Why is $\lim_{n\to \infty} n\chi_{(0,\frac{1}{n})} = 0$?

Question

Why is $\lim_{n\to \infty} n\chi_{(0,\frac{1}{n})} = 0$? Can you provide a short (trivial) explanation?

The sequence in the limit only have two possible values: $0$ and $n (\to \infty)$ so it is a little confusing.

Answer 1

Let $x \leq 0$. Then, $n\chi_{(0,1/n)}(x) = n \cdot 0 = 0$ for each $n \in \mathbb{N}$ as $x \not\in (0,1/n)$ for all $n \in \mathbb{N}$.

Let $x > 0$. There exists some $n_{0} \in \mathbb{N}$, such that $0 < 1/n_{0} < x$. For all $n \geq n_{0}$, we have that $n\chi_{(0,1/n)}(x) = n \cdot 0 = 0$. This is because $x \not\in (0,1/n)$ for each $n \geq n_{0}$.

So for all $x \in \mathbb{R}$ and for all $\varepsilon > 0$, there exists an $n_{0} \in \mathbb{N}$ such that for all $n \geq n_{0}$, we have that $n\chi_{(0,1/n)}(x) = 0 < \varepsilon$. Therefore, $\lim_{n\to\infty}n\chi_{(0,1/n)}(x) = 0$ pointwise.

Answer 2

conan gave a formal proof. Here's a more informal way to understand why this is true. Set $f_n(x) = n \chi_{(0, 1/n)}$.

Pointwise convergence to 0 means that, for any fixed $x$, the sequence $f_1(x), f_2(x), f_3(x), \dots$ ought to be a sequence which converges to 0. Let's try this for some specific values of $x$.

• When $x=0.1$, the sequence looks like $1,2,3,4,5,6,7,8,9,0,0,0,0,\dots$. This converges to 0, of course.

• When $x = 0.01$, the sequence looks like $1,2,3,\dots, 98, 99, 0, 0, 0, 0, \dots$. This also converges to $0$.

• When $x = 0.001$... you get the idea.

• When $x=0$, the sequence looks like $0,0,0,0,0,\dots$, which also converges to $0$.

So you can see that no matter what $x$ is, we get a sequence that converges to $0$.

It's true that, the smaller $x$ is, the slower the resulting sequence converges to 0. But this doesn't matter when we are talking about pointwise convergence. For uniform convergence, it does, and so this sequence of functions does not converge uniformly.