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Let ${ f_n\left(x\right):[0,1] \to \mathbb{R} }$ uniformly convergent to function $f(x)$ which is bounded on the interval $[0,1]$. Prove the following: ${ \lim_{n\to\infty} \sup_{[0,1]} f_n\left(x\right) = \sup_{[0,1]} f\left(x\right) }$.

I tried to show that $f$ is bounded and therefore it has a supremum $S$ for some $x_0 \in [0,1]$. In addition, for every $\epsilon > 0$ there exist $n_0$ such that for every $n > n_0$ and for every $x \in [0,1]$, $|f_n(x) - f(x)| < \epsilon$. I tried to put the $x_0$ instead of $x$, and replace the $f(x_0)$ with $S$, but it doesn't helped much...

How can I use the boundedness wisely?

ViktorStein
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Nave Tseva
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    does this help? https://math.stackexchange.com/questions/226392/interchanging-limit-with-infimum-supremum – Locally unskillful May 12 '19 at 20:04
  • Not exactly, because in the link the functions are continues (and he use it in his answer). Here the functions are not necessarily continues. – Nave Tseva May 13 '19 at 06:22

1 Answers1

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i) For $x_n\in [0,1]$, assume that $\sup\ f = \lim\ f(x_n)$

Assume that $|\sup\ f-f(x_n)|<\varepsilon$ for some $n$. By pointwise convergence, $i>i_0$ implies $|f_i(x_n)-f(x_n)|<\varepsilon$ so that $$\sup\ f_i\geq f_i(x_n)\geq \sup\ f -2\varepsilon$$ for all $i>i_0$

Hence $\lim\ \sup\ f_i\geq \sup\ f$

ii) $|\lim\ \sup\ f_i - \sup\ f_i|<\varepsilon$ for all $i>i_0$.

Further $| f_i(x_i)- \lim\ \sup\ f_i |<2\varepsilon$ for all $i>i_0$

By uniform convergence, when $i$ is large, then $|f(x_i)-f_i(x_i)| < \varepsilon$ so that $|f(x_i)-\lim\ \sup\ f_i |<3 \varepsilon$ for large $i$.

HK Lee
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  • Thanks! but why does $ \sup\ f = \lim\ f(x_n) $? The function is not necessarily continues. – Nave Tseva May 13 '19 at 06:59
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    $\sup\ f=\sup\ {f(x)|x\in [0,1]}$ so that we have a sequence $x_n\in [0,1]$ – HK Lee May 13 '19 at 07:01
  • Thanks, but I still don't some assumptions in the proof: Why does the $sup f_i$ even exist? we don't know anything about $f_i$ – Nave Tseva May 14 '19 at 18:38
  • Since $f$ is bounded, so we can let $|f|_\infty =\sup\ { |f(x)| |x\in [ 0,1] }\leq 1$

    Since $f_i$ goes to $f$ uniformly, then there is $N$ s.t. $i>N$ implies $|f_i|_\infty\leq 2$.

     – HK Lee May 14 '19 at 22:58