Question is to check uniform convergence:
If $f_n \rightrightarrows f$ and $g_n \rightrightarrows g$ on $A\subset \mathbb{R}$ then
$f_n+g_n \rightrightarrows fg$ on $A\subset \mathbb{R}$
$f_ng_n \rightrightarrows fg$ on $A\subset \mathbb{R}$
I could see that :
$$|(f_n+g_n)(x)-(f+g)(x)|$$ $$=|(f_n(x)-f(x))+(g_n(x)-g(x))|$$ $$\leq |f_n(x)-f(x)|+|g_n(x)-g(x)|$$
So, once I fix $\epsilon >o$ I have $N_1\in \mathbb{N}$ and $N_2\in \mathbb{N}$ such that :
$$|f_n(x)-f(x)|<\frac{\epsilon}{2}\text{ for all $n\geq N_1$ and for all $x\in A$}$$ and $$|g_n(x)-g(x)|<\frac{\epsilon}{2}\text{ for all $n\geq N_2$ and for all $x\in A$}$$
So, $$|(f_n+g_n)(x)-(f+g)(x)|\leq |f_n(x)-f(x)|+|g_n(x)-g(x)|<\frac{\epsilon}{2}+\frac{\epsilon}{2} =\epsilon$$
for all $n\geq N=Max \{N_1,N_2\}$ and for all $x\in A$
Thus, I conclude that $f_n+g_n\rightrightarrows f+g$ on $A\subset \mathbb{R}$.
Now, I want to imitate same kind of logic here :
$$|(f_ng_n)(x)-(fg)(x)|$$ $$=|f_n(x)g_n(x)-f(x)g(x)-f_n(x)g(x)+f_n(x)g(x)|$$ $$=|f_n(x)[g_n(x)-g(x)]+g(x)[f_n(x)-f(x)]|$$
$$\leq |f_n(x)||[g_n(x)-g(x)]|+|g(x)||[f_n(x)-f(x)]|$$
$$\leq |f_n(x)|\frac{\epsilon}{2}+|g(x)|\frac{\epsilon}{2}$$
$$=(|f(x)|+|g(x)|).\frac{\epsilon}{2}$$
With this I believe that if i can make sure $f(x)$ and $g(x)$ are bounded then I would have no problem and we would have :
$$f_ng_n \rightrightarrows fg\text{ on } A\subset \mathbb{R}$$
Now, I will have problem if $f(x)$ and/or $g(x)$ is unbounded.
I believe for sure that in this case we do not have uniform convergence.
I have two requests :
- Please make sure if my idea is correct
- Please suggest some "natural" convergence in which limit function is not bounded.
EDIT : Just after asking for "natural" choice I got some idea :
If I take $f_n(x)=x+\frac{1}{n}$ then I would have limit function $f(x)=x$.
If i consider this convergence on whole $\mathbb{R}$ then :
I have uniform convergence but limit function is not bounded.... :D :D :D
Now, I do not want to make it so complicated so, I want to take my $g_n(x)=f_n(x)$
I have $f_n(x)g_n(x)=(x+\frac{1}{n})^2$ we would have limit function as $f(x)g(x)=x^2$
But the convergence is not uniform when i see this convergence on whole of $\mathbb{R}$
SO, I would like to conclude that :
$f_n \rightrightarrows f$ and $g_n \rightrightarrows g$ does not necessarily imply $f_ng_n \rightrightarrows fg$