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0
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0
answers
320
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Unit sphere of a norm is a submanifold implies the norm is smooth?
Let us call a norm on $\mathbb{R}^n$ smooth if its restriction $\| \cdot \|:\mathbb{R}^n\setminus \{ 0 \} \to \mathbb{R}$ is a smooth map.
Suppose the unit sphere of a norm $\| \cdot \|$ is an embedd …
6
votes
Accepted
Under what conditions a linear automorphism is an isometry of some norm?
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$(1)\iff (4)$ leads to an interesting point: The union of all isometries of all norms equals the union of all isometries of all inner products. … Since all the norms are equivalent This implies that $||A^kv||$ diverges. But since $v=x+iy$ with $x$ and $y$ in $\mathbb{R}^n$, and $A^kv=A^kx+iA^ky$, either $||A^kx||$ or $||A^ky||$ diverge. …
9
votes
1
answer
1k
views
Under what conditions a linear automorphism is an isometry of some norm?
By properties of norms $B$ must be a bounded open set containing the origin. … (That is we allow arbitrary norms, not just those that come from inner products). …
12
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2
answers
955
views
which norms can be realized as operator norms?
Are there any other obstructions for a norm on ${\rm Hom}(V,W)$ to be realized as an operator norm for some suitable norms on $V,W$?
(The main interest is in the case where $\dim V>1,\dim W>1$. …
17
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1
answer
1k
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How many values determine a norm?
Then, if we indeed assume $V$ is a finite-dimensional space, every two norms on $V$ are equivalent, and in particular are continuous w.r.t each other. …