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Q. Find an example of a sequence of functions $\{f_n\}$ and $\{g_n\}$ that converge uniformly to some $f$ and $g$ on some set $A$, but such that $\{f_ng_n\}$ (the multiple) does not converge uniformly to $fg$ on $A$. Hint: Let $A=R$, let $f(x)=g(x)=x.$ You can even pick $f_n=g_n.$

As it is stated in Hint, let $A=R,$ and $f(x)=g(x)=x,$ and $f_n=g_n $

Since $f_n$ converges uniformly to $x$. $\forall\varepsilon>0, \exists M\in s.t. $ for $n\ge M, |f_n-x|<\varepsilon \hspace{0.5cm}\forall x \in R.$

But, $|f_n^2-x^2| = |f_n+x||f_n-x|=|f_n+x|\varepsilon\not < \varepsilon$ for some $x$.

I don't know how to pick $f_n$ satisfying these conditions. Could you give some hint?

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    $\begingroup$ It is extremely misleading to write $f_n=g_n=x_n$, since $f_n$ and $g_n$ are functions. $\endgroup$ – Eric Wofsey May 26 '18 at 6:09
  • $\begingroup$ You are right. I fixed it. $\endgroup$ – Sihyun Kim May 26 '18 at 6:12
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I would take $f_n(x)=g_n(x)=x+\frac{1}{n}$. Then they converge uniformly over $\mathbb{R}$ to $f(x)=g(x)=x$.

On the other hand, if $A$ is any unbounded set of $\mathbb{R}$ then $$\sup_{x\in A}|f_n(x)g_n(x)-f(x)g(x)|=\sup_{x\in A}\left|\left(x+\frac{1}{n}\right)^2-x^2\right|=\sup_{x\in A}\left|\frac{2x}{n}+\frac{1}{n^2}\right|=+\infty$$ which means that $f_ng_n$ does not converge uniformly to $fg$ over $A$.

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Take $f_n(x)=g_n(x)=x+\frac{1}{n}$. You can show that these sequences converge uniformly. Now take a look at the product of these, it will definitely converge pointwise to $x^2$, but not uniform. To show that the product sequence does not converge uniformly to $x^2$,

$$\sup\limits_{x\in\mathbb{R}}||f_ng_n(x)-x^2||=\sup\limits_{x\in\mathbb{R}}||x^2+\frac{2x}{n}+n^2-x^2||=\sup\limits_{x\in\mathbb{R}}||\frac{2x}{n}+n^2||\to\infty$$

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