# Let $f: \mathscr{B}(\mathbb{N};\mathbb{R})\rightarrow \mathbb{R}$ given by $f(x) = \lim\inf x_n.$ Prove that $f$ is continuous.

Let $f: \mathscr{B}(\mathbb{N};\mathbb{R})\rightarrow \mathbb{R}$ given by $f(x) = \lim\inf x_n.$ Prove that $f$ is continuous.

Here $\mathscr{B}(\mathbb{N};\mathbb{R})$ means the set of all bounded real sequences with the supremum norm (uniform norm).

I think that the best procedure is to show that if $x_n$ is a sequence of functions such that $x_n \rightarrow x$, then $f(x_n) = f(x) = \lim\inf x_n$. But i'm not very familiar with sequence of functions, so i dunno how to proceed very well. Any help would be appreciated.

• What topology do you use on $\mathscr B(\mathbb N;\mathbb R)$? Supremum norm? Oct 13, 2017 at 1:34
• exactly. I will add that Oct 13, 2017 at 1:35

Be careful, an element $x$ is actually a sequence $x = (x_k)$. So by a sequence $(x^n)$ in $B(\mathbb N, \mathbb R)$, it is really a sequence of sequences

$$x^n = (x^n_1, x^n_2,x^n_3,\cdots).$$

Now consider $x^n \to x$ with the supremum norm. Then for all $\epsilon>0$, there is $N$ so that

$$\| x^n - x\|_{\infty} < \epsilon$$

whenever $n\ge N$. That is,

$$|x^n _ k - x_k|<\epsilon$$

for all $k\in \mathbb N$ and $n\ge N$. The above inequality can be written as

$$x^n_k -\epsilon < x_k < x^n_k + \epsilon.$$

Fix $n$ and take liminf with respect to $k$, so

$$\liminf_{k\to \infty} x^n_k -\epsilon < \liminf_{k\to \infty} x_k < \liminf_{k\to \infty} x^n_k + \epsilon.$$

By definition of $f$, this is the same as

$$f(x^n)-\epsilon < f(x)<f(x^n)+\epsilon \Rightarrow |f(x^n)-f(x)|<\epsilon.$$

That is, for all $\epsilon>0$, there is $N$ so that $|f(x^n) - f(x)|<\epsilon$ when $n\ge N$. Thus

$$\lim_{n\to \infty} f(x^n) = f(x)$$

and so $f$ is continuous.

• Thank you for the answer. When i was thinking about this question i have taken the care that you talked about, since i was considering $x_n$ a sequence of functions. I just didn't managed to use a notation for that xD Oct 13, 2017 at 2:01