Search Results

. $(1)\iff (4)$ 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. …

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 $\dim V>1,\dim W>1$. …

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). …

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. …

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 …