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Let $f_n: \Omega \rightarrow \mathbb R$ be a sequence of real valued functions that converges uniformly to a function $f$.

Suppose there exists a sequence $K_n \in \mathbb R^+$ such that $\sup\limits_{x \in \Omega} \vert f_n(x) \vert < K_n$ for each $n\in \mathbb N$.

Can $f$ be unbounded ? If not, does that mean that $f_n$ is uniformly bounded (I mean that there exists a $K$ such that $\sup\limits_n K_n < K$)?

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No. By uniform convergence, there is an $N$ such that $|f(x)-f_n(x)|<1$ for all $x\in \Omega$ and $n\geq N$. By the triangle inequality, we have $$ |f(x)|<K_N+1 $$ This also means that the $f_n$ are uniformly bounded, by the maximum of $$ K_1,K_2,\ldots,K_{N-1}, K_N+2 $$

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No. There is a $p\in\mathbb N$ such that$$(\forall n\in\mathbb{N})(\forall x\in\Omega):n\geqslant p\implies\bigl|f(x)-f_n(x)\bigr|<1.$$But then $f_p$ cannot be bounded, because$$(\forall x\in\Omega):f_p(x)=f_p(x)-f(x)+f(x)\geqslant f(x)-1$$and $f$ is unbounded.

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