functional analysis - What is the implication that $| cdot |_2$ and $| cdot |_infty$ are equivalent norms on $mathbb{R^2}$

Given $\mathbb{X}$ = $\mathbb{R^2}$, consider $\| \cdot \|_2$ and $\| \cdot \|_\infty$

We can show that

$\| x \|_\infty \leq \| x \|_2 \leq \sqrt2 \| x \|_\infty$

Hence $\| \cdot \|_2$ and $\| \cdot \|_\infty$ are equivalent norms

Is there some deeper implication regarding this particular relationship? Why do we care if two norms are equivalent in this sense?

Answer

As pointed out by Clement C in the comments: Equivalent norms induce the same topology. Also the other direction of implication is true: When two norms induce the same topology then they are equivalent.

Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space and you ask yourself: When are the topologies from those norms the same? This is the case when open sets for $\|\cdot\|_1$ are also open in $\|\cdot\|_2$ and vice versa. This is the case when in each open ball in $\|\cdot\|_1$ contains an open ball of $\|\cdot\|_2$ and the other way around. From this you can prove that there are constants $c$ and $C$ such that $c\|\cdot\|_1 \le \|\cdot\|_2 \le C \|\cdot\|_1$.

So the answer to your question is: Two norms are equivalent iff their induced topologies are the same.

What is the benefit if two topologies are the same? If a "topological property" is valid for one norm, then it is valid for the other norm. For example:

• Open, closed and compact sets are the same.


• If a sequence converges in one norm, it converges also in the other norm.


• If a sequence is a Cauchy sequence in one norm it is Cauchy in the other.


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