So the questions says, let $a_n$ be a sequences of real numbers such that $\limsup |a_n| = 0$. Let $X = [0, 1]$ and for each $n \in \mathbb{N}$ the function $\space$ $f_n :$
$X \mapsto \mathbb{R}$: $\space$ $f_n(x) = (x + a_n)^2$. Does this sequence of functions converge uniformly?
Here is my work:
So since $\limsup |a_n| = 0$ we know that $\lim a_n$ = 0. Now we have:
$|(x + a_n)^2 - x ^2 | = | (a_n)^2 + 2x a_n |$ By triangle inequality we have $| (a_n)^2 + 2x a_n | \leq |(a_n)^2| + |2xa_n|$.
Since $\lim a_n = 0$ we have $ | a_n - 0| < \epsilon^2 $ and $|2x||a_n - 0| < 2x \epsilon$. Since $X$ is bounded above by 1, $|2x||a_n - 0| < 2 \epsilon$. Thus, $| (a_n)^2 + 2x a_n | < 2\epsilon + \epsilon^2$. Which means these sequence of functions converges uniformly. Is my approach right?
You can make your answer a bit more rigorous. Note that the $\lim\sup$ condition is equivalent to saying that $\lim a_n=0$ because $$0\le\lim\inf |a_n|\le\lim\sup|a_n|=0.$$ So for every $\epsilon>0$ there is an $N_\epsilon$ such that for all $n\ge N_\epsilon$ we have $|a_n|<\epsilon$.
Fix $\epsilon>0$. Then for all $n\ge N_\epsilon$ and all $x\in[0,1]$ we have
\begin{align} |f_n(x)-x^2|&=|a_n^2+2xa_n|\\ &\le |a_n|^2+2x|a_n|\\ &\le \epsilon^2+2\epsilon \end{align}
as in your argument.
Consequently, for every $\delta>0$, there is an $M_\delta$ such that for all $n\ge M_\delta$ and all $x\in[0,1]$
\begin{align} |f_n(x)-x^2|\le \delta. \end{align}
Indeed, we can take always find an $\epsilon_\delta>0$ such that $\epsilon_\delta^2+2\epsilon_\delta\le\delta$; for example, take $\epsilon_\delta=\delta/3$ in case $\delta\le 1$, and $\epsilon_\delta=1/3$ if $\delta>1$. Then $M_\delta:=N_{\epsilon_\delta}$ will do the trick.