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Let $f(x)=\sin x^2$.

1.) Give an $\varepsilon$-$\delta$ proof that $ f$ is continuous at every point $x_0\in \mathbb{R}$.

2.) Is this function uniformly continuous on $\mathbb{R}$? Explain why or why not.

Definition: $f$ is continuous at $x_0$ if $\forall\epsilon>0$ there $\exists\delta > 0$ such that $|f(x)-f(x_0)| <\epsilon$ whenever $|x-x_0|<\delta$.

All I got is "Let $\epsilon > 0$ be given.".

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It's a good start. – Antonio Vargas Dec 9 '12 at 10:09

3 Answers 3

up vote 4 down vote accepted

1.) Let $x_0 \in \mathbb{R}$. Recall that $\sin \alpha - \sin \beta = 2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$. Then as $x\rightarrow x_0$ we have: \begin{align*} \left|f(x) - f(x_0)\right| &= \left|\sin x^2-\sin x_0^2\right|\\ &= \left|2 \cos \frac{x^2+x_0^2}{2} \sin \frac{x^2-x_0^2}{2}\right|\\ &\le 2 \left|\sin \frac{x^2-x_0^2}{2}\right| \rightarrow 0. \end{align*} Just choose your $\delta$ so that the last term is small enough. If you can cite the continuity of $\sin$ and $x^2,$ that's simple. Otherwise, you might have to go through their proofs as well.

2.) No, $f$ is not uniformly continuous on $\mathbb{R}$. We can, for any $\delta$, choose $x_0$ so large that $x_0^2 = n\pi$ for some $n\in \mathbb{N}$, $x^2 = (n +\frac{1}{2}) \pi$, and $|x-x_0|<\delta$ (this might also require some additional proof). But then $|f(x) - f(x_0)| = 1$, so $f$ cannot be uniformly continuous.

You can fill in the details, but that's a general approach.

Edit: the other responses are much cleaner, but feel free to get your hands dirty with this one. ;)

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Would $\delta=\min[1,\frac{\varepsilon}{1+2x_0}]$ work? – Joakim Dec 13 '12 at 4:43

Use the handy fact that $\lvert\sin x-\sin y\rvert≤\lvert x-y\rvert$ to get $$ \lvert\sin x^2-\sin y^2\rvert≤\lvert x^2-y^2\rvert=\lvert x+y\rvert\cdot\lvert x-y\rvert. $$ Now start playing around with your $\epsilon$s and $\delta$s. For the second question, look in particular for choices of $x$ and $y$ with $\lvert x-y\rvert$ small, but where the inequality above is not very sharp (i.e., it should be close to an equality).

I hope this is enough to get your thinking unstuck.

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If you are allowed to use (or able to prove) that $$|\sin(a)-\sin(b)|\le|a-b|\qquad(1)$$ you are done, with $a=x^2$ and $b=(x+\varepsilon)^2$ (you choose $\delta:=|2x\varepsilon+\varepsilon^2|$)

Proving (1) can be tedious, depending on what you are allowed to assume. Given the addition theorems, it can be way easier that without, using complex numbers, it is rather trivial. Mind giving more information?

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What's the complex proof for (1)? – Josh Keneda Dec 9 '12 at 11:58
@JoshKeneda: It follows from the intermediate value theorem: it says that there exists some $c$ such that $\sin(a)-\sin(b)=\cos ^\prime (c)(a-b) \leq a-b$ for some $ c \in (b,a)$. – Fredrik Meyer Dec 9 '12 at 12:05
A more fundamental proof: $\lvert b-a\rvert$ is the distance betwwn $(\cos a,\sin a)$ and $(\cos b,\sin b)$ along the unit circle, hence ≥ the straight line distance, which in its turn is at least the difference in the $y$ coordinates of the two points. – Harald Hanche-Olsen Dec 9 '12 at 12:49

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