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Let $(g_n(x))$ a sequence of function such that $g_n :\mathbb{R}_+\to \mathbb{R}$ and $g_n(x)=\dfrac{x}{n^2}e^{-\frac{x}{n}}$

The target is to show $(g_n(x))$ converges uniformly.

Attempt :

  1. I show the pointwise convergence.

For $x=0$, $\quad g_n(x)=0$ $\quad \forall n\in \mathbb{N}^*$

For $x\in ]0,+\infty),$ $\quad g_n(x)\underset{n\to+\infty}{\longrightarrow}0$

  1. I show the uniform convergence to $f:x\to f(x)=0\qquad$

    $\underset{x\in\mathbb{R}_+}\sup\big\{|g_n(x)-f(x)|\big\}<\varepsilon \iff \underset{x\in\mathbb{R}_+}\sup\bigg\{\bigg|\frac{x}{n^2}e^{-\frac{x}{n}}\bigg|\bigg\}<\varepsilon$

$g_n'(x)=\dfrac{1}{n^2}\dfrac{(1-x)e^{x/n}}{e^{2x/n}}$ thus $g_n$ is increasing on $[0,1]$ and decreasing on $[1,+\infty]$ as $g_n(0)=0$ and $g_n(x)\underset{x\to +\infty}{\longrightarrow}0$ then $\underset{x\in\mathbb{R}_+}\sup\bigg\{\bigg|\frac{x}{n^2}e^{-\frac{x}{n}}\bigg|\bigg\}=\dfrac{1}{n^2e^{1/n}}\underset{n\to +\infty}{\longrightarrow}0$

Hence $\boxed{g_n\overset{\text{unif.}}{\longrightarrow}f}$

Is that correct, and are there other methods more efficient ??

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    $\begingroup$ Alternatively notice that $h(y)=y e^{-y}$ is a bounded function and therefore $g_n(x) = 1/n h(x/n)\le C/n\to 0$ with $C>0$ some constant independent of $x$. $\endgroup$ – J.R. May 25 '18 at 21:09
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You can shorten it. Consider $f : [0, +\infty\rangle \to \mathbb{R}$ defined as $f(t) = te^{-t}$.

$f$ is continuous and $\lim_{t\to+\infty} f(t) = 0$ and so $f$ is bounded on $[0, +\infty\rangle$, i.e. there exists $M > 0$ such that $f(t) \le M, \forall t \ge 0$.

Now we have

$$g_n(x) = \frac1n \cdot \frac{x}{n}e^{-\frac{x}n} = \frac1n f\left(\frac{x}n\right) \le \frac{M}n \xrightarrow{n\to\infty} 0$$ uniformly in $x \in [0, +\infty\rangle$.

Hence $g_n \xrightarrow{\text{unif.}} 0$.

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It is correct, but computing $g_n'$ is easier if you do it like this:$$g_n'(x)=\frac1{n^2}\left(x'e^{-\frac xn}+x\left(e^{-\frac xn}\right)'\right)=\frac1{n^2}\left(e^{-\frac xn}-\frac xne^{-\frac xn}\right).$$

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