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I asked this question before but I did not get a clear explanation for my problem with this question:

Define the function $f(x)=\frac{2x}{1+x^2}$ and let $f_n(x)=f(nx)$.

a) Show $f_n$ converges to 0 uniformly on $[1,\infty)$.
b) Show $f_n$ converges to 0 on $[0,1]$? is it uniform convergence.

for part a, I did the calculations and I came to $|f_n(x)|<\frac{1}{2}$ so there is no $n$ here to be restricted such that I can construct an epsilon from? why would I conclude part "a" is correct with that inequality.
for part b, I ended up with $lim sup [|f_n(x)-f(x)|]=\frac{1}{2}\neq 0 $ so in interval of part "b", I must say the sequence of functions is not uniformly convergent. right?

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BesMath
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  • I have taken the liberty to modify your title into something more attractive... – Jean Marie Apr 03 '20 at 06:52
  • You should cite your previous identical question : https://math.stackexchange.com/q/3606786. More generally it is not good acting to repost - only some hours later - the same question. Imagine everybody does that : this site would become a mess... – Jean Marie Apr 03 '20 at 06:55

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For a) show that $\frac {2t} {1+t^{2}}$ is decreasing function on $[1,\infty)$ (by showing that the derivative is negative). Hence the maximum value of $f_n(x)$ is $f_n(1)=\frac {2n} {1+n^{2}}$ which tends to $0$ as $ n \to \infty$.

For b) use the fact that $\sup |f_n(x)| \geq |f_n(\frac 1 n)|=1$ so the convergence is not uniform on $[0,1]$.

Kavi Rama Murthy
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