# Pointwise/Uniform Convergence of Sequence of Functions (and continuity!)

Define a sequence of functions on $\mathbb{R}$ by:

$$f_n(x)=\begin{cases}1, & \text{if x=1, 1\over 2 ,1\over 3,...,1\over n} \\ 0, & \text{otherwise} \end{cases}$$

and let $f$ be the point wise limit of $f_n$. Is each $f_n$ continuous at $0$? Does $f_n \rightarrow$f uniformly on $\mathbb{R}$? Is $f$ continuous at zero?

So basically this is what I know/where I am stuck:

+Having trouble computing $\lim_{n\to \infty}f_n(x)$. Would the limit be a piecewise function $$f(x)=\begin{cases}1, & \text{if 0\lt x\le 1 where x=1, 1\over 2 ,1\over 3,...,1\over n} \\ 0, & \text{otherwise} \end{cases}$$ +I'm supposed to define the sequence on the set of real numbers but does that mean I should consider negative $x$ values, or $x$ values greater than $1$? Clearly when $x\lt 0$ or $x\gt 1$, $f(x)=0$, but I am having trouble determining whether or not $f_n$ is continuous at zero.

+I'm wondering if the Density of Rational in Real property comes into play here, Although $1\over n$ is a small subset of the rational.

• Note: I've seen a lot of examples online where f(x)=1 if 0< x $\le$ $1\over n$, however I more specifically want f(x)=1 if x=$1\over n$. – Carly Aug 8 '15 at 4:52

(1) If $0< x< \frac{1}{n}$, then $$|f_n(x)-f_n(0)|=0$$ Hence $f_n$ is continuous at $0$.

(2) If $f$ is a pointwise limit, then $f(\frac{1}{n})=1$ for all $n\in \mathbb{N}$ and $f(x)=0$ if $x$ is not in $\{ \frac{1}{n} \mid n\in \mathbb{N}\}$

: For large $m\ (>n)$, we have $$| f(\frac{1}{n} ) - f_m (\frac{1}{n}) | < \epsilon$$

(3) Since $|f(0)- f(\frac{1}{n})| =1$ for all $n$, $f$ is not continuous at $0$

(4) Not uniform : ${\rm sup}_{x\in [0,1]} |f-f_n|(x) =1$

• Thank you for your answer! Would you mind specifying what $f$ is exactly? I think that's where I'm getting a little confused. – Carly Aug 8 '15 at 4:56

The limit function is what is called the characteristic function of the set {$1/n : n=1,2,..$} often written as $\chi_{1/n}$, i.e., the function that takes the value $1$ at each number of the form $1/n$ and it is $0$ otherwise.

For continuity of the $f_n$ , we can use a $\delta - \epsilon$ argument: we want to see if we can find, for each $\epsilon >0$ a Real $\delta >0$ so that $|f_n(x)-f_n(y)| < \epsilon$ when $|x-y| < \delta$ . Notice that $f_n(x)$ will be $0$ in the interval $(-\infty,1/n)$. Can you use this to find a value $\delta$ for fixed $\epsilon$? Look at the other answer below to help yourself.

For the continuity of $f=\chi_{1/n}$ , we also can use the $\delta- \epsilon$ approach . Can we find, for any $\epsilon >0$, a $\delta >0$ so that $|\chi_{1/n}(x) - \chi_{1/n}(y) | <\epsilon$ when $|x-y| < \delta$ ? Notice that $\chi_{1/n} (x)$ just takes the values in the set {$0,1$} , but notice that , as $n$ grows, the valus of {$1/n$ } become very close to $0$, and notice what happens when the difference $|\chi_{1/n}(x) - \chi_{1/n}(0)|$ when $n$ becomes very large.

Can you take it from there?