Am I proving correctly that the uniform convergence of the product is the product of the limits when each sequence is uniformly bounded?

Question

Let $(X,d)$ be a metric space, and for every positive integer $n$, let $f_{n}:X\to\textbf{R}$ and $g_{n}:X\to\textbf{R}$ be functions. Suppose that $(f_{n})_{n=1}^{\infty}$ converges uniformly to another function $f:X\to\textbf{R}$, and that $(g_{n})_{n=1}^{\infty}$ converges uniformly to another function $g:X\to\textbf{R}$. Suppose also that the functions $(f_{n})_{n=1}^{\infty}$ and $(g_{n})_{n=1}^{\infty}$ are uniformly bounded, i.e., there exists an $M > 0$ such that $|f_{n}(x)|\leq M$ and $|g_{n}(x)|\leq M$ for all $n\geq 1$ and $x\in X$. Prove that the functions$f_{n}g_{n}:X\to\textbf{R}$ converge uniformly to $fg:X\to\textbf{R}$.

My solution

Let us start with the definition of uniform convergence: for every $\varepsilon > 0$, there are a natural numbers $N_{1}\geq 1$ and $N_{2}\geq 1$ such that for every $x\in X$ we have that \begin{align*} \begin{cases} n\geq N_{1} \Rightarrow |f_{n}(x) - f(x))| < \varepsilon\\\\ n\geq N_{2} \Rightarrow |g_{n}(x) - g(x)| < \varepsilon \end{cases} \end{align*}

Consequently, due to the triangle inequality, for every $\varepsilon >0$, there is a natural number $N\geq\max\{N_{1},N_{2}\}$ such that for every $x\in X$ one has that \begin{align*} |f_{n}(x)g_{n}(x) - f(x)g(x)| & \leq |f_{n}(x)g_{n}(x) - f_{n}(x)g(x)| + |f_{n}(x)g(x) - f(x)g(x)|\\\\ & \leq |f_{n}(x)||g_{n}(x) - g(x)| + |g(x)||f_{n}(x) - f(x)|\\\\ & \leq M\varepsilon + G\varepsilon = (M + G)\varepsilon \end{align*} where $|g(x)| \leq G$.

Thence we conclude that $f_{n}g_{n}$ converges uniformly to $fg$, and we are done.

The result which allows us to tell that $g$ is bounded is the following:

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces and $(f_{n})_{n=1}^{\infty}$ be a sequence of functions from $X$ to $Y$ which converges uniformly to $f:X\to Y$. If each term $f_{n}$ is bounded, then the function $f$ is also bounded.

Could someone please tell me if I am missing any step?

Answer 1

Yes, your proof is right. As a side remark, your hypothesis contains a uniform boundedness assumption. This is actually unnecessary, because one can prove the following:

Let $X$ be a (non-empty) set, and $(f_n:X \to \Bbb{R})_{n=1}^{\infty}$ a sequence of functions, which are bounded, and which converge uniformly to a function $f:X \to \Bbb{R}$. Then, $(f_n), f$ are uniformly bounded. i.e there is an $M>0$ such that for all $x \in X$ and all $F \in \{f_n\}_{n=1}^{\infty} \cup\{f\}$, we have $|F(x)| \leq M$.