# Determine which of the following functions are uniformly continuous on the open unit interval (0,1)

Could someone help me through this problem? Determine which of the following functions are uniformly continuous on the open unit interval $(0,1)$ :

a) $1/(1-x)$

b) $1/(2-x)$

c) $\sin{x}$

d) $\sin{1/x}$

e) $x^3$

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What have you tried? –  user21436 Apr 23 '12 at 13:05
If this is homework, you should add the homework tag, please. –  Gerry Myerson Apr 23 '12 at 13:08
A nice exercise: $f : (0,1) \to \mathbb{R}$ is uniformly continuous if and only if it is continuous on $(0,1)$ and $\lim_{x \downarrow 0} f(x)$ and $\lim_{x \uparrow 1}f(x)$ both exist. –  Nate Eldredge Apr 24 '12 at 13:46

One way in which I check uniform continuity of function is the following: Given an interval $[a,b]$ look at points where the function increases rapidly, for example consider $f(x)=\frac{1}{x}$ on $(0,1)$. This function can't be uniformly continuous because as $x \to 0, \frac{1}{x} \to \infty$ and hence you can't have $|f(x)-f(y)| < \epsilon$ for all $x \in (0,1)$.

Another useful way of checking uniform continuity of differentiable functions to look at:"$\text{their derivative and see if they are bounded in that given interval.}$" For example, $f(x)=x^{3}$ on $(0,1)$ has derivative $3x^{2}$ and attains a maximum of $3$ as $x \to 1$. So $f$ here is bounded by $3$. So this function has to be uniformly continuous. Similarly functions such as $\sin{x},\cos{x}$ are uniformly continuous on any given interval. But functions such as $f(x)=x^{2},x^{3}$ aren't uniformly continuous over $\mathbb{R}$ but they are $\text{uniformly continuous}$ over any interval.

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The test of derivatives is actually a test of "lipschitzianity". A lipschitz map is always uniformly continuous. –  Siminore Apr 23 '12 at 13:18
@Siminore: Well I am not actually familiar with such terms. –  user9413 Apr 23 '12 at 13:19
A function $f$ is liptschiz when there exists a universal constant $L>0$ such that $|f(x)-f(y)|\leq L |x-y|$ for every $x$ and $y$ (in the domain of $f$, of course). This happens, as you suggest, if the first derivative $f'$ is bounded. It is immediate to check that such a function is uniformly continuous, by choosing $\delta=\varepsilon/L$ in the definition. –  Siminore Apr 23 '12 at 13:24
@Siminore. I'm pretty sure that Chandrasekhar is familiar with the definition of a Lipschitz map. Your terminology, in particular the word "lipschitzianity", was likely what caused the confusion. –  Thomas E. Apr 23 '12 at 13:49
Well, $x \mapsto \sin x$ can be extended continuously to $[0,1]$, and therefore it is uniformly continuous on $(0,1)$. The same for $x \mapsto x^3$. The function in (a) has a vertical asymptote $x=1$, and it is very easy to check that it cannot be uniformy continuous. Formally, choose $x_n=1-\frac{1}{n}$ and $y_n=1-\frac{1}{n+1}$. Then $x_n \to 1$, $y_n \to 1$, but $$\frac{1}{1-x_n}-\frac{1}{1-y_n}=n-(n+1)=-1.$$ Try to understand what happens in case (b). More generally, it is a nice exercise to prove the following: let $f \colon [a,b) \to \mathbb{R}$ a continuous function. If $\lim_{x \to b-} f(x) = \pm \infty$, then $f$ can't be uniformly continuous. Hint: otherwise, there would exist a continuous extension $\tilde{f} \colon [a,b] \to \mathbb{R}$ of $f$, namely $\tilde{f}(x)=f(x)$ for every $x \in [a,b)$.