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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\|\| …
Nik Weaver's user avatar
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6 votes
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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 …
Nik Weaver's user avatar
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5 votes
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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| …
Nik Weaver's user avatar
  • 42.8k