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Math Help - Uniformly continuous?

  1. #1
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    Uniformly continuous?

    Let f(x) be defined on an interval I (not assumed compact), and assume that its secants have bounded slope, i.e., for any two distinct points on the graph of f(x) over I, the slope \lambda of the line joining them is bounded: |\lambda| < K, where K is some fixed number not depending on the two points selected.
    (a) Prove f(x) is uniformly continuous on I.
    (b) Does \sqrt{x} on [0,1] satisye the above hypothesis on f(x)? Is it uniformly continuous on the interval?
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  2. #2
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    For the first one we have that for all x,y \in I with x \neq y \frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } \leq K so \vert f(x) - f(y) \vert \leq K \vert x-y \vert and so f is Lipschitz (taking \delta = \epsilon /K it follows that f is uniformly continous).

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