1
$\begingroup$

We have a sequence of functions $\{ f_n \}$ defined by $$\begin{cases} 0, \ \text{if} \ x<0 \\ \frac{\exp\big\{\frac{-x}{n}\big\}}{n}, \ \text{if} \ x \geq 0 \\ \end{cases}$$ converges uniformly to $0$, and want to prove this.

I've put together what I think is a reasonable proof, but I'm hoping someone could look it over and let me know how it looks.

We consider two separate cases: (i) $x < 0$ and (ii) $x \geq 0$.

(i) If $x < 0$, then rather trivially we have $\lvert f_n - 0 \rvert = \lvert 0 - 0 \rvert = 0 < \epsilon$. This is true for any $n \in \mathbb{N}$, and thus will work for a subset of $n$, as defined by the $N$ that secures uniform convergence for $x \geq 0$.

(ii) If $x \geq 0$, we have $f_n = \frac{exp\big\{\frac{-x}{n}\big\}}{n}$. We have \begin{align*} \lvert f_n - 0 \rvert & = \lvert f_n \rvert \\ & = \Bigg \lvert \frac{\exp\big\{\frac{-x}{n}\big\}}{n} \Bigg \rvert \\ & = \frac{\exp\big\{\frac{-x}{n}\big\}}{n} \\ & = \frac{1}{n \exp\big\{\frac{x}{n}\big\}} \\ & \leq \frac{1}{n} \end{align*} Take $N > \frac{1}{\epsilon}$. Thus, $\lvert f_n - 0 \rvert \leq \frac{1}{n}$ and $\frac{1}{n} < \epsilon$ holds for $n < \frac{1}{\epsilon}$, we get $\lvert f_n - 0 \rvert < \epsilon$ for $n > N$.

I'm having some difficulty explaining the logic of choosing $N$, which really just came from taking $\frac{1}{n} < \epsilon$ and isolating $n$. Since $\frac{1}{n}$ is strictly decreasing, any $n > N$ should guarantee a quantity less than $\epsilon$.

How does this look?

$\endgroup$
1
$\begingroup$

It looks fine.

You proved correctly that$$(\forall x\in\mathbb{R})(\forall n\in\mathbb{N}):\bigl|f_n(x)\bigr|\leqslant\frac1n.$$So, given $\varepsilon>0$, you can take (as you did take) $N>\frac1\varepsilon$. Then$$(\forall n\in\mathbb{N})(\forall x\in\mathbb{R}):n\geqslant N\implies\bigl|f_n(x)-0\bigr|\leqslant\frac1n\leqslant\frac1N<\varepsilon.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy