5

In a normed vector space $X$, when can we say:

$\lim\|x_n\|=\|\lim x_n\|$

and further, if $f\in X^{*}$, when can we say:

$\lim fx_n=f(\lim x_n)$?

37edude73
  • 111
  • 1
  • 5

2 Answers2

11

You can say that if $\lim x_n= x$, which is a particular case of the following

Theorem: Let $X$ and $Y$ be normed spaces. If $\lim x_n= x$ in $X$ and $T:X\to Y$ is continuous, then $$\lim T(x_n)=T(\lim x_n).$$

Proof: Let $\varepsilon>0$. As $T$ is continuous, there exists $\delta>0$ such that $$\|x-y\|<\delta\quad\Longrightarrow\quad\|T(x)-T(y)\|<\varepsilon.$$ As $\lim x_n=x$, there exists $n_0\in\mathbb{N}$ such that $$n>n_0\quad\Longrightarrow\quad\|x_n-x\|<\delta.$$ It follows that $$n>n_0\quad\Longrightarrow\quad\|T(x_n)-T(x)\|<\varepsilon$$ and thus $$\lim T(x_n)=T(x)=T(\lim x_n).$$

Corollary 1: Let $X$ and $Y$ be normed spaces. If $\lim x_n= x$, then $$\lim \|x_n\|=\|\lim x_n\|.$$

Proof: The mapping $$\begin{align} T:X&\longrightarrow \mathbb{R}\\ x&\longmapsto \|x\| \end{align}$$ is continuous.

Corollary 2: Let $X$ and $Y$ be normed spaces. If $\lim x_n= x$ and $f\in X^*$, then $$\lim f(x_n)= f(\lim x_n).$$

Proof: By definition of $X^*$, $f$ is continuous.

Remark: The theorem above is also true for more general spaces (e.g. metric spaces).

Pedro
  • 19,965
  • 9
  • 70
  • 138
  • Another remark: Any norm which is equivalent to $| \cdot |$ is continuous, too. – Roland Dec 03 '15 at 12:29
  • @Roland Good observation. Just to clarify (what is implicit in the statement and is obvious from the proof, but was not explicitly stated), the limit and the continuity refers to the topology induced by the original norm $|\cdot|$ on $X$. In the context of your comment, I think it is worth bearing this fact in mind. – Pedro Dec 03 '15 at 12:54
  • Linking Related Question: https://math.stackexchange.com/questions/265284/why-are-norms-continuous – texmex Oct 16 '24 at 05:21
0

Assuming that the definition of $\lim_n x_n=x$ is: $\lim_{n}\|x_n-x\|=0$. Yes, this follows from the fact that limits can be exchanged with continuous functions. Specifically, $f(x)=\|x\|$ is continuous in $x$:

$$|f(x)-f(y)|=|\|x\|-\|y\||\leq \|x-y\|,$$

by the reverse triangle inequality. You can show this follows in a similar way for $f\in X^*$ in general.

Alex R.
  • 33,289