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Two norms $F,G$ are equivalent when there are constants $a,b$ such that $aF \le G \le bF$. I'm reading about this idea, and so far I've seen that equivalence of norms implies that the underlying space $X$ has the same topology with respect to either norm. Maybe it preserves even more properties than this too.

But I'm finding it very difficult to use this property when doing proofs or problems because although it's very simple to state, I don't immediately see what it is saying. In comparison, when you define 'equivalence' in other settings, like in the definition of an isomorphism of abelian groups or a continuous map, it's very clear that a certain operation or object is being preserved as you pass across a map.

To be concrete, here are my questions:

(1) Is there another way to characterize when norms are equivalent that might provide more intuition for what it says about is being preserved

and (2) is there a way to show that this definition is the one you want by starting with something more fundamental (like saying that the norms induce the same topology) and then proving that it's equivalent to the stated definition?

Any intuition for the definition would be helpful for either of these questions.

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    $\begingroup$ It says that a convergent sequence say $\{x_n\}$ in the one norm is also convergent in the other norm, but I think you know that. If $\|x_n-x\|_F\rightarrow 0$, then $\|x_n-x\|_G\leq b\|x_n-x\|_F\rightarrow 0$. Maybe you can try to verify that if for each sequence that is convergent in the $F$-norm it follows that it is also convergent in the $G$-norm implies that there is a constant $b$, such that $\|.\|_G\leq b\|.\|_F$ i.e $F$ is stronger than $G$. $\endgroup$ – Svetoslav Jul 31 '15 at 12:12
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    $\begingroup$ In addition to what others have said, my 4 October 2006 sci.math post Lipschitz, uniformly, and topologically equivalent metrics might be of use. $\endgroup$ – Dave L. Renfro Jul 31 '15 at 21:38
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    $\begingroup$ @Freeze_S I will and thanks for doing it I appreciate it. $\endgroup$ – user3281410 Aug 1 '15 at 1:26
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    $\begingroup$ What you said sounds correct, but keep in mind that not every metric defined on a vector space can be a metric induced by some norm defined on that vector space. For instance, every metric induced by a norm is compatible with the algebraic structure in a way that an arbitrary metric on the vector space might not be. But it's certainly true that the metrics induced from your "equivalent norms" wind up being what I called "Lipschitz equivalent metrics". Incidentally, the terminology I used is not (entirely) standard, but it does avoid the ambiguity one often encounters in this topic. $\endgroup$ – Dave L. Renfro Aug 1 '15 at 13:10
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    $\begingroup$ @DaveL.Renfro: Adding: That's all nice result on Lipschitz-continuity but the question is rather what one wants at the end than what would be a technical implementation. Equivalent metrics, as you certainly know, define precisely the same uniform structure while weakly equivalent metrics only the same topology. $\endgroup$ – C-Star-W-Star Aug 1 '15 at 18:38
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Maybe it will be useful to consider an example of two norms $F$ and $G$ of a vector space $X$ not being equivalent to each other. What it means is that at least one of the quantities $\sup\limits_{x \in X}\frac{F(x)}{G(x)}$ or $\sup\limits_{x \in X}\frac{G(x)}{F(x)}$ is unbounded, i.e. there is a sequence $(x_n)_{n \geq 0}$ of vectors in the space such that $\frac{F(x_n)}{G(x_n)}$ or $\frac{G(x_n)}{F(x_n)}$ diverges to $+\infty$ when $n \to \infty$.

Intuitively, if a norm $G$ is equivalent to a norm $F$, $G$ neither stretches too much, nor shortens too much the lengths of the vectors already assigned by $F$. The "new" $G-$length of every vector is within a bound of its "old" $F-$length: $\forall x \in X, \quad aF(x) \leq G(x) \leq bF(x)$. There are not parts of the space stretched or shrunk arbitrarily much.

For an example of two nonequivalent norms, consider the space $C^2([0,1],\mathbb{R})$ of real twice continuously differentiable functions defined on $[0,1]$. Define the two norms $F(x(t)) = \sup\limits_{t \in [0,1]}|x(t)|$ and $G(x(t)) = |x(0)| + \sup\limits_{t \in [0,1]}|x'(t)|$. Now, if we look at the sequence $x_n(t) = \sin(n\pi t)$, then the $F-$norm of every $x_n$ is $1$ while $G(x_n) = n\pi$. The unit $F-$sphere is torn away to infinity when the $G$ norm is used.

