# Not Uniform Continuous?

• Oct 12th 2008, 12:08 PM
Not Uniform Continuous?
Consider the function $\displaystyle f(x)=xsinx$

Now, $\displaystyle f'(x)=xcosx+sinx$, which is not bounded, so it is not uniformly continuous.

So pick $\displaystyle \epsilon = 1$, let $\displaystyle \delta > 0$, for every $\displaystyle x,y \in \mathbb {R} , |x-y| < \delta$, we need $\displaystyle |xsinx-ysiny| \geq \epsilon$, but how should I go about that? Thanks.
• Oct 12th 2008, 12:29 PM
Laurent
"$\displaystyle f$ is not uniformly continuous" means:
$\displaystyle \exists\varepsilon_0>0, \forall \delta>0, \exists x,y$ such that $\displaystyle |x-y|<\delta$ and $\displaystyle |f(x)-f(y)|>\varepsilon_0$.

I let you get convinced yourself that a proof would consist in finding a sequence a couples $\displaystyle (x_n,y_n)_{n\geq 1}$ such that $\displaystyle |x_n-y_n|\to_n 0$ and $\displaystyle |f(x_n)-f(y_n)|>\varepsilon_0$ for some $\displaystyle \varepsilon_0$ (or even $\displaystyle |f(x_n)-f(y_n)|\to_n+\infty$, if possible).

In order to choose those couples, look at (or just imagine) what the graph looks like: the $\displaystyle x_n,y_n$ must be close but the $\displaystyle f(x_n),f(y_n)$ must be way apart from each other.

$\displaystyle x_n=n\pi$ is a possible start.
• Oct 12th 2008, 04:50 PM
Thanks for the hint. So for the choice that $\displaystyle x_n = n \pi$, we would have $\displaystyle f(x_n) = 0 \ \ \ \forall n$, but any point right afterward would get larger as n increases.

Pick $\displaystyle y_n = x_n + \frac { \delta}{2}$, then $\displaystyle |x_n - y_n | < \delta$, but $\displaystyle |f(x_n)-f(y_n)| \rightarrow \infty$

Is this right?
• Oct 12th 2008, 11:59 PM
Laurent
Quote:

Pick $\displaystyle y_n = x_n + \frac { \delta}{2}$, then $\displaystyle |x_n - y_n | < \delta$, but $\displaystyle |f(x_n)-f(y_n)| \rightarrow \infty$

Is this right?

No, because for every $\displaystyle \delta>0$ you must have $\displaystyle |x_n-y_n|<\delta$ for some $\displaystyle n\geq 0$. This is why I said you must have $\displaystyle |x_n-y_n|\to_n 0$.

You can write $\displaystyle y_n=x_n+u_n=n\pi+u_n$ where $\displaystyle u_n\to_n 0^+$ is to be determined.

You have $\displaystyle |f(y_n)-f(x_n)|=|f(y_n)|=|(n\pi+u_n)\sin(n\pi+u_n)|=|(n\pi +u_n)(-1)^n\sin(u_n)|$$\displaystyle \sim n\pi\sin(u_n)\sim n\pi u_n$ as $\displaystyle n\to\infty$. So that you must choose $\displaystyle u_n\to_n 0$ such that $\displaystyle \liminf_n n u_n>0$. For instance, $\displaystyle u_n=\frac{1}{n}$ (hence $\displaystyle \varepsilon_0$ must be small enough), or $\displaystyle u_n=\frac{1}{\sqrt{n}}$ (and any $\displaystyle \varepsilon_0$ is adequate).

You can do a more explicit proof using the convexity inequality $\displaystyle \sin(u_n)\geq\frac{2u_n}{\pi}$ which holds as soon as $\displaystyle 0\leq u_n\leq \frac{\pi}{2}$. This yields to the same possibilities for the choice of $\displaystyle (u_n)_n$.
• Oct 16th 2008, 09:32 AM
Thanks for the help. But one thing I'm not clear is, why is it enough to show that $\displaystyle |x_n-y_n|\to_n 0$ implies $\displaystyle |f(x_n)-f(y_n)|>\varepsilon_0$?
You know that "$\displaystyle f$ is not uniformly continuous" means:
$\displaystyle \exists\varepsilon_0>0, \forall \delta>0, \exists x,y$ such that $\displaystyle |x-y|<\delta$ and $\displaystyle |f(x)-f(y)|>\varepsilon_0$.
Suppose we found a sequence of couples $\displaystyle (x_n,y_n)_{n\geq 1}$ and an $\displaystyle \varepsilon_0>0$ such that $\displaystyle |x_n-y_n|\to_n 0$ and, for all $\displaystyle n\geq 1$, $\displaystyle |f(x_n)-f(y_n)|>\varepsilon_0$.
Then, for every $\displaystyle \delta>0$, there is an $\displaystyle n$ such that $\displaystyle |x_n-y_n|<\delta$ (because of the convergence to 0). And for this same $\displaystyle n$, we have $\displaystyle |f(x_n)-f(y_n)|>\varepsilon_0$. As a consequence, we have realized the above definition of "not uniformly continuous" with $\displaystyle x_n,y_n$ in place of $\displaystyle x,y$.