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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?

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    $\begingroup$ How do you define $b$? $\endgroup$ Commented Jul 25, 2018 at 20:33
  • $\begingroup$ those were placeholder values that I forgot to edit in, my bad. Fixing now $\endgroup$
    – Peetrius
    Commented Jul 25, 2018 at 20:34
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    $\begingroup$ Are you given that $f_n$ are continuous or $f$ is continuous? $\endgroup$
    – B. Mehta
    Commented Jul 25, 2018 at 20:34
  • $\begingroup$ I am not given either those pieces of information, just that $f_n\to f$ uniformly on $[0,1]$. $\endgroup$
    – Peetrius
    Commented Jul 25, 2018 at 20:35
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    $\begingroup$ 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 $n$. $\endgroup$
    – B. Mehta
    Commented Jul 25, 2018 at 20:38

2 Answers 2

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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 $[0,1]$.

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

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    $\begingroup$ 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 $\delta>0$? $\endgroup$
    – Robert Z
    Commented Jul 25, 2018 at 21:31
  • $\begingroup$ 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? $\endgroup$
    – Peetrius
    Commented Jul 25, 2018 at 21:41
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    $\begingroup$ There is no need of $\delta$. Given $\epsilon>0$ there is $N$ such that for $n\geq N$, $\sup_{x\in[0,1]}|f_n(x)-f(x)|<\frac{\epsilon}{C}$. $\endgroup$
    – Robert Z
    Commented Jul 25, 2018 at 21:48
  • $\begingroup$ Got it, my book used $\delta$ for a similar proof so I tried using that as a baseline for mine. $\endgroup$
    – Peetrius
    Commented Jul 25, 2018 at 21:57

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