4
$\begingroup$

Definition. A norm $\|\cdot\|$ in a vector space $X$ is said to be equivalent to a norm $\|\cdot\|_0$ on $X$ if there are positive numbers $a$ and $b$ such that for all $x \in X$ we have

$$ a\| x \|_0 \leq \|x\| \leq b\|x\|_0 $$

My question. If two norms $\|\cdot\|$ and $\|\cdot\|_{0}$ on a vector space $X$ are equivalent, then $\|x_{n} - x\| \rightarrow 0$ if and only if $\|x_n - x\|_0 \rightarrow 0$.

I know that two equivalents norms induce same the topology. How can I use it to prove the sentence.

Ref: (Kreyszig) Introductory Functional Analysis with Applications.

|cite|improve this question
$\endgroup$
  • $\begingroup$ What is the definition of equivalent norms that you are using? If posed in the right way the proof is really short. $\endgroup$ – Umberto P. Jun 14 '16 at 16:58
  • 2
    $\begingroup$ Yes to both questions. For the first one, look up the topological definition of convergence. For the second one, that's the definition of convergence. ($x_n \to x$) $\endgroup$ – Nigel Overmars Jun 14 '16 at 17:00
  • $\begingroup$ @UmbertoP. I put the definition in the question. $\endgroup$ – Jonathan G. Jun 14 '16 at 17:09
  • $\begingroup$ @NigelOvermars How can I use (1) and (2) to answer my question? $\endgroup$ – Jonathan G. Jun 14 '16 at 17:11
  • $\begingroup$ What is your question? (1) and (2) obviously answer your question provided your question is what you say your question is. (1) is precisely the definition of convergence. You are saying $(x_n)$ converges to $x$ in the topology induced by the norm $||\cdot||$ if and only if $(x_n)$ converges to $x$ in the topology induced by the norm $||\cdot||_0$. $\endgroup$ – D1X Jun 14 '16 at 17:18
1
$\begingroup$

You only need to use the definition of convergence for Topological Spaces. Since you are dealing with norms and every norm induces a metric you may use:

For a sequence $(x_n)_{n \in \mathbb{N}} \subset X$, $x_n \to x$ if for every $\varepsilon >0$ there exists $n_{\varepsilon} \in \mathbb{N}$ such that for every $n \geq n_{\varepsilon}$ $x_n \in B(x, \varepsilon)$, where $B(x, \varepsilon)$ represents the open ball centered on $x$ and of radius $\varepsilon$.

If both norms induce the same topology over $X$ then the result follows.

Let's suppose $\mathcal{T}_1 \subset \mathcal{T}_2$. If $x_n \to x$ in $\mathcal{T}_2$ then, as every open ball in $\mathcal{T}_1$ is an open ball in $\mathcal{T}_2$ then $x_n \to x$ in $\mathcal{T}_1$.

Recall that equivalent topologies means $\mathcal{T}_1 \subset \mathcal{T}_2$ and $\mathcal{T}_2 \subset \mathcal{T}_1$ (i.e. $\mathcal{T}_1 =\mathcal{T}_2$).

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.