Can you, please, check if the following proof is correct? Thanks for your time and effort.

Suppose that $f_n\to f$ and $g_n\to g$, as $n\to \infty,$ uniformly on $E\subseteq \Bbb{R}.$ Then, $f_n g_n\to fg,$ as $n\to \infty,$ pointwise on $E$.

Let $\epsilon>0$ be given and $x\in E$ be fixed. Then,

\begin{align} \left|(f_n g_n)(x)-(f g)(x)\right| &= \left|f_n(x) g_n(x)-f(x) g(x)\right| \\&= \left|f_n(x) g_n(x)-f_n(x) g(x)+f_n(x) g(x)-f(x) g(x)\right| \\&\leq \left|f_n(x) g_n(x)-f_n(x) g(x)\right| + \left|f_n(x) g(x)-f(x) g(x)\right| \\&= \left|f_n(x)\right| \left|g_n(x)- g(x)\right| + \left|g(x)\right| \left|f_n(x)-f(x) \right| \end{align} The sequences $\{f_n(x) \}_{n\in\Bbb{N}}$ and $\{g_n(x) \}_{n\in\Bbb{N}}$ are real sequences that converge to $f(x)$ and $g(x)$, respectively. So, they are bounded, i.e., for fixed $x\in E,$ there exists $M>0$ and $K>0,$ such that \begin{align}\left|f_n(x)\right| \leq M\;\text{and}\;\left|g_n(x)\right| \leq K,\;\forall \;n\in\Bbb{N}.\end{align} Therefore, \begin{align}\lim\limits_{n\in\Bbb{N}}\left|g_n(x)\right| =\left|\lim\limits_{n\in\Bbb{N}}g_n(x)\right|=\left|g(x)\right|\leq K.\end{align} So, we have \begin{align} \left|(f_n g_n)(x)-(f g)(x)\right| &\leq \left|f_n(x)\right| \left|g_n(x)- g(x)\right| + \left|g(x)\right| \left|f_n(x)-f(x) \right|\\&\leq M \left|g_n(x)- g(x)\right| + K \left|f_n(x)-f(x) \right|\end{align} Since $f_n(x)\to f(x)$ and $g_n(x)\to g(x)$, as $n\to \infty,$ for fixed $x\in E$, then there exists $N_1(\epsilon),N_2(\epsilon)$ such that \begin{align} \left|f_n(x)-f(x) \right|<\dfrac{\epsilon}{2K} ,\;\;\forall\;n\geq N_1(\epsilon)\end{align} and \begin{align} \left|g_n(x)-g(x) \right|<\dfrac{\epsilon}{2M } ,\;\;\forall\;n\geq N_2(\epsilon)\end{align} Let $N(\epsilon)=\max\{N_1(\epsilon),N_2(\epsilon)\}.$ Then, \begin{align} \left|(f_n g_n)(x)-(f g)(x)\right| &\leq M \left|g_n(x)- g(x)\right| + K \left|f_n(x)-f(x) \right|\\&<\dfrac{\epsilon}{2 } +\dfrac{\epsilon}{2 }=\epsilon,\;\;\forall\;n\geq N(\epsilon)\end{align}

  • $\begingroup$ Your notation can be eased tremendously if you give $\sup|f_n(x)|$ a name, and avoid the big fractions by worrying about shrinking $\epsilon$ towards the end. $\endgroup$ – John Jiang Dec 29 '18 at 20:19
  • $\begingroup$ @ John Jiang: Thanks for that advice. Will fix it! $\endgroup$ – Omojola Micheal Dec 29 '18 at 20:20
  • $\begingroup$ Since $f_n$ and $g_n$ converge uniformly, your $|\cdot|$ should be the sup-norm w.r.t. $x$. Thus your $N$ should only depend on $\epsilon$ not $x$. $\endgroup$ – induction601 Dec 29 '18 at 20:22
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    $\begingroup$ I think that you don't need the uniform convergence of $f_n$ and $g_n$ for $(f_n\times g_n)(x)$ to converge to $(f \times g)(x)$ pointwise. Moreover, isn't the important difference between pointwise and uniform convergence the fact that you can choose $N$ such that it depends only on epsilon and not on $x$? In other words, you haven't used uniform convergence in your argument. $\endgroup$ – stressed out Dec 29 '18 at 20:22
  • $\begingroup$ @stressed out: Thanks! $\endgroup$ – Omojola Micheal Dec 29 '18 at 20:26

As I mentioned in the comments, you have not used the uniform convergence of $f_n$ or $g_n$ in your argument. And in fact, it is not necessary for your statement to hold. The important difference between pointwise and uniform convergence is that the later is global in the sense that the $N$ you find depends only on $\epsilon$ and not on $x$. Geometrically, suppose that $f_n \to f$ uniformly. Then, consider the graph of $f$ engulfed by $f+\epsilon$ and $f-\epsilon$. Picture this like a cylinder of radius $\epsilon$ wrapped around $f$ at each point of its domain entirely.

When the convergence is uniform, after a certain $N$, all $f_n$'s will be contained in this tubular (cylindrical) neighborhood globally for $n \geq N$. When the convergence is not uniform, no matter what $N$ is, some $f_n$'s will not be completely contained in this tubular (cylindrical) neighborhood. This is the geometric idea. In other words, the cylinder wrapped around $f$ might get wider or narrower around some points to contain $f_n$'s. It won't look uniform to us anymore.

In your case, your theorem is implied by the following simple theorem about real sequences:

$$\lim_{n\to\infty} a_n \times \lim_{n\to\infty} b_n = \lim_{n\to\infty} a_n\times b_n$$

where $a_n$ and $b_n$ are real sequences. Note that for any $x$, $f_n(x)$ and $g_n(x)$ are real sequences. Now try to continue your reasoning from here.

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  • $\begingroup$ Alright, I'll work on that! $\endgroup$ – Omojola Micheal Dec 29 '18 at 20:42
  • $\begingroup$ @Mike Yes. It's better now. Why don't you use the same boundedness argument for $g_n$ as well? Your argument is correct anyway, but since the question asks you to assume that $f_n$ and $g_n$ are uniformly convergent, you might want to emphasize that $N$ can be chosen as a function of only $\epsilon$. Even though your argument is correct anyway. $\endgroup$ – stressed out Dec 29 '18 at 21:01
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    $\begingroup$ All $N$'s now depend only on $\epsilon.$ Thanks for your comments and help! $\endgroup$ – Omojola Micheal Dec 29 '18 at 21:17
  • $\begingroup$ @Mike It's correct now. You're welcome. $\endgroup$ – stressed out Dec 29 '18 at 21:20

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