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  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Sequence of function.

    Let (f_n)^\infty_{n=1} be a sequence of functions which are defined on [a,b], and uniformly converges there to bounded function f. Let \Phi\in C(\mathbb{R}) which is continuous function on all \mathbb{R}.
    Prove that the sequence (\Phi(f_n))^\infty_{n=1} uniformly converges and find the limit function.


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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let (f_n)^\infty_{n=1} be a sequence of functions which are defined on [a,b], and uniformly converges there to bounded function f. Let \Phi\in C(\mathbb{R}) which is continuous function on all \mathbb{R}.
    Prove that the sequence (\Phi(f_n))^\infty_{n=1} uniformly converges and find the limit function.


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    What have you tried? Are you going to just use the definition? If so, maybe seeing what \lim\text{ }\Phi(f_n) is equal to would help. First, just as a matter of formality we really are considering \Psi=\Phi\mid_X where X=\bigcup_{n\in\mathbb{N}}f_n\left([a,b]\right) and the restriction of a continuous function is continuous. So, let \varphi(x)=\lim\text{ }\Psi\left(f_n(x)\right). Then, for a fixed x_0\in [a,b] we have that \varphi(x_0)=\lim\text{ }\Psi\left(f_n(x_0)\right)\overset{{\color{red}*}}  {=}\Psi\left(\lim\text{ }f_n(x_0)\right)=\Psi\left(f(x_0)\right) so that we see \varphi(x)=\lim\text{ }\Psi\left(f_n(x)\right)=\Psi\left(f(x)\right). Now, tell me, what did I use at \color{red}*?
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