Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
1
vote
Adjoint of an operator-valued linear operator
For fixed $u \in U$, the map that takes $(z,k)$ to $\langle z,(Bk)u\rangle_Z$ is a bounded sesquilinear form on $Z\times K$. (Bounded because $|\langle z,(Bk)u\rangle| \leq \|z\|\|(Bk)u\| \leq \|B\|\| …
6
votes
Accepted
Non-standard tensor products of inner product spaces
Assuming complex scalars, no, any inner product which satisfies (2) for all $v \in \mathcal{V}$ and $w \in \mathcal{W}$ has the given form. To see this, let $[\cdot,\cdot]$ be any inner product which …
5
votes
Accepted
An inner product on the vector space $\mathbb{R}[x_1,\cdots,x_n]_m$
It looks like you have rediscovered the Bargmann-Segal space! Take $n = 1$ for simplicity. Define $BS$ to be the set of analytic functions in $L^2(\mathbb{C},\mu)$ where $\mu$ is $\frac{1}{\pi}e^{-|z| …