# Proof involving uniform convergence of a sequence and functions.

Suppose $f_n\to f$ uniformly on $[0,1]$ and $g$ is continuous on $[0,1]$. Prove that the sequence of functions $f_ng$ converges uniformly to the product $fg$ on $[0,1]$.

My draft of a proof is the folloing:

Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. Since $f$ is bounded and $\{f_n\}$ is uniformly bounded, there is an $M>0$ such that $\text{max}\{|f_n(x)-f(x)|:x\in[0,1],n\in\mathbb{N}\}\leq M$. Given $\epsilon>0$ choose $\delta>0$ so small that $0<x<0+\delta$ or $1>x>1-\delta$ inplies $|g(x)|<\frac{\epsilon}{M}$. By hypothesis, $f_n\to f$ uniformly on $[0+\delta,1-\delta]$. Thus choose $N$ so large that $x\in[0+\delta,1-\delta]$ and $n\geq N$ imply $|f_n(x)-f(x)|<\frac{\epsilon}{C}$. If $n\geq N$ and $x\in[0,1]$, then $$|f_n(x)g(x)-f(x)g(x)|=|f_n-f(x)||g(x)|<\begin{cases} (\frac{\epsilon}{C})\cdot C=\epsilon \hspace{.5cm} \text{for} \hspace{.5cm} x\in[0+\delta,1-\delta]\\ (\frac{\epsilon}{M})\cdot M=\epsilon \hspace{.5cm} \text{for} \hspace{.5cm} x\not\in[0+\delta,1-\delta].\end{cases}$$

Therefore, $f_ng\to fg$ uniformly on $[0,1]$

Is there anywhere I can improve or are incorrect?

• How do you define $b$? Commented Jul 25, 2018 at 20:33
• those were placeholder values that I forgot to edit in, my bad. Fixing now Commented Jul 25, 2018 at 20:34
• Are you given that $f_n$ are continuous or $f$ is continuous? Commented Jul 25, 2018 at 20:34
• I am not given either those pieces of information, just that $f_n\to f$ uniformly on $[0,1]$. Commented Jul 25, 2018 at 20:35
• In that case, it may not be true that $f$ is bounded: for instance consider $f(0) = 0$ and $f(x) = 1/x$ for $x > 0$, and $f_n = f$ for all $n$. Commented Jul 25, 2018 at 20:38

The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous): $$\sup_{x\in[0,1]}|f_n(x)g(x)-f(x)g(x)|\leq\sup_{x\in[0,1]}|f_n(x)-f(x)|\cdot \sup_{x\in[0,1]}|g(x)|.$$ Therefore, if $f_n\to f$ uniformly in $[0,1]$ then $\sup_{x\in[0,1]}|f_n(x)-f(x)|\to 0$ and, by the above inequality, $\sup_{x\in[0,1]}|f_n(x)g(x)-f(x)g(x)|\to 0$ and we may conclude that $f_ng\to fg$ uniformly in $[0,1]$.

• That might be true only for $n$ large enough if you limit yourself to finite quantities. Commented Jul 25, 2018 at 20:42
• $g$ is always bounded in $[0,1]$ as it is continuous. Commented Jul 25, 2018 at 20:45
• @mathcounterexamples.net Do you think that the inequality is not true? Why? Commented Jul 25, 2018 at 20:50
• I’m just saying that $\sup \vert f(x)-f_n(x) \vert$ may not be defined for « small » integers. Commented Jul 25, 2018 at 20:58
• The $\sup$ can be $+\infty$. I don't see any problem with that. Commented Jul 25, 2018 at 21:02

Thanks to Robert and B. Mehta I think I am confident in my proof.

Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem.

In other words, there is a $C>0$ such that $|g(x)|\leq C$ for all $x\in[0,1]$.

Additionally, there is an $M>0$ such that $\sup_{x\in[0,1]}\{|f_n(x)-f(x)|,n\in\mathbb{N}\}\leq M$.

Given $\epsilon>0$ choose $\delta>0$ so small that $0<x<0+\delta$ or $1>x>1-\delta$ implies $|g(x)|<\frac{\epsilon}{M}$.

By hypothesis, $f_n\to f$ uniformly on $[0+\delta,1-\delta]$.

Thus choose $N$ so large that $x\in[0+\delta,1-\delta]$ and $n\geq N$ imply $|f_n(x)-f(x)|<\frac{\epsilon}{C}$.

If $n\geq N$ and $x\in[0,1]$, then $$\sup|f_n(x)g(x)-f(x)g(x)|\leq\sup|f_n-f(x)|\cdot\sup|g(x)|\leq\sup|g(x)|\frac{\epsilon}{C}<\epsilon\hspace{.2cm}\text{for}\hspace{.2cm}x\in[0+\delta,1-\delta]$$

Therefore, $f_ng\to fg$ uniformly on $[0,1].$

• Your proof shows that $f_ng\to fg$ in $[0+\delta,1-\delta]$ for any $\delta>0$ which does not inply that the convergence is uniformly also in $[0,1]$. But why do you need $\delta>0$? Commented Jul 25, 2018 at 21:31
• Line 4 of my proof describes $\delta$. Isn't $\delta$ necessary since this problem has a $\epsilon$ for function values, it would need $\delta$ for input values? Commented Jul 25, 2018 at 21:41
• There is no need of $\delta$. Given $\epsilon>0$ there is $N$ such that for $n\geq N$, $\sup_{x\in[0,1]}|f_n(x)-f(x)|<\frac{\epsilon}{C}$. Commented Jul 25, 2018 at 21:48
• Got it, my book used $\delta$ for a similar proof so I tried using that as a baseline for mine. Commented Jul 25, 2018 at 21:57