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Thread: Sequence of function.

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

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


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