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I recently read this theorem in real analysis:(Actually a corollary to a theorem)

{$f_m$} is a sequence of continuous functions defined on $D$ such that $f_m$$\to$$f$ uniformly on $D$ then for every sequence {$x_m$} in $D$ satisfying $x_m$$\to$$x$ then $f_m(x)$$\to$$f(x)$.

please correct me if I wrote this wrong. The doubt is that I am unable to use it in a problem: $$f_m(x)=\dfrac{mx}{1+mx}\qquad x\in[0,1].$$ I want to prove that this is not uniformly convergent, but I am not able to understand what is $x_m$. How does it depend on $m$? Does it even depend on $m$? If yes then how to prove that $f_m(x)$ does not converge to $f(x)$ when $x_m$ tends to $x$(to show that it is not uniformly convergent).(also, is there $x_m$ or $x$ inside $f_m(\cdot)$ . Basically, how to approach this problem, please explain stepwise because I need to get the feeling of this theorem.

Rohan Marwah
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    Are you sure the original statement is correct? I suspect that it should be about a sequence {$x_n$} where $x_n\to x$ and $f_m(x_n)\to f(x)$. – herb steinberg Nov 23 '18 at 21:41
  • I am not really sure about it. You might be right here. I'll change it when someone else points it out too.Thanks! – Rohan Marwah Nov 23 '18 at 21:44
  • @herbsteinberg I believe that n=m, because if they are not, then finding $f_m(x_n)$ will be vague since it will be a sequence of functions and so we will have to vary both $n$ and then $m$ to find out what $f_m(x_n)$ converges to. Correct me if i am wrong. Also take a look at the comment by Yadati. – Rohan Marwah Nov 23 '18 at 22:26
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    Since I am not familiar with this theorem, the comment by Yadati may be correct. – herb steinberg Nov 23 '18 at 22:36

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Another way to show $f$ is not uniformly convergent. $$f_m=\begin{cases}0 &x=0\\\dfrac{mx}{1+mx} &x\neq0\end{cases}$$ $\:$For $x\neq0\:$ $f_m(x)=\dfrac{mx}{1+mx}=\dfrac{x}{\frac1m+x}$. $\:$ Now if $m\to\infty\implies f_m(x)\to1$.

So $$f_m\to\begin{cases}0 &x=0\\1 &x\neq0\end{cases}=f$$

and $f$ is clearly not continuous. Hence the convergence is not uniform.

$\rule{17cm}{1pt}$

$0\neq x_m=\dfrac1m\to0\implies\:\forall\:m\quad f_m(x_m)=\dfrac12\not\to f(0)=0$. So the convergence is not uniform.

Yadati Kiran
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