Let $X$ be an $n$-dimensional vector space with a basis $\{e_1,\dots,e_n\}$. Consider the norm $\|\sum_{i=1}^n \alpha_ie_i\|=\max_{i\leqslant n} |\alpha_i|$ for $x=\sum_{i=1}^n\alpha_ie_i\in X$.

We say that two normed spaces $(X,\|\cdot\|_X$ and $(Y,\|\cdot\|_Y)$ (over the same field) are isometrically isomorphic if there is a bijective linear map $\psi\colon X\to Y$ such that $\|\psi(x)\|_Y=\|x\|_X$ for all $x\in X$.

We want to prove that $(X,\|\cdot\|)$ is isometrically isomorphic to $(\mathbb R^n, \|\cdot\|_\infty)$, where $\|\cdot\|_\infty$ is defined by $\|\mathbf x\|=\max_{i\leqslant n}|x_i|$ for $\mathbf x=(x_1,\dots,x_n)\in\mathbb R^n.$

Firstly, I think the lecturer forgot to say that $X$ is a vector space over $\mathbb R$, otherwise the result does not hold. Is this correct?

Then, the proof we were given considers the map $\psi(\sum_{i=1}^n\alpha_ie_i)=(\alpha_1,\dots,a_n)$. This is clearly linear, injective and an isometry.

However, I do not see how it is surjective. How can we prove that it is surjective when we are not given its co-domain? Which co-domain would make it surjective?

As you see this is not homework, so a complete answer would be appreciated. Thank you in advance.


The co-domain here is $\mathbb{R}^n$, because we are trying to construct an isomorphism $\psi: X\to \mathbb{R}^n$. To see that the map is surjective, let $(x_1, \dots, x_n)$ be a vector in $\mathbb{R}^n$. Now, $\psi(x_1 e_1 + \dots + x_n e_n) = (x_1, \dots, x_n)$, so every vector in $\mathbb{R}^n$ is in the image of $\psi$.

  • $\begingroup$ You are right. That was silly of me. Thanks! $\endgroup$ – Ryuky Dec 26 '15 at 20:58

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.