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Thread: Show that fn + gn uniformly converges to f + g

  1. #1
    Member thaopanda's Avatar
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    Show that fn + gn uniformly converges to f + g

    Let $\displaystyle f_{n}, g_{n}$ : R $\displaystyle \rightarrow$ R for n $\displaystyle \in$ N be functions such that $\displaystyle f_{n} \rightarrow f$ and $\displaystyle g_{n} \rightarrow $ g as n $\displaystyle \rightarrow \infty$ uniformly on the set E $\displaystyle \subset$ R, where f, g : R $\displaystyle \rightarrow$ R are functions. Show that $\displaystyle f_{n} + g_{n} \rightarrow f + g $ as n $\displaystyle \rightarrow \infty$ uniformly on the set E.
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  2. #2
    Member thaopanda's Avatar
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    Monotonically increasing

    For each n $\displaystyle \in$ N let $\displaystyle f_{n}$ : [a,b] $\displaystyle \rightarrow$ R where a,b $\displaystyle \in$ R are such that a < b. Show that if F : [a,b] $\displaystyle \rightarrow$ R is such that $\displaystyle F_{n} \rightarrow $ F as n $\displaystyle \rightarrow \infty$ in a pointwise fashion for each n $\displaystyle \in$ N the function $\displaystyle F_{n}$ is monotonically increasing, then F is monotonically increasing.
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  3. #3
    Super Member Deadstar's Avatar
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    limit of $\displaystyle f_n = f$
    limit of $\displaystyle g_n = g$

    Fix $\displaystyle \epsilon > 0$, There exist positive integers $\displaystyle n_1$, $\displaystyle n_2$ such that for all $\displaystyle n \geq n_1$, $\displaystyle |f_n - f| < \frac{\epsilon}{2}$ and for all $\displaystyle n \geq n_2$, $\displaystyle |g_n - g| < \frac{\epsilon}{2}$. Define $\displaystyle n_0 = \textrm{max}(n_1, n_2)$. Then for all $\displaystyle n \geq n_0$,

    $\displaystyle |(f_n + g_n) - (f + g)| \leq |f_n - f| + |g_n - g| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ which proves it.
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