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  • $\begingroup$ Thank you for the example. Does one of two in-equivalent norms always diverge on the unit ball of the other? $\endgroup$ – user3281410 Jul 31 '15 at 16:10
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    $\begingroup$ Yes. Once you find a sequence of vectors $(x_n)_{n \geq 0}$ such that $F(x_n)/G(x_n) \to \infty$, say, you can always consider the sequence $(x_n/F(x_n))_{n \geq 0}$ which lies on the unit $F-$sphere and still has the property that $F(x_n/F(x_n))/G(x_n/F(x_n)) = F(x_n)/G(x_n) \to \infty$. $\endgroup$ – Nocturne Jul 31 '15 at 16:15
  • $\begingroup$ That's very helpful. I'm trying to picture this as well as I can. If you don't mind, do you know if it's also true that one norm has to diverge on any ball of the other, not necessarily at the origin? $\endgroup$ – user3281410 Jul 31 '15 at 16:23
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PROBLEM

Answer

Equivalent norms define the same uniform vector space.

Explanation

Completeness is a concept by uniform vector spaces.

Demonstration

Given the real line $\mathbb{R}$.

Consider the metrics: $$d(x,y):=|y-x|$$ $$d'(x,y):=\arctan|y-x|$$

Then one obtains: $$\mathcal{N}=\mathcal{N}'\quad\mathcal{U}\neq\mathcal{U}$$

Concluding problem.


PROOF

Identification

Given normed spaces $\Omega$ and $\Omega'$.

Regard the category: $$\mathrm{UVS}:\quad\mathrm{Hom}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')$$

Identification: $$\Phi:\Omega\leftrightarrow\Omega':x\mapsto x$$

Linearity follows: $$\Phi(x+y)=x+y=x+'y=\Phi(x)+'\Phi(y)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot x=\lambda\cdot'x=\lambda\cdot'\Phi(x)$$

But by the below: $$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_L(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{B}(\Omega,\Omega')$$

Explicitely that is: $$\|x\|'=\|\Phi x\|'\leq\|\Phi\|\cdot\|x\|$$ $$\|x'\|=\|\Phi^{-1}x'\|\leq\|\Phi^{-1}\|\cdot\|x'\|'$$

Concluding proof.


CATEGORIES

Topological Vector Spaces

Note for linear maps: $$\Phi\in\mathcal{L}(\Omega,\Omega')\implies\Phi^{-1}\in\mathcal{L}(\Omega',\Omega')$$

Continuous at zero: $$\mathcal{C}_0(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega:\Phi^{-1}(\mathcal{N}_{\Phi0})\subseteq\mathcal{N}_0\}$$

It holds equality: $$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_0(\Omega,\Omega')$$

Homomorphisms: $$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}(\Omega,\Omega')$$

Isomorphic spaces: $$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\Phi(\mathcal{N})=\mathcal{N}'$$ $$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$

Basic entourages: $$B_N:=\{(x,y):(y-x)\in N\}\subseteq\Omega\times\Omega$$

Uniform structure: $$\mathcal{U}:=\uparrow\{B_N: N\in\mathcal{N}_0\}$$

Going first step up:

Uniform Vector Spaces

Neighborhoods: $$\mathcal{N}_z:=\{U[z]:U\in\mathcal{U}\}$$

Uniform maps: $$\mathcal{C}_U(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':\Phi^{-1}(\mathcal{U}')\subseteq\mathcal{U}\}$$

It holds equality: $$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}(\Omega,\Omega')$$

Homomorphisms: $$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')$$

Isomorphic spaces: $$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\Phi(\mathcal{U})=\mathcal{U}'$$ $$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$

Suppose one finds: $$\text{Locally Convex Base}$$

Induced seminorms: $$\mu_U(x):=\inf\{r\geq0:x\in rU\}$$

Going next step up:

Locally Convex Spaces

Basic entourages: $$B_{\mu\varepsilon}:=\{(x,y):\mu(y-x)<\varepsilon\}\subseteq\Omega\times\Omega$$

