the extended real numbers.

We can extend $f$ by continuity to a function $f^*$ defined on the interval $[0, +\infty]$. That is, define

$$ f^*(x) = \begin{cases} f(x) & x < +\infty \\ \lim_{y \to +\infty} f(y) & x = +\infty \end{cases} $$

Since $[0, +\infty]$ is compact, and $f^*$ is continuous, we can conclude $f^*$ is uniformly continuous on $[0, +\infty]$. And thus $f^*$ is uniformly continuous on $[0, +\infty)$ as well, from which we conclude $f$ is uniformly continuous.

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  • $\begingroup$ (incidentally, there is not a standardized notation for the continuous extension of a function; the use of this decoration to mean that applies only within the context of this answer) $\endgroup$
    – user14972
    Feb 16, 2014 at 13:29
  • $\begingroup$ is $[0, +\infty]$ closed and bounded? Maybe I'm being naive, but it currently appears to me that any interval containing $\infty$ is by definition not closed nor bounded. $\endgroup$
    – Jubbles
    May 28, 2015 at 21:09
  • $\begingroup$ @Jubbles: $[0, +\infty]$ is not a subset of $\mathbb{R}$, so the characterization of compact subsets of $\mathbb{R}$ doesn't apply. $\endgroup$
    – user14972
    May 29, 2015 at 5:31
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    $\begingroup$ @Jubbles: Every open cover of $[0, +\infty]$ must include some open set containing an interval $(a, +\infty]$. The remaining open sets cover $[0,a]$ and so there is a finite collection that cover $[0,a]$. Together with $(a, +\infty]$ we get a finite subcover of $[0, +\infty]$. $\endgroup$
    – user14972
    Jul 23, 2016 at 18:29
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    $\begingroup$ @Jubbles: Or a short proof: $x/(1-x)$ extends to a homeomorphism from $[0,1]$ to $[0, +\infty]$. The former is compact, and thus so is the latter. $\endgroup$
    – user14972
    Jul 23, 2016 at 18:32
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This result is deeper and has a simpler proof in general Euclidean spaces or general metric spaces. In fact let's the result for Euclidean spaces.

Lemma If $f:\mathbb R^d\to \mathbb R$ is a continuous function such that $\lim_{|x|\to\infty } f(x) = L$ with $L\in\mathbb R$, then $f$ is uniformly continuous on $\mathbb R^d.$

Proof of the Lemma
Let $\varepsilon >0$. We want to show there is $\delta>0$ such that $ |f(x)-f(y)|\le \varepsilon$ whenever $|x-y|\le \delta$.

But since $\lim_{|x|\to\infty } f(x) = L$ we know that there is $R>0$ such that $$ \begin{equation}\label{eq1}\tag{I}|f(x)-L|\le \varepsilon/2 \qquad\text{whenever}\qquad |x|\ge R\end{equation} $$ But since $f$ is continuous it is uniformly continuous on the compact $$\bar{B}(0,R+1) = \{x\in\mathbb R^d: |x|\le R+1\}$$

Hein Borel result Every function $f$ continuous on compact set is therein uniformly continuous.

hence there exists $0<\delta<1$ such that for every \begin{equation}\label{eq2}\tag{II} \text{$x,y\in \bar{B}(0,R+1)$ with $|x-y|<\delta$ we have $ |f(x)-f(y)|\le \varepsilon /2$} . \end{equation}

Now let $x,y\in \mathbb R^d$ such that $|x-y|<\delta$

  1. If $x,y\notin \bar{B}(0,R)$ i.e $|x|>R$ and $|y|>R$ by $\eqref{eq1}$ it readily follows that $$|f(x)-f(y)|\le|f(x)-L|+|f(y)-L|\le \varepsilon .$$
  2. If $x,y\in \bar{B}(0,R)\subset\bar{B}(0,R+1)$ then \eqref{eq2} yields $$|f(x)-f(y)|\le \varepsilon .$$
  3. If $x \in \bar{B}(0,R)$ and $y\notin \bar{B}(0,R)$ then $y\in \bar{B}(0,R+1)$ indeed $$|y|\le|x|+|x-y|\le R+\delta <R+1$$ thus \eqref{eq2} yields $$|f(x)-f(y)|\le \varepsilon .$$

In any case we have $ |f(x)-f(y)|\le \varepsilon$for all $x,y\in\mathbb R^d$ such that $|x-y|\le \delta$. Which prove the uniform continuity.

The case where $ \mathbb R^d$ is replaced by a locally compact metric space $(X,d)$ analogious reason lead to same conclusion. except that, $ f:(X,d)\to \mathbb R$ vanish at infinty means: for a given $ \varepsilon > 0$ there exists a Compact subset $K$ of $X$ such that $|f(x)|\le \varepsilon $ for every $x\in X\setminus K.$ Here the ball $\bar{B}(0,R+1)$ is replaced by the compact set . $$K_1 = \{x\in X: d(x,K)\le 1\}$$

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