# Pointwise and uniform convergence of a series of functions

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?

• Have you tried to plot $f_n$ for large consecutive $n$'s ? Apr 7, 2016 at 6:46
• Consider for example what happens when $x=4.1$ Apr 7, 2016 at 6:49
• In that case, the function only converges to 1 when $n=4$, and $0$ at any other value. It seems that this function only converges to 1 when $n \leq x \leq n+1/n$. Apr 7, 2016 at 6:54
• But you have that $n \to \infty$, while $x$ is fixed. Now, $f_n(x)=0$ for all but one index, so that $\lim_{n \to \infty} f_n(x)=0$. For example, the sequence $f_n(4.1)$ goes like this: $0,0,0,1$ (yey!) $, 0,0,0,0...$ and then it is always zero. So what is the limit in your opinion? Apr 7, 2016 at 7:03
• Choose some $x\geq 0$, there exists some $n\in \mathbb{N}$ such that $x<n$. For all $m\geq n$, we have that $x<m$, hence $f_m(x)=0$. This more or less solves the pointwise limit. As for uniform convergence, for each $n$, $f_n$ is zero everywhere except that it's one on a small interval. This insight should suffice to prove that there is no uniform convergence. Apr 7, 2016 at 7:15

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