# Thread: Uniformly continuous?

1. ## Uniformly continuous?

Let $\displaystyle f(x)$ be defined on an interval $\displaystyle I$ (not assumed compact), and assume that its secants have bounded slope, i.e., for any two distinct points on the graph of $\displaystyle f(x)$ over $\displaystyle I,$ the slope $\displaystyle \lambda$ of the line joining them is bounded: $\displaystyle |\lambda| < K,$ where $\displaystyle K$ is some fixed number not depending on the two points selected.
$\displaystyle (a)$ Prove $\displaystyle f(x)$ is uniformly continuous on $\displaystyle I.$
$\displaystyle (b)$ Does $\displaystyle \sqrt{x}$ on $\displaystyle [0,1]$ satisye the above hypothesis on $\displaystyle f(x)?$ Is it uniformly continuous on the interval?

2. For the first one we have that for all $\displaystyle x,y \in I$ with $\displaystyle x \neq y$ $\displaystyle \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } \leq K$ so $\displaystyle \vert f(x) - f(y) \vert \leq K \vert x-y \vert$ and so $\displaystyle f$ is Lipschitz (taking $\displaystyle \delta = \epsilon /K$ it follows that $\displaystyle f$ is uniformly continous).

For the second one take $\displaystyle x=0$ then $\displaystyle \frac{ \sqrt{y} }{y} = \frac{1}{ \sqrt{y} }$ and it follows that the slope of a line through $\displaystyle 0$ and $\displaystyle f(y)=\sqrt{y}$ can be made arbitrarily large taking $\displaystyle y$ close to $\displaystyle 0$ so $\displaystyle \sqrt{x}$ doesn't satisfy the property in $\displaystyle [0,1]$, although it is unif. cont. in $\displaystyle [0,1]$ since it's continous on a compact interval.