- #1
ericm1234
- 73
- 2
I have read a definition of isomorphism as bijective isometry. I was also showed a definition that isomorphism is a bijective map where both the map and its inverse are bounded (perhaps only for normed spaces??). This does not seem to be the same thing as an isometry.
For example, the poisson problem, from H^1_0 to H^-1 (dual space of H^1_0), is bijective by Lax Milgram, and I can show both maps (the original and the inverse) are bounded. But showing an isometry doesn't seem possible.
For example, the poisson problem, from H^1_0 to H^-1 (dual space of H^1_0), is bijective by Lax Milgram, and I can show both maps (the original and the inverse) are bounded. But showing an isometry doesn't seem possible.