# Show that $f$ is Uniformly Continuous

Let $f : [0,1] \to \mathbb{R}$ be defined by

$$f(x) = \begin{cases} {x \sin \frac{1} {x}}, & \text{ }x \in (0,1], \\ 0, & \text{ }x = 0. \end{cases}$$

How can I show that $f$ is uniformly continuous on $[0,1]$. Also is $f$ differentiable at $x = 0$?

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For homework-type questions, you should tell us what you have done so far, and where you got stuck. – GEdgar Apr 3 '12 at 0:04

Well $f(x)$ is continuous on $[0,1]$ and since $[0,1]$ is compact then $f(x)$ is uniformly continuous on $[0,1]$.

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...by Cantor's theorem. – Salech Alhasov Apr 3 '12 at 0:17
It is a nice answer. – Paul May 22 '13 at 1:04

Since $\sin(\text{anything})$ is between $-1$ and $1$ (inclusive), the function $x\sin\dfrac 1 x$ is between $-x$ and $x$.

Given $\varepsilon>0$, the function $f$, being continuous, is uniformly continuous on the compact set $[\varepsilon/2,1]$. On the set $[0,\varepsilon/2)$, any two values differ by $< \varepsilon$ because of what was noted in the first paragaph above, i.e. setting $\delta=\varepsilon/2$ works.

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$f(x)$ uniformly continuous on $[0,1]$:

Note that the function is continuous as long $x\ne 0$. Also, we have that $|x\sin\frac{1}{x}|\le x$, so we have $\lim_{x\to 0}=0$(Why?). We just proved that $f(x)$ continuous on $x=0$, and by Cantor's theorem, we can conclude that $f(x)$ is uniformly continuous on $[0,1]$

$f(x)$is not differentiable on $[0,1]$:

By definition,

$$\frac{f(x)-f(0)}{x-0}=\frac{x\sin\frac{1}{x}}{x}=\sin\frac{1}{x}$$

Now, what you can say about $\lim_{x\to 0}\sin\frac{1}{x}$?

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For a somewhat more basic proof of uniform continuity:

The "trick" is to consider two cases: 1) both $x$ and $y$ are near the origin; and 2) $x$ and $y$ are bounded away from the origin. In case 1), we can make a crude estimate of $|f(x)-f(y)|$, since both $f(x)$ and $f(y)$ will be small (the graph of $f$ is sandwiched between the graphs of $y=x$ and $y=-x$). In case 2), we can use the fact that the derivative of $f$ is bounded on $[\delta,1]$ for $\delta>0$ to make our estimate of $|f(x)-f(y)|$.

So, on with the show:

Let $\epsilon>0$.

For $x$ and $y$ in the interval $[0,\epsilon/ 2)$, we have $$\tag{1} |f(x)-f(y)|\le x+y<\epsilon.$$

For $x$ in the interval $[\epsilon/4,1]$, we have $$f'(x)=\sin{1\over x}-{1\over x}\cos{1\over x};$$ and so for $x\in[\epsilon/4,1]$ $$|f'(x)|\le 1+{1\over x}\le 1+{4\over\epsilon}={\epsilon+4\over\epsilon}.$$ Thus, by the Mean Value Theorem, if $x$ and $y$ are both in $[\epsilon/4,1]$, we have $$\tag{2} |f(x)-f(y)|\le {\epsilon+4\over\epsilon}|x-y|.$$

Now, given $\epsilon>0$, choose $\delta=\min\{\epsilon/4, \epsilon^2/(4+\epsilon)\}$.

Then if $|x-y|<\delta$, either

$\ \ \ 1)$ $x$ and $y$ are both in the interval $[0,\epsilon/2)$

or

$\ \ \ 2)$ $x$ and $y$ are both in the interval $[\epsilon/4,1 ]$.

In case $1)$, we have by $(1)$ that $|f(x)-f(y)|<\epsilon$.

In case $2)$ we have by $(2)$ that $|f(x)-f(y)|\le\textstyle{4+\epsilon\over\epsilon} |x-y| < ( \textstyle{ \epsilon+4\over\epsilon})\cdot{\epsilon^2\over 4+\epsilon}=\epsilon$.

For differentiability at 0, just consider the definition of $f'(0)$: $$\lim_{h\rightarrow 0}{f(h)-f(0)\over h}= \lim_{h\rightarrow 0} {h\sin{1\over h}-0\over h} =\lim_{h\rightarrow 0} \sin{1\over h}.$$ Does this limit exist?

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The new answer notification system is somewhat lacking... – David Mitra Apr 3 '12 at 0:48