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Question: Let the sequence of functions $f_n:X\subset\mathbb{R}\to\mathbb{R}$ be uniformly convergent and $f_n$ is bounded, that is, exists $K_n>0$ such that $|f_n(x)|\leq K_n$ for all $x\in X$. Prove that exists a unique $K\geq 0$ such that $|f_n(x)|\leq K$ for all $x\in X$ and $n\in\mathbb{N}$.

My attempt: I tried to take the $\sup\limits_n K_n$, but I think that's not work cause exists infinites $K_n$, so I can't guarantee that supremum is a constant and not the $\infty$.

Ps.: Sorry for my bad English, please someone correct me.

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  • $\begingroup$ This is definitely false as stated, for if the inequality is true for $K$ then it's also true for $K+1$, say. $\endgroup$ Apr 27, 2021 at 1:49

1 Answer 1

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As posted in the comments, you can show that a bound $K$ exists, but not that it is unique.

Hint Look at the limit $f(x)=\lim_n f_n(x)$. Use the uniform convergence and boundedness of $f_n$ to show that $f$ is bounded.

Hint 2 Deduce from here that there exists some $N$ such that, for all $n >N$ and all $x$ you have $$ |f_n(x)| \leq |f(x)|+1 \leq A+1 $$ where $A$ is a bound for $f$.

Can you finish from here?

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  • $\begingroup$ Yes, thanks a lot, i used your hint and finished the question. You rock man! $\endgroup$
    – Law
    Apr 27, 2021 at 2:11

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