The problem is, if $f_n$ is a sequence of continuous (on real) functions that converge pointwise to a function, then we are asked to prove that there exists a constant $M$ and an interval such that $\sup|f_n| Now we are studying Baire Category theorem, so this problem must be related to that. I tried to prove that the set contains all $x \in \mathbb{R}$ such that $\sup|f_n|=\infty$ is meagre, but even so I cannot prove the existence of a interval. Anyone can give a hint on how to approach? Thanks!
There exists an interval on which the supreme of a point-wise covergent fuctions is bounded?
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real-analysis
functional-analysis
baire-category
1 Answers
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The set $A_m = \{ x \in X \mid \lvert f_n(x) \rvert \leq m \text{ for all }n \in \mathbb{N}\} = \bigcap_{n \in \mathbb{N}} \{ x \in X \mid \lvert f_n(x) \rvert \leq m\}$ is closed. Since each sequence $((f_n(x))_{n \in \mathbb{N}}$ is bounded, we have $\mathbb{R} = \bigcup_{m \in \mathbb{N}} A_m$.
It follows that there is some $M_0$ such that $A_{M_0}$ has non-empty interior…