I need to study the uniform convergence of
$f_n(x) = n(x-1)e^{-nx}$
on the interval $[0,+\infty)$
I've shown that :
at $x =0$ $f_n(0)=-n \xrightarrow{} -\infty$
at $x =1$ $f_n(1)=0 \xrightarrow{} 0$
for every $x \neq 0$ we have that $f_n(x)\xrightarrow{} 0$ for n $\rightarrow$ + $\infty$
It remains to prove the uniform convergence on the interval $[0,+\infty)$. But at $0$ there isn't pointwise convergence and the function $f(x)=0$ is defined for every $x \neq 0$. So in $[0,\infty)$ there isn't uniform convergence?
Thanks in advance for any help.
You are right, it does not converge uniformly on $[0,\infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,\infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $\displaystyle{\lim_{n\to\infty}\sup_{x>0}|f_n(x)|=0}$. But $\displaystyle{\sup_{x>0}|f_n(x)|\geq\lim_{x\to0^+}|f_n(x)|=n}$, hence the limit above is equal to $+\infty$. Therefore the sequence does not converge uniformly on $(0,\infty)$ either. Let's take some small $\varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[\varepsilon,\infty)$. We have that $\displaystyle{\sup_{x\geq\varepsilon}|f_n(x)|=|f_n(\varepsilon)|}$ (take $\varepsilon\leq1$, it is easy to verify it). Now it is $|f_n(\varepsilon)|\to0$, therefore the convergence is uniform on $[\varepsilon,\infty)$ for any $\varepsilon>0$, but not on $[0,\infty)$, nor $(0,\infty)$.