# How does the existence of a limit imply that a function is uniformly continuous

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?

Any help is greatly appreciated.

-
You can cut off the function on some $[0, N]$ for a large $N$ and then use the limit condition. – Jonas Teuwen Oct 24 '11 at 19:21
To whomever first edited my question by including the LaTeX, thanks. I now see that the LaTeX expressions require both an opening 'dollar_sign' and a closing 'dollar_sign' (I tried expressions using only opening 'dollar_sign'). – Jubbles Oct 24 '11 at 19:33

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 $[0,N]$, of ir 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.

-
I don't have enough reputation points to edit, but shouldn't $\delta_0$ be $\delta_2$? – Jubbles Oct 24 '11 at 20:05
@Jubbles: Yes; I changed it midstream. I'll fix it. Thanks. – Arturo Magidin Oct 24 '11 at 20:07
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 $\delta = min(\delta_1\ \text{,} \delta_2)$. – Jubbles Oct 25 '11 at 0:23
@Jubbles: Yes, that will do it. – Arturo Magidin Oct 25 '11 at 2:54
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). – Jubbles Oct 25 '11 at 3:07

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?

-
Replace $|f(x)|$ by $|f(x)-\ell|$ twice, with $\ell=\lim\limits_{x\to+\infty}f(x)$. – Did Oct 24 '11 at 19:31
Oh, I had assumed that the limit was $0$. I'll modify. – Jonas Teuwen Oct 24 '11 at 19:32
@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 $f$. – Beni Bogosel Oct 25 '11 at 6:28
@BeniBogosel Yes, that's true. I will leave it this way I suppose, people can read your comment. – Jonas Teuwen Oct 25 '11 at 9:11

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.

-

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.

-
(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) – Hurkyl Feb 16 '14 at 13:29
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. – Jubbles May 28 '15 at 21:09
@Jubbles: $[0, +\infty]$ is not a subset of $\mathbb{R}$, so the characterization of compact subsets of $\mathbb{R}$ doesn't apply. – Hurkyl May 29 '15 at 5:31