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$f_n[0,2 \pi] \to \mathbb{R}$ ,$f_n\overset{u}{\rightarrow} \sin $, I need to prove that $\sup _{x \in [0, 2 \pi]}f_n(x)$ and $\inf _{x \in [0, 2 \pi]}f_n(x)$ converges and find their limits.

I know that due to uniformly convergence, for $\epsilon=1$ there's $N$ that for every $n>N$ to all $x \in [0, 2\pi]$, $f_n(x)-f(x)|<1$, so $|f_n(x)|<2$ and bounded so $\sup_{[0,2\pi]}$ is finite. same case to the infimum. How should I go on in order to find their limits? ( 1 and 0?)

Thanks!

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1 Answer 1

up vote 1 down vote accepted

Let $\varepsilon > 0$. Then for $n \ge N$ large enough and every $x \in [0,2\pi]$, $\sin x - \varepsilon \le f_n(x) \le \sin x + \varepsilon$ (by uniform convergence).

In particular, $\sup_x f_n(x) \le 1+\varepsilon$. Choosing $x = \pi/2$ then shows that $\sup_x f_n(x) \ge 1-\varepsilon$. Make a similar argument for the infimum (which is not $0$).

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