Let $(f_n)$ be a function sequence on $D \subset \mathbb R$ with $f_n \to f^*$ uniform on $D$. Assume $f_n$ bounded for all $n$.
I want to prove the following:
$f^*$ is bounded and $$ \| f_n \|_\infty \to \| f^* \|_\infty, \quad \sup_{x \in D} f_n(x) \xrightarrow{n\to\infty} \sup_{x \in D} f^*(x)\,. $$
What I have done thus far;
We have that $f_n(x)$ is bounded for each $n \in \mathbb N$. Therefore, for all $n \in \mathbb N$ we can find a $M \in \mathbb R$ (possibly depending on $n$) such that for all $x \in D$ we have $$ |f_n(x)| < M_n\,. $$ As $f_n$ converges uniformly to $f^*$ on $D$, we have $$ \forall_{\epsilon > 0}\exists_{n^* \in \mathbb N}\forall_{n\geq n^*}\forall_{x\in D}: |f_n(x) - f^*(x)| < \epsilon\,. $$
We shall first show that $f^*$ is bounded, therefore we need to prove that there exists an $M^* \in \mathbb R$ such that for all $x \in D$ we have $$ |f^*(x)| < M^*\,. $$
To this end, take $\epsilon = 1$, then for $n \geq n^*$ we have that for all $x \in D$ \begin{align*} |f_n(x) - f^*(x)| < 1 \iff |f^*(x) - f_n(x)| &< 1\\ \therefore -1 < f^*(x) - f_n(x) &< 1 \\ \therefore -1 + f_n(x) < f^*(x) &< 1 + f_n(x) \qquad \mbox{but $|f_n(x)| < M_n$, hence} \\ \therefore -1 - M_n < f^*(x) &< 1 + M_n \iff |f^*(x)| < 1 + M_n\,. \end{align*}
Now we have that $f^*$ is bounded by a quantity depending on $n \geq n^*$, which I assume is not enough. At this point I did some research and found this link, which basically says this is sufficient. However, when looking at this link, I would suspect my proof is not enough and we want a value in $\mathbb R$ that bounds $f^*$ independent of any $x$ or $n$.
N.B. I haven't started the second part of the proof yet.
