# Prove that exists a unique K>0 such that $|f_n(x)|\leq K$ for all $x\in X$ and $n\in\mathbb{N}$.

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 $$\infty$$.

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

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 $$f$$.