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(Disclaimer: This is a homework question.) $$\text{For }n\text{ in }\mathbb{N}\text{ let }f_{n}:\mathbb{R} \rightarrow \mathbb{R} \text{ be defined by } f_{n}(x)=\frac{nx}{1+n^2x^2} $$

Show that the sequence $(f_{n})_{n\in\mathbb{N}}$ is convergent. Is this convergence uniform?

My attempt:
as $n \rightarrow \infty\quad f_{n} \rightarrow \frac{1}{nx} \implies f_{n} \rightarrow f = \left\{\begin{matrix} &0 &x\neq0 \\ &?? &x = 0 \end{matrix}\right. $
It seems to me $f$ is discontinous at $x=0$, unless I misunderstood.
For uniform convergence it's required that:$$\forall \varepsilon>0\quad \exists K_{\varepsilon}\in\mathbb{N} \quad s.t.\quad \forall n\geq K_{\varepsilon}\\||f_{n}(x)-f(x)||<\varepsilon$$ if we pick $n=k$ and $x_{k} = \frac{1}{k}$ then $$f_{n} = 1\quad \forall n\in\mathbb{N}\\\implies||f_{n}(x)-f(x)|| = ||1-0||=1$$ Does this imply $f_{n}$ is not uniformly convergent?

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    $\begingroup$ What is the domain of this sequence of functions? $\endgroup$
    – SystematicDisintegration
    Commented Dec 12, 2017 at 1:07
  • $\begingroup$ you have $f_n(0)=0$ for all $n$ so $f(0)=0$. $\endgroup$
    – zwim
    Commented Dec 12, 2017 at 1:11
  • $\begingroup$ Right, your limit is funky. You wrote as $n \to \infty$ $f_{n} \to \frac{1}{nx}$, but if you had to actually do this limit in say a calculus course, would you not write that $f_{n} \to 0$? $\endgroup$
    – Boots
    Commented Dec 12, 2017 at 1:12
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    $\begingroup$ Instead of saying "convergent", it would be good to say " convergent pointwise everywhere on $\mathbb R." $\endgroup$
    – zhw.
    Commented Dec 12, 2017 at 1:15
  • $\begingroup$ That's how it was worded, but I'm assuming that's what the question meant by "convergent". $\endgroup$
    – Rahul Chowdhury
    Commented Dec 12, 2017 at 1:17

1 Answer 1

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Recall that a sequence $(f_n)_{n=1}^\infty$ converges pointwise to $f$ on $\mathbb{R}$ if for every $x\in \mathbb{R},$ the sequence of scalars $(f_n(x))_{n=1}^\infty$ converges to $f(x).$

Also, the sequence $(f_n)_{n=1}^\infty$ converges uniformly to $f$ on $\mathbb{R}$ if for every $\varepsilon>0,$ there exists a natural number $N_\varepsilon$ such that for any $n\geq N_\varepsilon$ and any $x\in \mathbb{R},$ we have $$| f_n(x)-f(x) |<\varepsilon.$$

If $f_n(x) = \frac{nx}{1+n^2x^2},$ then the sequence $(f_n)_{n=1}^\infty$ converges pointwise to $f=0$ on $\mathbb{R}.$ Indeed, fix $x\in \mathbb{R}.$ Then $$\lim_{n\to\infty}f_n(x) = \lim_{n\rightarrow\infty}\frac{nx}{1+n^2x^2} = \lim_{n\to\infty}\frac{\frac{x}{n}}{\frac{1}{n^2}+x^2} = 0 = f(x).$$ So the sequence $(f_n)_{n=1}^\infty$ converges pointwise on $\mathbb{R}.$

Note that the sequence $(f_n)_{n=1}^\infty$ does not converge uniformly to $f=0$ on $\mathbb{R}.$ Let $\varepsilon=\frac{1}{4}.$ Fix $N_\varepsilon\in\mathbb{N}.$ Choose $n = N_\varepsilon$ and $x = \frac{1}{N}.$ Then $$| f_n(x) - f(x) | = \bigg| \frac{1}{1+1} - 0 \bigg| = \frac{1}{2} > \frac{1}{4} = \varepsilon.$$ So the sequence $(f_n)_{n=1}^\infty$ does not converge uniformly to $f=0$ on $\mathbb{R}.$

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