# Uniform convergence of $\frac{(n+1)x}{1+(n+1)^2x^2}$ on $(-\delta , \delta)$

I want to determine whether $f_n=\displaystyle\frac{(n+1)x}{1+(n+1)^2x^2}$ converges uniformly or not on a $\delta$ neighbourhood of $0$ .

I know it converges point-wise to zero. I'm trying to prove its not uniformly convergent but I don't know where to start

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What if you take $x_n=1/(n+1)$ and consider $f_n(x_n)$? –  David Mitra Jan 8 '13 at 16:51
@DavidMitra Then $f_n(x_n)$ is $1$ , could you explain what that would mean ? –  user10444 Jan 8 '13 at 16:55
It's $1/2$ for each $n$. So, take $\epsilon=1/2$. Is it true that there is an $N$ so that for all $n\ge N$ and all $x\in(-\delta,\delta)$, we have $|f_n(x)-0|<\epsilon$? –  David Mitra Jan 8 '13 at 16:57
@DavidMitra Sorry for the calculation mistake. For every $n$ we have $|f_n(x_n)-0|\ge \epsilon$ then its not uniformly convergent –  user10444 Jan 8 '13 at 17:01
Yes, that's correct. This is usually a good approach, when you can do it, for showing a sequence is not uniformly convergent: select $x_n$ so that $f_n(x_n)$ is bounded away from the pointwise limit. –  David Mitra Jan 8 '13 at 17:04

It's a good idea to look at the graphs of the terms in the sequence. Below is a plot of the graphs of $\color{maroon}{f_4}$, $\color{blue}{f_7}$, $\color{darkgreen}{f_{11}}$, and $\color{pink}{f_{24}}$:

One may surmise from these plots that $(f_n)$ converges to $0$ pointwise on $(-\delta,\delta)$; but that the convergence is not uniform there, since every $f_n$ has a "peak" of height $1/2$.

And indeed, one can easily show that $(f_n)$ converges pointwise to the zero function on $(-\delta,\delta)$.

To disprove uniform convergence, let $x_n=1/(n+1)$. These values give the peaks: $f_n(x_n)=1/2$ for every $n$, and this implies that the convergence cannot be uniform (take $\epsilon=1/2$ in the definition of uniform convergence: it is not true that $|f_n(x)|<\epsilon$ for all $x\in(-\delta,\delta)$ for any $n$).

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The convergence is not uniform even in $(0,\delta)$: $$\left\|f_n-0\right\|=\sup_{x\in (0,\delta)}\left|\frac{(n+1)x}{1+(n+1)^2x^2}\right|=\sup_{x\in (0,\delta)}\frac{(n+1)x}{1+(n+1)^2x^2}\ge\frac{(n+1)\frac 1{n+1}}{1+(n+1)^2\frac 1{(n+1)^2}}=\frac 12$$ You might want to explain why the inequality is true (for large $n$)

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