Confused to determine $f_n(x)$ is uniformly convergent or not.

Question

Given the sequence of function $f_n(x)=\exp(-nx)$, for all $x\in\mathbb{R}$ and $n\in\mathbb{N}$. Determine $f_n(x)$ is uniformly convergent or not.

I try as follows.

Take $\varepsilon>0$. We want to show there is $N_\varepsilon\in \mathbb{N}$ such that for all $n\geq N_\varepsilon$ and all $x\in \mathbb{R}$, has $\vert f_n(x)-f(x)\vert<\varepsilon$.

The limit function is \begin{align} \lim\limits_{n\to\infty}f_n(x)= f(x)= \begin{cases} 1&x=0\\ 0&x>0 \end{cases}. \end{align} Let $x=0$. Then $\vert f_n(x)-f(x)\vert=|\exp(-n(0))-1|= |1-1|=0<\varepsilon$.

Let $x>0$. Then $\vert f_n(x)-f(x)\vert=|\exp(-nx)-0|= \exp(-nx)<\varepsilon$.

Now I confused to determine $f_n(x)$ is uniformly convergent or not because I can't find connection between "there is $N_\varepsilon\in \mathbb{N}$ such that for all $n\geq N_\varepsilon$ and all $x\in \mathbb{R}$" and "$\vert f_n(x)-f(x)\vert<\varepsilon$". Anyone can help me?

Answer 1

The convergence is not uniform (in $\Bbb R$), because each $f_n$ is a continuous function, but $(f_n)_{n\in\Bbb N}$ converges pointwise to a discontinuous function. Uniform convergence preserves continuity.

Answer 2

Note that $f_n(1/n)= 1/e= 0,36...$. let $\varepsilon =0,1$. Assume that there exists $n_0, \alpha$ such that if $n>n_0$, $\vert fn(x)-f(x)\vert <\varepsilon $, then $0<f_n(1/n) <\varepsilon$ contradiction.

Answer 3

With regard to your last question, the idea is that for $n>N_\epsilon$ the error must be bounded by $\epsilon$ at each $x$.

You arrived at the expression $\exp(-nx) < \epsilon$. Taking logs on both sides, this is the same as $N_\epsilon > -\log\epsilon/x$. So we see that as $x\to 0$, $N_\epsilon$ is unbounded!

Therefore there cannot be a finite $N_\epsilon$ such that $|f_n-f|<\epsilon$ for each $\epsilon>0$ and each $x$.