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Math Help - compact space and continuous function

  1. #1
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    compact space and continuous function

    Hey, I have a topology problem here:
    Let K be a compact subset in R and let f : K → R be a continuous function. Prove
    that for every ε > 0 there exists Lε > 0 such that
    |f (x) − f (y)| ≤ Lε |x − y| + ε , for every x, y ∈ K.
    I'm thinking of using the Lipschitz condition, but not quite sure how to handle it. Could you anyone please give me a hint? Any input is appreciated!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rain07 View Post
    Hey, I have a topology problem here:
    Let K be a compact subset in R and let f : K → R be a continuous function. Prove
    that for every ε > 0 there exists Lε > 0 such that
    |f (x) − f (y)| ≤ Lε |x − y| + ε , for every x, y ∈ K.
    I'm thinking of using the Lipschitz condition, but not quite sure how to handle it. Could you anyone please give me a hint? Any input is appreciated!
    You can't use the Lipschitz condition because every continuous function on a compact space isn't Lipschitz!

    So, let \varepsilon>0 be given. By assumption f is uniformly continuous and so there exists some \delta>0 such that |f(x)-f(y|<\varepsilon\leqslant \varepsilon+M|x-y| for |x-y|<\delta and M\geqslant 0.

    Now, assume that \delta\leqslant |x-y| by the boundedness of f there exists some M' such that \text{diam }f(K)\leqslant M'. So, let M=\frac{M'}{\delta}. Then,

    |f(x)-f(y)|\leqslant M'=\delta M\leqslant |x-y|M\leqslant M|x-y|+\varepsilon.

    Thus, if |x-y|\geqslant \delta we have that |f(x)-f(y)\leqslant M|x-y|+\varepsilon, but this also works for |x-y|<\delta since the extra non-epsilon term is superfluous.
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  3. #3
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    I got it. I didn't think much about the properties of continuous function in the first place. Thanks a lot for your help!
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