For finite dimension vector spaces, all norms are equivalent

[ShayanJ]

I searched for a proof of the statement in the title and found this document. But it just proves that for two norms $\rho(x)$ and $||x||$, we have $m\rho(x)\leq ||x|| \leq M \rho(x)$ for some m and M. But how does it imply that the two norms are equivalent?

Thanks

 

[fresh_42]

What do you mean by equivalence of norms? $mρ(x)≤||x||≤Mρ(x)$ implies $M^{-1} ||x|| \leq ρ(x) \leq m^{-1} ||x||$.

 

[ShayanJ]

I assume it means all of them give the same number for the same vector.

 

[Stephen Tashi]

But how does it imply that the two norms are equivalent?

I notice the current Wikipedia article https://en.wikipedia.org/wiki/Norm_(mathematics) simply defines "equivalent" to mean the existence of that bound.

To me, it would be nicer to define "equivalent" to mean something like "A sequence converges in one of the norms if and only if it converges in the other norm. It that an obvious consequence of the bound ?

 

[fresh_42]

I assume it means all of them give the same number for the same vector.

This cannot be since e.g. $||(v_1,v_2)||_2 = \sqrt{v_1^2+v_2^2} \neq \max\{|v_1|,|v_2|\} = ||(v_1,v_2)||_\infty$.
It only means the two (four) relations above, i.e. it is qualitatively the same, not quantitatively.

 

[ShayanJ]

It that an obvious consequence of the bound ?

Yeah, It makes sense to me.
But isn't there a standard definition of equivalent norms?

It only means the two (four) relations above, i.e. it is qualitative the same, not quantitative.

What do you mean by qualitatively the same?

 

[fresh_42]

What do you mean by qualitatively the same?

What Stephen has said. Switching between equivalent norms doesn't change the general behavior of convergence, boundedness and so on, it only changes numbers: the quantity, not the quality.

 

[S.G. Janssens]

I notice the current Wikipedia article https://en.wikipedia.org/wiki/Norm_(mathematics) simply defines "equivalent" to mean the existence of that bound.

Indeed. Two norms are equivalent if, by definition, the estimate given in the OP holds.

As you can easily show, this implies that for every $x \in X$ every open $\rho$-ball centered at $x$ contains an open $\|\cdot\|$-ball centered at $x$ and vice versa. Since these balls form bases for the norm topologies generated by $\rho$ and $\|\cdot\|$, respectively, we conclude that the two norm topologies are identical.

In this sense equivalent norms are "qualitatively the same".