Examine the uniform continuity of f : $\displaystyle [0,\infty) \rightarrow $ R given by f(x) := $\displaystyle e^{-x}$ on $\displaystyle [0,\infty)$.

I know it's not uniformly continuous but how do I prove that?

Printable View

- Nov 15th 2009, 06:08 PMthaopandaProve it's not uniformly continous
Examine the uniform continuity of f : $\displaystyle [0,\infty) \rightarrow $ R given by f(x) := $\displaystyle e^{-x}$ on $\displaystyle [0,\infty)$.

I know it's not uniformly continuous but how do I prove that? - Nov 15th 2009, 06:18 PMredsoxfan325
- Nov 15th 2009, 06:24 PMJose27
It is uniformly continous:

By the mean value theorem we have $\displaystyle \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } = \vert f'(c)\vert$ where $\displaystyle c\in [x,y]$. Taking the supremum over all $\displaystyle x,y\in [0, \infty )$ and noting that $\displaystyle \vert f'(x) \vert \leq 1$ we have that $\displaystyle \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } \leq 1$ for all $\displaystyle x\neq y$ and so $\displaystyle \vert f(x)-f(y) \vert \leq \vert x-y\vert$. Now taking $\displaystyle \delta < \epsilon$ in the definition of unif. cont. the result follows.

Actually you can prove that for a continously diff. function it's equivalent:

1) $\displaystyle f'$ is bounded

2) There exists $\displaystyle m>0$ such that $\displaystyle \vert f(x)-f(y)\vert \leq m\vert x-y\vert$