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Math Help - Uniformly Cont. Func

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    Uniformly Cont. Func

    1.) Let f(x) = \frac{x}{x-1}. Prove that f(x) is uniformly continuous on the interval [ 1.5,\infty).

    2.) Suppose that f and g are uniformly continuous on some interval A. Either prove or disprove f+g is uniformly continous on A.
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    This website should help.
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    Quote Originally Posted by fifthrapiers View Post
    1.) Let f(x) = \frac{x}{x-1}. Prove that f(x) is uniformly continuous on the interval [ 1.5,\infty).
    Theorem: If f is continuous on a closed interval [a,b], then f is uniformly continuous on [a,b]
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    Quote Originally Posted by fifthrapiers View Post
    2.) Suppose that f and g are uniformly continuous on some interval A. Either prove or disprove f+g is uniformly continous on A.
    go by the definition. there may be a shorter way, but i can't remember now.


    Defintion: Let f be a real valued function defined on a set S \subset \mathbb{R}. Then f is uniformly continuous on S if

    for each \epsilon > 0 there exists \delta > 0 such that for all x,y \in S and |x - y|< \delta we have |f(x) - f(y)|< \epsilon.



    I will start you off in the proof.

    Proof: Let f,g be uniformly continuous on some interval I \subset \mathbb{R}.

    Then we have that for each \epsilon > 0 there exists \delta_1 > 0 such that for all x,y \in I and |x - y|< \delta_1 we have |f(x) - f(y)|< \frac {\epsilon}2.

    A similar thing holds true for g if we choose some \delta_2 > 0.

    Now let h(x) = f(x) + g(x), fix such an \epsilon > 0 and choose \delta = \mbox{min} \{ \delta_1, \delta_2 \}.

    Then, for all x,y \in I and |x - y|< \delta, we have:

    |h(x) - h(y)| = |(f(x) + g(x)) - (f(y) + g(y))| = ...

    finish it off
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    Quote Originally Posted by Jhevon View Post
    Theorem: If f is continuous on a closed interval [a,b], then f is uniformly continuous on [a,b]
    But [1.5,\infty) is not a finite closed interval. How can you use the theorem?
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    Quote Originally Posted by ThePerfectHacker View Post
    But [1.5,\infty) is not a finite closed interval. How can you use the theorem?
    the theorem didn't say anything about the interval being finite, so i thought it was ok. blame the author

    EDIT: ...oh, i see [a,b] implies finite...ok
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    is up to his old tricks again! Jhevon's Avatar
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    could we use the theorem about the derivative being bounded on the interval?
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    Quote Originally Posted by Jhevon View Post
    could we use the theorem about the derivative being bounded on the interval?
    That is how I would do it.
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    Quote Originally Posted by ThePerfectHacker View Post
    That is how I would do it.
    that's what i originally thought of using, but i saw that the first theorem (if applicable) would get the job done a lot faster, with a lot less work.

    for the benefit of the class, here's the new theorem we are proposing:

    Theorem: Let f be a continuous function on an interval I ( I may be bounded or unbounded, hehe, something my last theorem was missing). Let I^o be the interval obtained by removing from I any endpoints that I may have. If f is differentiable on I^o and if f' is bounded on I^o, then f is uniformly continuous on I.
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    I'm still very stuck on #1. I hate proving unif. cont.

    I attempted #2, and I'd appreciate if you could look at it for me:

    Let \epsilon > 0 be given. Then, \frac{\epsilon}{2} > 0, thus, \exists R \in \mathbb{N} \ni r \geq R \implies ||f_r - f|| < \frac{\epsilon}{2} since (f_n) is unif. convergence to f on A. Similarly for g, \exists P \in \mathbb{N} \ni p \geq P \implies ||g_p - g|| < \frac{\epsilon}{2}. Let M = max{R,P}. Now, for n \geq N, we have:

    ||(f_n + g_n) - (f + g)|| = ||(f_n - f) + (g_n - g)|| \leq ||f_n - f|| + ||g_n - g|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

    Thus, f + g is unif. cont. on A.
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    Quote Originally Posted by fifthrapiers View Post
    I'm still very stuck on #1. I hate proving unif. cont.

