Prove that $f_n(x) = \frac{nx}{1+n^2x^2}$ is not uniformly convergent on any interval containing $0$.

Question

Let $\{f_n\}$ be a sequence of real valued functions defined by $$f_n(x) = \frac{nx}{1+n^2x^2}\ \text{for all}\ x\in \mathbb{R}$$ is not uniformly convergent on any interval containing $0$.

$\textbf{My attempt:}$ We see that $$\text{lim}_{n\to \infty} f_n(x) =\text{lim}_{n\to \infty} \frac{nx}{1+n^2x^2}= \text{lim}_{n\to \infty} \frac{\frac{x}{n}}{\frac{1}{n^2}+x^2} = 0$$ Hence the pointwise limit of $f_n (x)\to f(x)$, where $f(x)$ is defined by $f(x) = 0\ \text{for all}\ x$.

Let $0\in [a,b]$. Our aim is to prove that $f_n$ does not converge to $f$ uniformly on $[a,b]$. Assume $f_n \rightrightarrows f$ on $[a, b]$. Hence for each $\epsilon >0$ there exist a positive integer $N(\epsilon)$ such that $|f_n(x) - f(x)| <\epsilon$ for all $n\geq N(\epsilon)$ and $x\in [a,b]$.

Now choose $x =\frac{1}{n}$ $\color{red}{(\text{Can we choose? Why?)}}$ Then, $$|f_n(x)-f(x)| = \Big|\frac{nx}{1+n^2x^2} -0\Big| = \Big|\frac{nx}{1+n^2x^2}\Big| = \frac{1}{2}$$ If we take $\epsilon =\frac{1}{4}$, then $f_n$ does not converge uniformly to $f$ on $[a,b]$.

Is my proof correct? Can you please advise me explicitly that why I can choose $x =\frac{1}{n}$. I am little bit confused about it.Thanks.

Answer 1

Differentiate it

It is enough to show the derivative:

1. . . . is continuous away from zero.

2. . . . goes to infinity at zero.

Then you can find a suitable interval $[a,b]$ close to zero and use the mean value theorem to get $ |f(a)-f(b)| = |f'(c)||a-b|$. Show we can make $|f'(c)|$ big as we like by going close enough to the origin.

Answer 2

Uniform Limits

Assume the sequence is uniformly convergent. Then the restriction to the nonnegative axis is also uniformly convergent. The Uniform Limit Theorem says the uniform limit exists and is continuous.

Every uniform limit is also the pointwise limit. Hence (the restriction of) $f_n$ has a continuous pointwise limit.

However you can show (the restriction of) the sequence $f_n$ tends pointwise to $1/2$ times the indicator function $f$ of $0$.

This function is not continuous. Contradiction.