Suppose that functions $f_{k}$ defined on $\mathbb{R}^n$ converge uniformly to a function $f$. Suppose that each $f_{n}$ is bounded, say by $A_{k}$. Prove that $f$ is bounded.
Attempt
Want to Show: $\exists$ $A > 0$ such that $|f| \leq A$.
What we know
Given that the $f_{k}$ converge uniformly to $f$, means that for all $\epsilon > 0$ there exists $N > 0$ such that $\|f_{n} - f \| < \epsilon$ for all $n \geq N$ and $\forall x \in \mathbb{R}^n$.
We also know that $\exists A_{k} > 0$ such that $|f_{k}| \leq A_{k}$.
Proof:
Let $\epsilon > 0$. By uniform convergence:
$$\epsilon > \|f_{n} - f\| = \sup_{x \in \mathbb{R}^n} |f_{n} - f| \geq \Bigg| |f_{n}| - |f| \Bigg| = \Bigg| |f| - |f_{n}| \Bigg| \\ \Rightarrow |f| < |f_{k}| + \epsilon \leq A_{k} + \epsilon$$.
And I can designate $A_{k} + \epsilon = A$ as my bound because we just want a bound and the bound can change depending on which sequence term we choose from uniform convergence.
Thoughts: This feels like what needed to be done, but I am concerned on whether I'm being asked to find one uniform bound for $f$ or if the bound that I found is sufficient. Feedback ?
There is no need to deal with an arbitrary $\varepsilon$. Taking $\varepsilon=1$ will do. Then, there is a $N\in\mathbb N$ such that$$k\geqslant N\implies\sup_{x\in\mathbb R^n}\bigl\lvert f(x)-f_k(x)\bigr\rvert<1$$and, in particular,$$\sup_{x\in\mathbb R^n}\bigl\lvert f(x)-f_N(x)\bigr\rvert<1.$$But then, as you explained, you can take $A=A_N+1$.