Equivalent norms and isometries Let $X$ be a vector space, $\|\cdot\|_1$ and $\|\cdot\|_2$ two equivalent norms on $X$. Under what further assumptions can we prove there is an isometry between $(X,\|\cdot\|_1)$ and $(X,\|\cdot\|_2)$?
In particular, are $(\mathbb{R}^2,\|\cdot\|_1)$ and $(\mathbb{R}^2,\|\cdot\|_2)$ isometric (where the two norms are the maximum norm and the usual euclidean norm respectively)?
 A: These two spaces are not isometric. Unit sphere of the first space contains straight segment while unit sphere of second space doesn't. You can draw them to see this. But property of unit sphere to contain segments preserved under linear isometries, so there is no isometry between these two normed spaces. You can also check similar question asked earlier.
A: Normally, one proves that two normed linear  spaces are isometric in a constructive™ way, by exhibiting an isometry between them. One common exception: since it's widely known that any two separable Hilbert spaces are isometrically isomorphic to each other, an explicit isometry is unnecessary in this case. 
One elementary way to show that two spaces $X,Y$ are not isometrically isomorphic is to find a subset of $X$ that cannot be isometrically embedded into $Y$. For example, consider the subset $$A=\{(\pm 1,0), (0,\pm 1)\}\subset (\mathbb R^2,\|\cdot \|_1)$$
Any two points of $A$ are at distance $2$ from each other. Such a configuration is impossible in $(\mathbb R^2,\|\cdot \|_2)$. Indeed, three points at distance $2$ from one another must lie at the vertices of an equilateral triangle, and there is nowhere to put the fourth one. 