    I attempted #2, and I'd appreciate if you could look at it for me:

    Let \epsilon > 0 be given. Then, \frac{\epsilon}{2} > 0, thus, \exists R \in \mathbb{N} \ni r \geq R \implies ||f_r - f|| < \frac{\epsilon}{2} since (f_n) is unif. convergence to f on A. Similarly for g, \exists P \in \mathbb{N} \ni p \geq P \implies ||g_p - g|| < \frac{\epsilon}{2}. Let M = max{R,P}. Now, for n \geq N, we have:

    ||(f_n + g_n) - (f + g)|| = ||(f_n - f) + (g_n - g)|| \leq ||f_n - f|| + ||g_n - g|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon

    Thus, f + g is unif. cont. on A.
    Grr, I mis-read it. I proved that they are uniformly CONVERGENT, not uniformly CONTINUOUS!!

    Well, I know how to prove f is unif. continuous but not f + g...

    So proving f is unif. continuous...:

    Let (f_n) be the sequence of functions where f_n : A \rightarrow \mathbb{R} are unif. cont. on A and that [tex](f_n) converges unif. to f on A. Let \epsilon > 0 be given. Then, \exists a natural number N \in \mathbb{N} \ni \forall n \geq N, |f_{n}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A by unif. conv. of f_n to f on A. Particularly, |f_{N}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A.

    As f_N is unif. cont on A, \exists \delta > 0 \ni |f_{N}(x) - f_{N}(y)| < \frac{\epsilon}{3} whenever x,y \in A satisfy |x-y| < \delta. Then we have for this \delta, suppose x,y \in A satisfy |x - y| < \delta and consider:

    \begin{eqnarray}
    |f(x) - f(y)| &=& |f(x) - [f_{N}(x) - f_{N}(y)] - [f_{N}(y) - f_{N}(y)] - f(y)|\\
    &=& |[f(x) - f_{N}(x)] + [f_{N}(x) - f_{N}(y)] + [f_{N}(y) - f(y)]|\\
    &\leq& |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(y)| + |f_{N}(y) - f(y)|\\
    &<& \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{eqnarray}

    Thus, f is unif. cont. on A.

    (NOTE: I get LaTeX syntax error on the last part (for the eqnarray stuff, not sure why) so I left it without [tex] so you can see the general jist. I think it's easy to read just in latex format.

    But, I have no idea how to show f + g will be unif cont...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    Grr, I mis-read it. I proved that they are uniformly CONVERGENT, not uniformly CONTINUOUS!!

    Well, I know how to prove f is unif. continuous but not f + g...

    So proving f is unif. continuous...:

    Let (f_n) be the sequence of functions where f_n : A \rightarrow \mathbb{R} are unif. cont. on A and that [tex](f_n) converges unif. to f on A. Let \epsilon > 0 be given. Then, \exists a natural number N \in \mathbb{N} \ni \forall n \geq N, |f_{n}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A by unif. conv. of f_n to f on A. Particularly, |f_{N}(x) - f(x)| < \frac{\epsilon}{3} \forall x \in A.

    As f_N is unif. cont on A, \exists \delta > 0 \ni |f_{N}(x) - f_{N}(y)| < \frac{\epsilon}{3} whenever x,y \in A satisfy |x-y| < \delta. Then we have for this \delta, suppose x,y \in A satisfy |x - y| < \delta and consider:

    \begin{eqnarray}
    |f(x) - f(y)| &=& |f(x) - [f_{N}(x) - f_{N}(y)] - [f_{N}(y) - f_{N}(y)] - f(y)|\\
    &=& |[f(x) - f_{N}(x)] + [f_{N}(x) - f_{N}(y)] + [f_{N}(y) - f(y)]|\\
    &\leq& |f(x) - f_{N}(x)| + |f_{N}(x) - f_{N}(y)| + |f_{N}(y) - f(y)|\\
    &<& \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon \end{eqnarray}

    Thus, f is unif. cont. on A.

    (NOTE: I get LaTeX syntax error on the last part (for the eqnarray stuff, not sure why) so I left it without [tex] so you can see the general jist. I think it's easy to read just in latex format.

    But, I have no idea how to show f + g will be unif cont...
    firstly, you do not have to prove (nor do you have enough information to do so) that f is uniformly continuous, that was given. secondly, we are not dealing with a sequence of functions here. were that the case, we would be concerned with uniform convergence, not uniform continuity.

    your proof should start EXACTLY as mine did (you can neglect to define the new function as h(x) if you wish), all you have to do is complete my proof
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    I'm still very stuck on #1. I hate proving unif. cont.
    all you have to show is that the derivative of the function exists and is bounded on (1.5, oo). then apply the theorem i mentioned in post #9
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