Let $f_n = (\frac{1}{n})\chi_{[0,n]}$ and $f = 0$. Show that $(f_n)$ converges uniformly to $f$.

Question

Let $f_n = (\frac{1}{n})\chi_{[0,n]}$ and $f = 0$. Show that $(f_n)$ converges uniformly to $f$.

I have never done an example of convergence of sequences that have characteristic (indicator) functions and honestly have no idea how to work with them in limits. Can anyone please explain to me how to do this?

Answer 1

We have that $g_n$ converges uniformly to $g$ if and only if $$\sup\limits_{x\in\mathbb{R}}|g_n(x) - g(x)|\to 0$$ as $n\to \infty$. In your case we have that

$$\sup\limits_{x\in\mathbb{R}}\left|\left(\frac1n\right) \chi_{[0,n]}(x) - 0\right| = \frac1n\to 0$$ as desired.

Answer 2

Definition:$\forall \varepsilon>0 \ \exists N: \forall x \in \mathbb{R}, n\geq N: |f_n(x) - f(x)|<\varepsilon $.

So lets take $N: = [\frac{1}{\varepsilon}]+1$ and we have

$\forall \varepsilon>0 \ \exists N=[\frac{1}{\varepsilon}]+1: \forall x \in \mathbb{R}, n\geq N: |\frac{1}{n}\chi_{[0;n]} - 0|<\frac{1}{n} < \varepsilon $

as indicator is either $\frac{1}{n}$ or $0$.