consider the function $f_n(x)=n^2x^2e^{-xn}$ I am asked if it uniform converge on $A=(a,\infty) \quad a>0$
So it easy to see that in converge to $0$, but when I wanted to check uniform converge I noticed that the function getting maximum on $x=\frac{2}{n}$ pluging this value of $x$ to the function will give:
$f_n(\frac{2}{n})=\frac{4}{e^2}$, from here I concluded that $\sup|f_n(x)-f(x)|>0$
but this solution is wrong according to their solution, they actually used the same value of $x$ that I found to show it is uniform converge.they say for every $n>\frac{2}{a}$ u get that $\sup |f_n|=f_n(a)$ wich can indicates that is uniform converge, I actually not understand why we looking on $n>\frac{2}{a}$ I found specific $x$ that getting a value for the function so the supremum cannot be zero
It is false that $\sup_{x\in[a,\infty)}|f_n(x)|=4e^{-2}$. This is true if $\frac2n\geqslant a$, but this only occurs for finitely many $n$'s. If $\frac2n<a$, then $f_n$ is strictly decreasing and it attains its maximum at $a$. That maximum is $n^2a^2a^{-na}$ and, since $\lim_{n\to\infty}n^2a^2a^{-na}=0$, your sequence converges uniformly to $0$ indeed.