Pointwise bounded sequence is "uniformly bounded"?

Question

Consider the real valued functions of $x$, $F:\mathbb{R} \rightarrow \mathbb{R}$ and $F_n:\mathbb{R} \rightarrow \mathbb{R}$ and suppose I have shown that $$ |F_n(x)-F(x)|\leq A \text{ } \text{ }\forall x $$ Dos this mean that $\sup_{x} |F_n(x)-F(x)| \leq A$?

Answer 1

Yes, of course. If $l := \sup_{x}|F_{n}(x) - F(x)| > A$, then by definition (if you are using this definition; otherwise we can prove this) there is some $x$ such that $$ l \geq |F_{n}(x) - F(x)| > l - \frac{l-A}{2} > A, $$ a contradiction.

Answer 2

Yes. You have shown that $A$ is an upper bound for the set $\{|F_n(x)-F(x)|~:~ x \in \mathbb{R}\}$. Since $\sup_{x \in \mathbb{R}} |F_n(x)-F(x)|$ is the least upper bound of the set $\{F_n(x)-F(x)|~:~x\in\mathbb{R}\}$, it is no larger than $A$.

Answer 3

yes, since $A$ is an upper bound of $|F_n(x)−F(x)|$ its supremum, which is defined as the least upper bound will be

$\sup_{x \in \mathbb{R}} |F_n(x)−F(x)| \le A$