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Math Help - Another Unif. Cont.

  1. #1
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    Another Unif. Cont.

    Prove or disprove (by counterex.) f(x) = xlog(\frac{1}{x}) is unif. cont. on (0,1) and prove or disprove f(x) = \frac{cosx}{mx+b} is unif. cont. on (0,1) for all nonzer0 m,b \in \mathbb{R}

    So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given \epsilon > 0 set \delta = ?. If x,a \in (0,1), then |xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2 (it's always going to be < 2 because our interval is (0,1)). Then I say if (x,a) \in (0,1) and |x-a| < \delta then |f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}| .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
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  2. #2
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    Quote Originally Posted by Math2010 View Post
    Prove or disprove (by counterex.) f(x) = xlog(\frac{1}{x}) is unif. cont. on (0,1) and prove or disprove f(x) = \frac{cosx}{mx+b} is unif. cont. on (0,1) for all nonzer0 m,b \in \mathbb{R}

    So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given \epsilon > 0 set \delta = ?. If x,a \in (0,1), then |xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2 (it's always going to be < 2 because our interval is (0,1)). Then I say if (x,a) \in (0,1) and |x-a| < \delta then |f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}| .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
    The function f(x) = x\log(1/x) is not Lipschitz on (0,1), because its derivative becomes infinite as x\to0. But \textstyle\lim_{x\to0}f(x) does exist (it is 0), as is \textstyle\lim_{x\to1}f(x). So f extends to a continuous function on the closed interval [0,1] and is therefore uniformly continuous.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Math2010 View Post
    Prove or disprove (by counterex.) f(x) = xlog(\frac{1}{x}) is unif. cont. on (0,1) and prove or disprove f(x) = \frac{cosx}{mx+b} is unif. cont. on (0,1) for all nonzer0 m,b \in \mathbb{R}

    So I tried getting both into Lipschitz but was not successful. So all I know to do is for the first one is say: Given \epsilon > 0 set \delta = ?. If x,a \in (0,1), then |xlog\frac{1}{x}+alog\frac{1}{a}|\le |xlog\frac{1}{x}| + |alog\frac{1}{a}| < 2 (it's always going to be < 2 because our interval is (0,1)). Then I say if (x,a) \in (0,1) and |x-a| < \delta then |f(x) - f(a)| = |xlog\frac{1}{x}-alog\frac{1}{a}| .... How do I know what I should set delta to be? Also I am totally lost starting here. Let me know if these is even remotely on the right track because I don't know for sure. Thank you so much.
    The second one's kind of a pain in the ***. But, clearly both f(x)=\frac{1}{mx+b},g(x)=\cos(x) are uniformly continuous on I=(0,1) since \left|f'(x)\right|\leqslant\left|\frac{m}{b^2}\rig  ht|,\left|g(x)\right|\leqslant 1 and so they are Lipschitz. And, there is a theorem that says that since f,g are uniformly continuous and I is bounded that f\cdot g is uniformly continuous on I.

    That's the least painful way I can see.
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