# $\lim_{n\to\infty}f_n(s)=0$ , but $\lim_{n\to\infty}ρ(f_n,z)≠0$

Let $S = (0,1)$, and let $X$ be the set of bounded, real-valued functions on $S$. Define $ρ\colon X\times X \rightarrow \mathbb{R}$ by

$$ρ(f,g)=\sup\{|f(s)-g(s)|:s\in S\}$$

Give a sequence $(f_n)_{n\in\mathbb{N}}$ in $X$ such that $\lim_{n\to\infty}f_n(s)=0$ for all $s\in S$, but $\lim_{n\to\infty}ρ(f_n,z)≠0$, where $z$ denotes the function mapping every $s\in S$ to the real number zero.

I would like to know if $f_n(s)=\frac{s}{n}$ is ok or not? If not, in what way i can think of sequence like this. I just thought there should be an $n$ in the denominator.

Thanks

• Well, $\rho(f_n,z)=1/n$ – Siminore Mar 4 '13 at 16:00

You are looking for a sequence $(f_n)_{n \in \mathbb{N}}$ such that $f_n$ converges pointwise to $0$, but not uniformly. This is not fulfilled for $f_n(s) := \frac{s}{n}$:

$$\varrho(f_n,z) = \sup \{|f_n(s)-0|; s \in (0,1)\} = \frac{1}{n}$$

and therefore $$\lim_{n \to \infty} \varrho(f_n,z)=0$$ i.e. $f_n$ converges uniformly to $0$ on $S=(0,1)$.

The most popular example for a (bounded) sequence converging pointwise but not uniformly on $(0,1)$ is $$f_n(s) := s^n$$ Then $f_n(s) \to 0$ for all $s \in (0,1)$, but $$\varrho(f_n,z)=\sup\{|s^n|; s \in (0,1)\}=1$$ does not converge to $0$ as $n \to \infty$.

What if $f_n(s)=s^n$? It is easy to check that $\sup_{s \in S} s^n =1^n=1$ for each $n$.

The sequence $\phi_n(x) = \frac{x}{n}$ is not OK, as $|\phi_n(x)| < \frac{1}{n}$ for all $x \in S$, hence $\lim_n \rho(\phi_n,0) = 0$.

However, it is easy to construct a sequence with the desired properties:

Let $f_n(x) = 1_{\{\frac{1}{n}\}}(x)$. Then $\lim_n f_n(x) = 0$ for all $x \in S$. However, since $f_n(\frac{1}{n}) = 1$, we see that $\rho(f_n,0) = 1$ for all $n$, hence $\lim_n \rho(f_n,0) = 1$.

• sorry, may i ask what does this $f_n(x) = 1_{\{\frac{1}{n}\}}(x)$ mean ? – Kayne Mar 4 '13 at 23:20
• The notation is the indicator function of a set. So, $1_A(x)$ is one if $x \in A$, and zero if $x \notin A$. In this case, $f_n(x) = 1$ if $x = \frac{1}{n}$, and zero otherwise. – copper.hat Mar 5 '13 at 6:07