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Question: Let $X=X'=\mathbb{R}$ with the usual metric.

(a) Show that a polynomial function $p(x)$ on $\mathbb{R}$ is uniformly continuous if and only if $\deg(p(x))<2$.

So I thought all polynomial functions were continuous? Clearly all polynomial functions regardless of degree preserve limits so that if $x_n\rightarrow x$, $p(x_n)\rightarrow p(x)$, hence continuous?

(b) Show that $f(x)=\sin(x)$ is uniformly continuous on $\mathbb{R}$.

For this one, I have to show that given $\epsilon>0$, exists $\delta>0$ s.t. for any $x,y\in\mathbb{R}$, $d'(f(x),f(y))<\epsilon$ whenever $d(x,y)<\delta$. The important thing is that $\delta$ is independent of the point in $\mathbb{R}$. Not sure how to find this $\delta$ though.

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It's good to keep in mind that uniform continuity, while equivalent to continuity when your domain is compact, is much stronger than continuity in general. As a warm-up, it's good to prove that $x^2$ is not uniformly continuous on the real line: for a given $\varepsilon$, you have to choose smaller and smaller $\delta$s as you move away from $0$. –  Dylan Moreland Feb 6 '12 at 14:27
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3 Answers

Hint: Just check whether the derivative of the function is bounded. If the derivative is bounded then the function is uniformly continuous.

And if $f(x)$ is a polynomial with $\operatorname{deg}{f(x)} < 2$ then you can see that $f'(x)$ is bounded. Since the only polynomials of $\operatorname{deg} < 2$ are constant functions and linear polynomials.

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For (b): $$\eqalign{ |\sin (x+h)-\sin x|&= |( \sin x\cos h+\sin h\cos x) - \sin x|\cr &=|\sin x( \cos h-1)+\sin h\cos x|\cr &\le|\sin x|| \cos h-1 |+|\sin h||\cos x|\cr &\le | \cos h-1 |+|\sin h|. \cr } $$

Now note that $| \cos h-1|+|\sin h|$ can be made as small as you like, by taking $h$ sufficiently small, independent of $x$.

Part a) can get messy, I imagine, but you could appeal to the fact that for large $x$ a polynomial behaves like its leading term. Also, for $n>1$ fixed, $x>0$, and $h$ positive $$ |(x+h)^n-x^n| \ge|nh x^{n-1}|\quad\buildrel{x\rightarrow\infty}\over\longrightarrow \quad\infty. $$ Which shows that $f(x)=x^n$ is not uniformly continuous on $\Bbb R$.

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For all x,y, in domain,

$|\sin(x) - \sin(y)| = |x-y|\cos(c)$ (from mean value theorem, where $c$ is a point between $x$ and $y$) But since $\cos(c)$ is bounded by $1$, $|x-y|\cos(c) \leq |x-y|$

So that $|\sin(x) - \sin(y)| \leq |x-y|$. Now just pick $\delta = \epsilon$.

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