Pointwise and uniform convergence of a piece wise function

Question

$\forall n \in \mathbb{N}$ let $f_n:[0,1] \rightarrow \mathbb{R}$ be the function defined by

$f_n(x)= 1$ if $0<x \leq 1/n$

$f_n(x) =0 $ otherwise

Study the pointwise convergence and the uniform convergence of the sequence $(f_n)$ in $[0,1]$

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My guess is that it converges to $0$ but I don’t quite know how to show this formally.

About the uniform convergence, I thought on taking the sequence $$a_n = \sup \{|f_n(x)- f(x)|:x \in [0,1]\}$$ where $f(x)$ is the limit function of $f_n(x)$ which in this case would be the $0$ function, and showing that $$\lim (a_n) = 0$$ which would mean that the sequence converges uniformly (the supremum is not actually needed, it suffices with any superior bound $(b_n) \geq (a_n)$ and seeing that $\lim(b_n) = 0$ which would imply that the sequence converges uniformly to the $0$ function on [0,1].

I don’t really know if I am on the right track, if I am, any hints on how to write formally what I have just said would be appreciated. If I am not right then please show me what my mistakes are. Thank you.

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What about $\sum_{n=1}^\infty f_n$ in $[0,1]$? Don’t know how to think of a sum of a function defined piece wisely

Answer 1

There is only one way to show pointwise convergence: find the limit at each point. If $x=0$ then $f_n(x)=0$ for all $n$. If $x\in (0,1]$ then there is $n_0\in\mathbb{N}$ such that $\frac{1}{n_0}<x$. Then for all $n\geq n_0$ we have $f_n(x)=0$. So the limit at every point is zero, hence $f_n\to f$ when $f$ is the zero function.

There is no uniform convergence because for each $n\in\mathbb{N}$ you can take $x_n=\frac{1}{n}$. Then $\sup_{x\in [0,1]}|f_n(x)-f(x)|\geq |f_n(x_n)-f(x_n)|=|1-0|=1$. This is true for all $n$ and this of course implies that the limit of the sequences of supremums cannot be zero.