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Math Help - Is there something wrong with this proof?

  1. #1
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    Is there something wrong with this proof?

    Someone please tell me if this proof is wrong:

    QUESTION: Let f_n:X\to\mathbb{R} be a sequence of functions such that f_n(x)\neq 0 for all n\in\mathbb{N} and for all x\in X. If there exists c\in\mathbb{R} and N\in\mathbb{N} such that |\frac{f_{n+1}(x)}{f_n(x)}|\leq c<1 for all n\geq N and all x\in X then \sum |f_n(x)| and \sum f_n(x) converge uniformly on X.

    PROOF: Let x\in X and n\geq N. Then |\frac{f_{N+1}(x)}{f_N(x)}|\leq c, |\frac{f_{N+2}(x)}{f_{N+1}(x)}|\leq c, ... , |\frac{f_{n}(x)}{f_{n-1}(x)}|\leq c. Multiplying all these inequalities we get |\frac{f_{n}(x)}{f_N(x)}|\leq c^{n-N}\Rightarrow |f_n(x)|\leq c^{n-N}|f_N(x)|. If we put a_n=c^{n-N}|f_N(x)| then the conclusion follows from the Weierstrass M-Test, since \sum a_n is convergent.
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  2. #2
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    Quote Originally Posted by math9 View Post
    Someone please tell me if this proof is wrong:

    QUESTION: Let f_n:X\to\mathbb{R} be a sequence of functions such that f_n(x)\neq 0 for all n\in\mathbb{N} and for all x\in X. If there exists c\in\mathbb{R} and N\in\mathbb{N} such that |\frac{f_{n+1}(x)}{f_n(x)}|\leq c<1 for all n\geq N and all x\in X then \sum |f_n(x)| and \sum f_n(x) converge uniformly on X.

    PROOF: Let x\in X and n\geq N. Then |\frac{f_{N+1}(x)}{f_N(x)}|\leq c, |\frac{f_{N+2}(x)}{f_{N+1}(x)}|\leq c, ... , |\frac{f_{n}(x)}{f_{n-1}(x)}|\leq c. Multiplying all these inequalities we get |\frac{f_{n}(x)}{f_N(x)}|\leq c^{n-N}\Rightarrow |f_n(x)|\leq c^{n-N}|f_N(x)|. If we put a_n=c^{n-N}|f_N(x)| then the conclusion follows from the Weierstrass M-Test, since \sum a_n is convergent.
    In the last line of the proof, you can't apply the M-test in that way, because a_n depends on x. The M-test only applies if (a_n) is a sequence of constants. If the function f_N(x) is bounded, say |f_N(x)|\leqslant K for all x\in X, then you can take a_n = c^{n-N}K. In that case, the M-test does apply and the conclusion is correct. In particular, if the space X is compact, then f_N(x) will automatically be bounded and the result is correct.

    But suppose for example that X=\mathbb{R} and f_n(x) = 2^{-n}x. Then \sum_{n=1}^\infty f_n(x) = x for all x\in\mathbb{R}, but the convergence is not uniform.
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    Quote Originally Posted by Opalg View Post
    In the last line of the proof, you can't apply the M-test in that way, because a_n depends on x. The M-test only applies if (a_n) is a sequence of constants. If the function f_N(x) is bounded, say |f_N(x)|\leqslant K for all x\in X, then you can take a_n = c^{n-N}K. In that case, the M-test does apply and the conclusion is correct. In particular, if the space X is compact, then f_N(x) will automatically be bounded and the result is correct.

    But suppose for example that X=\mathbb{R} and f_n(x) = 2^{-n}x. Then \sum_{n=1}^\infty f_n(x) = x for all x\in\mathbb{R}, but the convergence is not uniform.
    I see.... Do you have any suggestions as to how I can fix my proof? Or maybe another proof?
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  4. #4
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    Quote Originally Posted by math9 View Post
    I see.... Do you have any suggestions as to how I can fix my proof? Or maybe another proof?
    No. Unless you have some extra condition to ensure that f_N(x) is bounded, the result is false (as the example at the end of my previous comment shows).
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    Quote Originally Posted by Opalg View Post
    No. Unless you have some extra condition to ensure that f_N(x) is bounded, the result is false (as the example at the end of my previous comment shows).
    Well.. the example you gave doesn't really show the result is false since your sequence of functions take on the value 0 at x=0 and one of the hypothesis of the problem is that f_n(x)\neq0 for all x \in X.
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  6. #6
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    Quote Originally Posted by math9 View Post
    Well.. the example you gave doesn't really show the result is false since your sequence of functions take on the value 0 at x=0 and one of the hypothesis of the problem is that f_n(x)\neq0 for all x \in X.
    I can easily fix that by restricting the domain to the interval X = [1,\infty). The hypotheses of the question are then all satisfied, but \sum f_n(x) does not converge uniformly on X.
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