# Uniformly convergent on compact interval

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.

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

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.

• 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. Apr 3, 2023 at 18:46
• 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 Apr 3, 2023 at 18:48

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.

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 $$f(x)=xe^{-x}.$$

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

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

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 $$n\to \infty$$. $$\Box$$