- #1

mathmari

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MHB

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Let $0<\alpha \in \mathbb{R}$ and $(f_n)_n$ be a sequence of functions defined on $[0, +\infty)$ by: \begin{equation*}f_n(x)=n^{\alpha}xe^{-nx}\end{equation*} - Show that $(f_n)$ converges pointwise on $[0,+\infty)$.

For an integer $m>a$ we have that \begin{equation*}0 \leq n^{\alpha}xe^{-nx} \leq n^{m}xe^{-nx}\end{equation*}

For $x> 0$ we have that \begin{equation*}\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}n^{m}xe^{-nx}=\lim_{n\rightarrow +\infty}\frac{n^{m}x}{e^{nx}}\ \overset{m\text{-times De L'Hopital}}{=}\ \lim_{n\rightarrow +\infty}\frac{m!}{x^{m-1}e^{nx}}=0\end{equation*} For $x= 0$ we have that \begin{equation*}\lim_{n\rightarrow +\infty}f_n(0)=\lim_{n\rightarrow +\infty}0=0\end{equation*} So we get \begin{equation*}\lim_{n\rightarrow +\infty}0 \leq \lim_{n\rightarrow +\infty}n^{\alpha}xe^{-nx} \leq \lim_{n\rightarrow +\infty}n^{m}xe^{-nx} \Rightarrow \lim_{n\rightarrow +\infty}0 \leq \lim_{n\rightarrow +\infty}n^{\alpha}xe^{-nx} \leq 0\end{equation*} It follows that $\displaystyle{\lim_{n\rightarrow +\infty}n^{\alpha}xe^{-nx}=0}$.

Therefore $f_n(x)$ converges pointwise to $0$.

Is everything correct? :unsure:

- Calculate $\max_{x\in [0, +\infty)}f_n(x)$ and conclude that $f_n$ converges uniformly on $[a, +\infty)$ for $a>0$.

We have that

\begin{align*}&f_n(x)=n^{\alpha}xe^{-nx}\\ &\rightarrow f_n'(x)=n^{\alpha}e^{-nx}-n^{\alpha+1}xe^{-nx}=\left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx} \\ &\rightarrow f_n'(x)=0 \Rightarrow \left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx}=0 \Rightarrow n^{\alpha}-n^{\alpha+1}x=0 \Rightarrow x=\frac{1}{n} \\ &f_n\left (\frac{1}{n}\right )=\frac{n^{\alpha-1}}{e}\end{align*}

Since we have the intervall $[\alpha, +\infty)$ we have to check also the value of the function at $x=\alpha$, right?

We get $f_n(\alpha)=n^{\alpha}\alpha e^{-n\alpha}$. But how can we compare it with $\frac{n^{\alpha-1}}{e}$ ? Or do we have to check if $f_n(x)$ is increasing or decreasing? :unsure:

- Show that $f_n$ converges uniformly on $[0, +\infty)$ iff $a<1$.

The maximum is $\frac{n^{\alpha-1}}{e}$ (since at $x=0$ we have $f_n(0)=0$ which is smaller) and for $a<1$ the limit goes to $0$, and this means that $f_n$ converges uniformly, right?

This shows that if $a<1$ then $f_n$ converges uniformly, or not? It is left to show that if $f_n$ converges uniformly then $a<1$, or not? :unsure: