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Let $f_n:=\frac{x}{n}\exp(-\frac{x}{n})$. I want to analyse, if $f_n$ is uniformly convergent on every compact interval. I could show that it converges pointwise to $f(x)=0$ and that for any $n \in \mathbb N$, we have that on the interval $[0,n]$ that $f_n(x)$ is bounded by $\frac{1}{e}$ and hence it is not uniformly convergent. But I miss now the last part, to conclude. Any help would be appreciated.

EDIT: Let $[a,b] \subset \mathbb R$ with $a\leq b$. I want to bound the following expression: $$\sup\{\|\frac{x}{n}\exp(-\frac{x}{n})\|:x \in [a,b]\}$$ If the following equality holds: $$\sup\{\|\frac{x}{n}\exp(-\frac{x}{n})\|:x \in [a,b]\}\leq \sup\{\|\frac{x}{n}\|:x \in [a,b]\} \times \sup\{\|\exp(-\frac{x}{n})\|:x \in [a,b]\}$$ we are done by taking the maximum in absolut value between $a$ and $b$, since then the first term goes to zero as n goes to infinity.

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  • $\begingroup$ try to bound with the max of $f_n$ $\endgroup$ Apr 3, 2023 at 17:18
  • $\begingroup$ Could you please have a look at my EDIT $\endgroup$ Apr 3, 2023 at 18:46
  • $\begingroup$ $|f_n(x)| \leq \frac{|a|+|b|}{n} e^{-a/n}.$ $\endgroup$
    – William M.
    Apr 3, 2023 at 19:18

5 Answers 5

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Choose any interval $[a, b]$.

To bound $f_n$ on $\mathbb{R}$ compute $$ \frac{\mathrm{d}}{\mathrm{d}x} \frac{x}{n}\exp\left(-\frac{x}{n}\right) = -\frac{(x-n)\exp\left(\frac{x}{n}\right)}{n^2}. $$ So there is a unique critical point at $x=n$. It is very easy to prove that this is a global maximizier on $\mathbb{R}$ for all $n \in \mathbb{N}$. So for very large $n$ the maximizer leaves $[a, b]$ and the maximum on $[a, b]$ is attained at either $a$ or $b$. Note that $$ \lim_{n \rightarrow \infty} \max_{x \in [a, b]} \lvert f_n(x) \rvert = \lim_{n \rightarrow \infty} \left \lvert \frac{y}{n}\exp\left(-\frac{y}{n}\right) \right \rvert = 0 $$ for $y \in \lbrace a, b \rbrace$ no matter the (fixed) values of $a$ or $b$.

So we indeed have uniform convergence on any compact subinterval.

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  • $\begingroup$ Thank you for this elegant way of showing this. However, I would like to prove it without derivation and wonder if my equation in the EDIT is correct. $\endgroup$ Apr 3, 2023 at 18:46
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    $\begingroup$ The equality that you stated does not hold in general. But what is true is that $\sup f_n \leq \sup\frac{x}{n} \cdot \sup \exp(x/n)$. But then your argument should work $\endgroup$ Apr 3, 2023 at 18:48
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Here is an elementary approach. If $[a,b]\subset[0,\infty)$, then the conclusion follows readily, since $e^{-x/n}\le 1$, and $x/n$ converges uniformly to zero on $[a,b]$. If $a < 0$, then we can write $[a,b]\subset [a,0]\cup[0,b]$, and all that's left is to check that $f_n(x)$ converges uniformly to zero on an interval of the form $[a,0]$ for $a < 0$.

Changing variables, let $x = -y$, for $y\in[0,|a|]$, and write $f_n(x) = f_n(-y) = (-\frac{y}{n})e^{y/n}$. Taking absolute values, $|f_n(-y)| = \frac{y}{n}e^{y/n} \le \frac{|a|}{n}e^{|a|/n}$. Given $\epsilon>0$, as long as $N$ is large enough depending only on $\epsilon$ and $|a|$, and $n\ge N$, we have $|f_n(-y)|\le \epsilon$. Therefore, changing variables back, and using our remarks in the first paragraph, $f_n(x)$ converges to zero for any interval $[a,b]$, uniformly in $x\in [a,b]$.


Added: This solution uses that $e^{y/n}$ and $y/n$ are both positive and increasing on $[0,\infty)$, hence so is their product.

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Let $f:\mathbb{R}\to\mathbb{R}$ be a function continuous at $x=0.$ Then the sequence of functions $f_n(x)=f(x/n)$ tends to $f(0)$ uniformly on any bounded interval. Indeed, consider $-a\le x\le a.$ Fix $\varepsilon>0.$ There exists $\delta>0$ such that $$|u|<\delta \implies |f(u)-f(0)|<\varepsilon$$ Then for $n>{a\over \delta}$ and $|x|\le a$ we get ${|x|\over n}<\delta,$ hence $|f_n(x)-f(0)|<\varepsilon.$

Apply the above to the function $f(x)=xe^{-x}.$

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If you want to be lazy with it (and steal ideas from 3 other answers in here):

Suppose $[a,b]\subset \mathbb{R}$ is a compact interval. Let $L=\max\{|a|,|b|\}$.

Notice that for any $x \in [a,b]$, and each $n \in \mathbb{N}$ we have $$0\leq \left| \frac{x}n \exp\left(-\frac{x}n\right) \right|\leq \frac{L}n\exp\left(\frac{L}n\right).$$

Since $g(x)=xe^{x}$ is continuous at $0$ and $\lim_{n \to \infty}L/n = 0$ ($L<\infty$), we have $\lim_{n \to \infty} g(L/n)=g(0)=0.$

Thus we have

$$0 \leq \lim_{n \to \infty} \max_{x \in [a, b]} \left| f_n(x) \right| \leq \lim_{n \to \infty} g(L/n) = 0.$$

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The sequence $$f_n(x)=\frac{x}{n}e^{-\frac x n}$$ indeed converges uniformly to $0$ on $[0, \infty)$ and there is a "soft" proof, without any computation.

SOFT PROOF. Note that $$ f_n(x)=f\left(\frac x n \right)\quad \text{ for }f(x)=xe^{-x}.$$ This function has a global maximum at some $x_0\in (0, \infty)$; indeed, it is manifestly continuous and satisfies $f(0)=0=\lim_{x\to \infty} f(x)$. It follows immediately that $$ f_n(x)\le f\left(\frac{x_0}{n}\right),\ \forall x\in[0, \infty)\quad \text{ i.e. }\quad \lVert f_n\rVert_\infty \le f\left( \frac{x_0}{n}\right),$$ thus $$ \limsup_{n\to \infty} \lVert f_n\rVert_\infty \le f(0)=0, $$ concluding the proof that $\lVert f_n\rVert_\infty \to 0$ as $n\to \infty$. $\Box$

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