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Let $f_n$ converges uniformly to $f$ on $S$. Show: $\lim_{n\to \infty}||f_n||_S=||f||_S$. $||f||_S=sup_S|S|$

My attempt: since $\{f_n\}$ converges uniformly to $f$, we have: $$\forall \epsilon>0, \exists N\in\mathbb{N}\text{ such that } |f-f_n|<\epsilon \forall n>N$$

Now, \begin{align} |||f_n||-||f|||&=|\sup_S|f_n|-\sup_S|f||\\ &\leq |\sup_S|f_n|-\inf_S|f||\\ &=|\sup_S|f_n|+\sup_S-|f||\\ &=|\sup_S|f_n|-|f||\\ &\leq|\sup_S|f_n-f||\\ &\leq|\sup_S\epsilon|=\epsilon, n>N\\ &\implies\lim_{n\to \infty}||f_n||_S=||f||_S \end{align}

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  • $\begingroup$ $$\sup_S \lvert f_n\rvert-\sup_S \lvert f\rvert\neq \sup_S (\lvert f_n\rvert-\lvert f\rvert) $$ $\endgroup$ – Michael Lee Dec 2 '17 at 20:02
  • $\begingroup$ @Michael Lee its wrong even for sets. I corrected my proof. $\endgroup$ – MAS Dec 2 '17 at 20:21
  • $\begingroup$ What is $\|\cdot\|_S$? $\endgroup$ – José Carlos Santos Dec 2 '17 at 20:26
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    $\begingroup$ Why would you assume $\lvert \sup_S \lvert f_n\rvert-\sup_S \lvert f\rvert\rvert\leq \lvert \sup_S \lvert f_n\rvert-\inf_S \lvert f\rvert\rvert $? Consider $f_n\equiv 0$ for some $n $. Then, your inequality becomes $\sup_S \lvert f\rvert\leq \inf_S \lvert f\rvert$, which is absurd unless $f $ is constant on $S $. $\endgroup$ – Michael Lee Dec 2 '17 at 20:30
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    $\begingroup$ Yes, but $a\leq b $ doesn't imply $\lvert a\rvert\leq \lvert b\rvert $. $\endgroup$ – Michael Lee Dec 2 '17 at 20:38
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Take $\varepsilon>0$. Since $(f_n)_{n\in\mathbb{N}}$ converges uniformly to $f$, there is a natural $p$ such that$$(\forall n\in\mathbb{N})(\forall x\in S):n\geqslant p\implies|f_n(x)-f(x)|<\varepsilon.$$Therefore$$(\forall n\in\mathbb{N}):n\geqslant p\implies\|f_n-f\|_S\leqslant\varepsilon.$$But $\bigl|\|f_n\|_S-\|f\|_S\bigr|\leqslant\|f_n-f\|_S$, since$$\|f_n\|_S=\|f_n-f+f\|_S\leqslant\|f_n-f\|_S+\|f\|_S\implies\|f_n\|_S-\|f\|_S\leqslant\|f_n-f\|_S$$and$$\|f\|_S=\|f-f_n+f_n\|_S\leqslant\|f-f_n\|_S+\|f_n\|_S\implies\|f\|_S-\|f_n\|_S\leqslant\|f_n-f\|_S.$$Therefore$$(\forall n\in\mathbb{N}):n\geqslant p\implies\bigl|\|f_n\|_S-\|f\|_S\bigr|\leqslant\varepsilon.$$What this means is that$$\lim_{n\to\infty}\|f_n\|_S=\|f\|_S.$$

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    $\begingroup$ Since $\lvert \| f_n\|_S-\|f\|_S\rvert \leq \|f_n-f\|_S $ was where the asker appeared to be stuck, it might be helpful to mention that this is a direct result of the triangle inequality and is often called the "reverse triangle inequality." $\endgroup$ – Michael Lee Dec 2 '17 at 20:44
  • $\begingroup$ @MichaelLee Excellent suggestion! I've edited my answer. $\endgroup$ – José Carlos Santos Dec 2 '17 at 20:52
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Follows of triangular inequalities, that $| |f(x)| - |f_n(x)|| \leq |f(x) - f_n(x)|$, $\forall x$, we have that

$$|f(x)| - |f_n(x)| \leq |f(x) - f_n(x)|$$

and therefore

$$ - |f_n(x)| \leq |f(x) - f_n(x)| + |f(x)|$$

and so

$$ - |f_n(x)| \leq \|f - f_n\|_S + \|f\|_S.$$ Then $$ \|f\|_S \leq \|f - f_n\|_S + \|f_n\|_S.$$

Therefore

$$ \|f\|_S - \|f_n\|_S\leq \|f - f_n\|_S .$$ Analogously, we can show that

$$ \|f_n\|_S - \|f\|_S\leq \|f - f_n\|_S $$ and therefore $$ 0\leq |\|f_n\|_S - \|f\|_S|\leq \|f - f_n\|_S $$

Doing $n \to \infty$, we heve proofed.

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