Pointwise and uniform convergence of a series of functions

Question

Define a sequence of functions on $[0,\infty)$ such that $\forall n\in\mathbb{N}$, $$ f_n(x)\triangleq \begin{cases} 1 & x\in[n,n+\frac{1}{n}]\\ 0 & \text{otherwise} \end{cases} $$

Does the pointwise limit exist? When $x=0$, this function converges to $0$, but I'm not sure how to tackle the case when $x\in(0,\infty)$. Any tips?

Answer 1

Pointwise convergence definition

$$ (\forall x \in [0,\infty)) (\forall \varepsilon>0) (\exists n_0 \in \mathbb{N}) (\forall n>n_0) \left | f(x) - f_n(x) \right |<\varepsilon $$

But for every $x$ we choose $ n_0 $ so that $ n_0>x $

That means $ \forall n>n_0, \ n>x \implies f_n(x)=0 $

We hence know that $$ \lim_{n\to\infty}f_n(x)=0 $$

Now for the uniform convergence

$$ \lim_{n\to\infty} \sup_{x\in[0,\infty)} \left | f_n(x) \right |= \lim_{n\to\infty} 1 \ne 0 $$

Note that in the upper part we have chosen $ x=n $

We have proven that the function does not converge uniformly on $ [0,\infty) $ but converges pointwise