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"Prove that if each $f_n$ is a bounded function and $\sum_0^\infty f_n$ converges uniformly on $D$ to $f$, then $f$ is a bounded function"

I don't know how to do this at all. Any help appreciated.

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    $\begingroup$ boundedness follows easily if a sequence of bounded functions $(g_n)$ converges uniformly to $f$ because 'converges uniformly' means that for any $\epsilon > 0$ there exists $N$ such that $n > N \implies$ ... $\endgroup$ – reuns Mar 30 '16 at 9:54
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    $\begingroup$ Use the triangle inequality $|f| \leq |f-f_n| + |f_n|$ in conjunction with the hint above. $\endgroup$ – Winther Mar 30 '16 at 10:03
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By Cauchy's Convergence Principle of uniformly convergence, $\forall\varepsilon>0\exists N>0$ s.t.$\forall n,m>N$,$|f_n-f_m|<\varepsilon,\forall x\in D$. In particular, $|f_n-f_{N+1}|<\varepsilon,\forall x\in D$. Thus $|f_n|<sup|f_{N+1}|+\varepsilon$ i.e. $f_n$ is uniformly bounded. Then you can use the definition of uniformly convergence on $f$.

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Let $\epsilon \gt 0$ be given. $f_n$ converges uniformly to $f$ so $\exists N \in \mathbb{N}$ s.t.

$|f_n - f| \lt \epsilon$ whenever $n \geq N$.

For a fixed $n = N$,

$|f_N - f| \lt \epsilon \implies$ $-\epsilon + f_N \lt f \lt \epsilon + f_N$ --- $(1)$

But $f_N$ is bounded so $|f_N| \leq M_N$ (for some $M_N \gt 0$) $\implies -M_N \leq f_N \leq M_N$

So $-\epsilon + f_N \geq -\epsilon - M_N$ and $\epsilon + f_N \leq \epsilon + M_N$

So $(1)$ becomes $-\epsilon - M_N \lt f \lt \epsilon + M_N$ $\implies |f| \lt \epsilon + M_N$ so $f$ is bounded.

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