$\{f_n\}_{n=1}^{\infty}$ (each bounded) converges uniformly to $f$ then $\{f_n\}_{n=1}^{\infty}$ is uniformly bounded

Question

Let $\{f_n\}_{n =1}^{\infty}$ be a sequence of bounded functions. Suppose $\{f_n\}_{n=1}^{\infty}$ converges uniformly to $f$. Prove $\exists M \gt 0$ s.t. $|f_n| \leq M \ \ \forall n \in \mathbb{N}$.

Proof:

Each $f_n$ is bounded so let $\{M_n\}_{n=1}^{\infty}$ be the sequence of upper bounds of each $f_n$. Then the sequence is bounded for if that were not the case, there would be an unbounded function in $\{f_n\}_{n =1}^{\infty}$ which would contradict the hypothesis.

Since $\{M_n\}_{n=1}^{\infty}$ is bounded it has a supremum. Let $S = \sup \{M_n\}_{n=1}^{\infty}$.

So $\forall n \in \mathbb{N}$, $|f_n| \leq S$ clearly. Therefore, $f_n$ is uniformly bounded.

Are there any mistakes in the proof above? Any room for improvements?

Edit: my question is specifically about uniform boundedness of $f_n$ and not abound $f$. Hence, I don't think it's a duplicate of the linked question.

Answer 1

It is not give that $(M_n)$ is bounded. The hypothesis is that for each $n$, $|f_n|$ has a bound but the sequence of upper bounds is not given to be bounded.

There exists a function $f$ and $n_0$ such that $|f_n(x)-f(x)| <1$ for ll $n > n_0$ for all $x$. This gives $|f(x)| \leq 1+M_{n_0}$, so $f$ is a bounded function. Hence$|f_n(x)|\leq |f(x)|+|f_n(x)-f(x)| <1+M_{n_0}+1$ for $n>n_0$ . Now let $M$ be the maximum of $2+M_{n_0}$ and the numbers $M_1,M_2,...,M_{n_0}$. The $|f_n(x)| \leq M$ for all $n$ and all $x$.

Answer 2

First mistake here:

Each fn is bounded so let {Mn}∞n=1 be the sequence of upper bounds of each fn. Then the sequence is bounded for if that were not the case, there would be an unbounded function in {fn}∞n=1 which would contradict the hypothesis.

Example $f_n(x) = n \cdot sin(x)$: each function is bounded, but sequence of M is not.