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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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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
up vote 2 down vote accepted

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. – T. Eskin 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 $x \in [a,b)$.

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