Uniform structure: $$\mathcal{U}:=\uparrow\{B_{\mu\varepsilon}:\mu\in\mathcal{S},\varepsilon>0\}$$

Isomorphic spaces: $$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\Phi(\mathcal{S})=\mathcal{S}'$$ $$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$

Suppose one finds: $$\text{Countable Base}$$

Induced Metric: $$d(x,y):=\sum_{k=1}^\infty\frac{1}{2^k}\frac{\sigma_k(y-x)}{1+\sigma(y-x)}$$

Going next step up:

Metrizable Vector Space

Induced seminorm: $$\mu(x):=d(x,0)=d(0,x)\geq0$$

Lipschitz maps: $$\mathcal{C}_L(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':d(\Phi\cdot,\Phi\cdot)'\leq L_\Phi d(\cdot,\cdot)\}$$

It holds equality: $$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_L(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')$$

Isometric maps: $$\mathcal{I}(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':d(\Phi\cdot\Phi\cdot)'=d(\cdot,\cdot)\}$$

Homomorphisms: $$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{I}(\Omega,\Omega')$$

Isomorphic spaces: $$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad d(\Phi\cdot,\Phi\cdot)'=d(\cdot,\cdot)$$ $$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$

By construction: $$d(x+a,y+a)=d(x,y)\quad(a\in\Omega)$$

Suppose one has: $$d(\lambda x,\lambda y)=|\lambda|d(x,y)\quad(\lambda\in\mathbb{C})$$

Induced norm: $$\|x\|:=d(x,0)\geq0$$

Going next step up:

Normed Spaces

Induced metric: $$d(x,y):=\|y-x\|\geq0$$

Lipschitz maps: $$\mathcal{B}(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':\|\Phi\cdot\|'\leq\|\Phi\|\cdot\|\cdot\|\}$$

It holds equality: $$\mathcal{L}(\Omega,\Omega')\cap\mathcal{B}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_L(\Omega,\Omega')$$

Isomorphic spaces: $$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\|\Phi(\cdot)\|'=\|\cdot\|$$ $$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$

Suppose one has: $$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$$

Induced scalar product: $$\langle x,y\rangle:=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|x+i^\alpha y\|$$

Going final step up:

Hilbert Spaces

Induced norm: $$\|x\|^2:=\langle x,x\rangle\geq0$$

Orthogonal maps: $$\mathcal{O}(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':\langle\Phi\cdot,\Phi\cdot\rangle'=\langle\cdot,\cdot\rangle\}$$

It holds equality: $$\mathcal{L}(\Omega,\Omega')\cap\mathcal{O}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{I}(\Omega,\Omega')$$

Homomorphisms: $$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{O}(\Omega,\Omega')$$

Isomorphic spaces: $$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad \langle\Phi\cdot,\Phi\cdot\rangle'=\langle\cdot,\cdot\rangle$$ $$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$ $$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$

Concluding categories.

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Suppose you have a sequence which converges in $G $. The lower bound implies it converges in $F $ to the same limit.

Suppose you have a sequence which ddoes not converge in $G $. The upper bound implies it does not converge in $F $.

That's all you need, since metric spaces are sequential spaces.

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You can start with "the norms induce the same topology". Then use the fact that a linear transformation is continuous if and only if it is bounded. And this is one of your inequalities. For the other direction, use the inverse of that linear transformation.

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  • $\begingroup$ This is very intriguing, but I can't figure out what the linear transformation you're referring to should be. If you view the other norm as a function on the space, isn't it sub-linear and not necessarily linear? $\endgroup$ – user3281410 Jul 31 '15 at 16:20
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    $\begingroup$ The linear transformation is the "identity". From the space with one norm to the space with the other. $\endgroup$ – GEdgar Jul 31 '15 at 16:22
  • $\begingroup$ Ok I understand now. So norms being equivalent is exactly the same as saying that the identity is a homeomorphism. $\endgroup$ – user3281410 Jul 31 '15 at 16:28
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    $\begingroup$ @GEdgar: Probably you know already: In general, boundedness does not imply continuity over a TVS. $\endgroup$ – C-Star-W-Star Jul 31 '15 at 19:07
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    $\begingroup$ I hope user knows that Freeze's comments are about non-norm situations. $\endgroup$ – GEdgar Jul 31 '15 at 20:03

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