Uniform convergence of the sequence of funtions for $x \in [a,\infty)$ with $a>0$

Question

Show that

$f_n= \frac {2n^2x}{(1+n^2 x^2) \ln(1+n)}$

does converges uniformly in $x\in [a,\infty)$ for $a>0$

How do I prove the uniform convergence for that interval?

How differently is from doing it for $x\in\mathbb R$ ?
Why for $x\in\mathbb R$ it does not converges uniformly?

Answer 1

For $x\geq 0$ then $f_n(x)\to 0$ as $n$ goes to infinity. Moreover, since $$f'_n(x)=\frac{2n^2}{\ln(1+n)}\cdot\frac{(1-n^2x^2)}{(1+n^2x^2)^2}$$ it follows that $f_n$ is a nonnegative continuous function which is increasing in $[0,1/n]$, it attains the maximum value at $1/n$ and it is decreasing in $[1/n,+\infty)$. Hence, for $a>0$, when $n\geq 1/a$, $$\sup_{[a,+\infty)}|f_n(x)-f(x)|=f_n(a)\to 0,$$ and $(f_n)_n$ converges uniformly in $[a,+\infty)$. On the other hand $$\sup_{[0,+\infty)}|f_n(x)-f(x)|=f_n(1/n)=\frac {n}{\ln(1+n)}\to +\infty$$ which implies that $(f_n)_n$ does not converge uniformly in $[0,+\infty)$.

Answer 2

$|f_n-f|= \frac {2n^2x}{(1+n^2 x^2) ln(1+n)}-0= \frac {2x}{(1/n^2+ x^2) ln(1+n)}\leq \frac {2x}{(x^2) ln(1+n)}\leq \frac {2}{a \ln(1+n)}$

So, for all $x$, you can bound this distance by $\epsilon$ for large enough $n$

For the last step of inequality, you definitely need $a>0$ and the bounded domain.

Answer 3

For a fixed $a>0$, when $x\geq a$ we have $0<1/(1+n^2 x^2)<1/n^2x^2$. So $x\geq a$ implies $$0<f_n(x)<2n^2 x/n^2 x^2\ln (1+n)=2/x\ln (1+n)\leq 2/a\ln (1+n).$$

So $\lim_{n\to \infty}\sup_{x\geq a}|f_n(x)|=0.$

But $f_n(1/n)=n/\ln (1+n)$ so $\lim_{n\to \infty}\sup_{x>0}|f_n(x)|=\infty.$

Try drawing the graph of $f_n$ for various $n$.