Examine the almost uniform convergence of the function with the formula
$$ f_n: \mathbb R \rightarrow \mathbb R \mbox{ such that } f_n(x) = \frac{nx}{e^{(nx-1)^2}} $$.
Can somebody check if my attempt is correct?
solution
point convergence is easy to show (I don't have problem with that. Let $$ c := \mbox{ argument for which $|f_n|$ reaches maximum value} $$ $c$ exists because in bounded interval monotonous $f_n$ reaches own extremes $$D = [a,b]$$ $$ \sup_{x \in D} \left|\frac{nx}{e^{(nx-1)^2}} \right| = \left|\frac{nc}{e^{(nc-1)^2}} \right|$$ Of course series with given asymptotic converges so we have local uniform convergance
Monotonicity is not true so your argument does not work. $f_n(x)=g(nx)$ where $g(y)=ye^{-(y-1)^{2}}$. It is easy to show that $g(y) \to 0$ so $f_n\to 0$ uniformly in around any point other than $0$. However the sequence does not converge uniformly in any neighborhood of $0$ since $f_n(x)=g(1) >0$ when $x =\frac 1 n$.