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Math Help - Uniform Convergence

  1. #1
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    Uniform Convergence

    (1) Assume that (f_n) converges uniformly to f on A and that each f_n is uniformly continuous on A. Prove that f is uniformly continuous on A

    (2) Assume (f_n) converges uniformly to f on a compact set K, and let g be a continuous function on K satisfying g(x) \not= 0. Show f_n/g) converges uniformly on K to f/g

    (3) Assume (f_n) and (g_n) are uniformly convergent sequences of functions.

    (a) Show that (f_n + g_n) is a uniformly convergent sequence of functions.
    (b)Give an example to show that the product (f_ng_n) may not converge uniformly.
    (c) Prove that if there exists an M>0 such that |f_n| \leq M and g_n \leq M for all n \in \mathbb{N}, then (f_ng_n) does converge uniformly.

    I need a walkthrough with these, thanks
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  2. #2
    Super Member PaulRS's Avatar
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    (2)

    Since g(x) is continuous on A it has a min., let m=\min_{x\in A} [|g(x)|]>0 we have 0<\frac{1}{|g(x)|}\leq \frac{1}{m} (1)

    By definition for all \epsilon>0 there is a natural number n_{\epsilon} such that if n\geq{n_{\epsilon}} we have: |f_n(x)-f(x)|<m\cdot{\epsilon} for all x\in A

    Now divide by |g(x)|: \left|\frac{f_n(x)}{g(x)}-\frac{f(x)}{g(x)}\right|<m\cdot{\epsilon}\cdot{\fr  ac{1}{m}}=\epsilon by (1)

    So by definition \frac{f_n}{g} converges uniformly to f/g on A

    (3)

    (a) Suppose f_n converges uniformly to f on a interval A and g_n converges uniformly to g also on A

    By definition for all \epsilon>0 there is a natural number n_{\epsilon1} such that if n\geq{n_{\epsilon1}} we have: |f_n(x)-f(x)|<\frac{\epsilon}{2} for all x\in A

    Again for all \epsilon>0 there is a natural number n_{\epsilon2} such that if n\geq{n_{\epsilon2}} we have: |g_n(x)-g(x)|<\frac{\epsilon}{2} for all x\in A

    Set n_{\epsilon}=\max[n_{\epsilon1},n_{\epsilon2}]

    Then for all \epsilon>0 there is a natural number n_{\epsilon} such that if n\geq{n_{\epsilon}} we have: |f_n(x)-f(x)|<\frac{\epsilon}{2} and |g_n(x)-g(x)|<\frac{\epsilon}{2} for all x\in A

    Summing |f_n(x)+g_n(x)-(f(x)+g(x))|\leq |f_n(x)-f(x)|+|g_n(x)-g(x)|<\epsilonfor all x\in A

    Thus (definition) we have that (f_n+g_n) converges uniformly to (f+g) on A
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  3. #3
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    Opalg's Avatar
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    Quote Originally Posted by shadow_2145 View Post
    (1) Assume that (f_n) converges uniformly to f on A and that each f_n is uniformly continuous on A. Prove that f is uniformly continuous on A
    You probably know the theorem that a uniform limit of continuous functions is continuous. (If not, look it up in a textbook.) You should be able to adapt that proof to show that a uniform limit of uniformly continuous functions is uniformly continuous.

    Quote Originally Posted by shadow_2145 View Post
    (3) Assume (f_n) and (g_n) are uniformly convergent sequences of functions.
    ...
    (b)Give an example to show that the product (f_ng_n) may not converge uniformly.
    Let f_n(x) = g_n(x) = x+(1/n)\;\;(x\in\mathbb{R}).
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by shadow_2145 View Post
    (1) Assume that (f_n) converges uniformly to f on A and that each f_n is uniformly continuous on A. Prove that f is uniformly continuous on A
    Let \varepsilon >0
    f_n unif cont on A implies \exists \delta >0 such that for u,v \in A then |f_n(u) - f_n(v)| < \frac{\varepsilon}{3} whenever |u - v| < \delta

    (f_n) converges uniformly to f on A implies \exists \, N such that for all n \geq N and for all x \in A, |f_n(x) - f(x)| < \frac{\varepsilon}{3}

    thus, if |u - v| < \delta, we have

    |f(u) - f(v)| = |f_n(v) + f(u) - f(v) - f_n(v) + f_n(u) - f_n(u)|

    \leq |f_n(v) - f(v)| + |f(u) - f_n(u)| + |f_n(u) - f_n(v)|

    < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon
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