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Thread: Uniform Convergence

  1. #1
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    Uniform Convergence

    I would appreciate any help that I can get with the following proof:

    Let $\displaystyle f_n,f:[0,1]\rightarrow \mathbb{R}$ and $\displaystyle x,x_n\in [0,1]$ such that each $\displaystyle f_n$ is continuous. Also, $\displaystyle f_n\rightrightarrows f$ and $\displaystyle x_n\rightarrow x$.
    Prove that $\displaystyle f_n(x_n)\rightarrow f(x)$.

    So i think my goal is to find an $\displaystyle N$ such that $\displaystyle n\geq N\Rightarrow |f_n(x_n)-f(x)|<\epsilon$.
    Also, we know that there exists an $\displaystyle N$ such that $\displaystyle n\geq N\Rightarrow |f_n-f|<\epsilon$ and also, $\displaystyle |x_n-x|<\epsilon$.
    Assuming I'm correct so far, I need some help putting it all together...
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  2. #2
    Super Member girdav's Avatar
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    $\displaystyle |f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$.
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  3. #3
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    So could we say that since $\displaystyle f_n\rightrightarrows f$ then $\displaystyle |f_n(x)-f(x)|<\epsilon$ for any $\displaystyle x\in [0,1]$. Thus, since $\displaystyle x_n\in [0,1]$ we have that $\displaystyle |f_n(x_n)-f(x_n)|<\epsilon$ must be true. Then, since we know $\displaystyle f$ is continuous and $\displaystyle x_n\rightarrow x$ we have that $\displaystyle |f(x_n)-f(x)|<\epsilon$ must also be true.
    Then, we have that $\displaystyle |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<2\epsilon$. Then, by the triangle inequality, $\displaystyle |f_n(x_n)-f(x)|\leq|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|<2\epsilon$. Thus, we have that $\displaystyle f_n(x_n)\rightarrow f(x)$.
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