Let $(f_{n})_{n}$ be a sequence of continuous functions on $[a,b] \to \mathbb C$ that converges uniformly to $f:[a,b] \to \mathbb C$ that is also continuous. Required to prove that $(f_{n}^2)_{n}$ converges uniformly to $f^2$. Steps taken: Since $f_{n}$ is continuous $\forall n \in \mathbb N$, then $f_{n}^2$ is also continuous $\forall n \in \mathbb N$. The same goes for $f$ and $f^2$ so $f^2$ is also continuous. The continuity of $f_{n}^2$ and $f^2$ means that all I need to prove is that $f_{n}^2$ converges to $f^2$ pointwise (I'm not sure if my assumption is correct here). I also assume that $f_{n}^2: [a,b] \to [a,b]$ as well as $f^2:[a,b] \to [a,b]$.
Let $\epsilon >0 \exists N \in \mathbb N \forall n >= N: \forall x \in [a,b]: |f_{n}(x)-f(x)|< \epsilon $ but that includes $f_{n}^2$ as $f_{n}^2(x)=f_{n}(f_{n}(x))$ where $f_{n}(x) \in [a,b]$.
Am I on the right track? Any help would be greatly appreciated.
Note that $f_n\to f$ uniformly means $$ \lim_n \sup_x |f_n(x)-f(x)| = \lim_n \|f_n-f\|_\infty = 0. $$ We can use the classical decomposition $$ a^2-b^2 = (a+b)(a-b) $$ to compute $$ \|f_n^2-f^2\|_\infty \leq \|f_n-f\|_\infty \|f_n+f\|_\infty \leq \|f_n-f\|_\infty (\|f_n\|_\infty+\|f\|_\infty) $$ Now pass to the $\limsup$ on both sides $$ 0\leq \limsup_n \|f_n^2-f^2\|\leq \limsup_n \|f_n-f\|_\infty (\|f_n\|_\infty+\|f\|_\infty) \leq 0\times 2\|f\| = 0, $$ i.e. $f_n^2 \to f^2$ uniformly.