All Questions
8 questions
8
votes
1
answer
446
views
Parallelogram law for vectors of equal length
Does the parallelogram law for vectors of equal length imply the full parallelogram law? That is,
if for all norm one vectors $x$ and $y$ in a Banach space $X$ it holds that $\lVert x-y\rVert^2+\lVert ...
- 1,361
5
votes
2
answers
642
views
Non-standard tensor products of inner product spaces
For two inner product spaces $(\mathcal{V}, (\cdot,\cdot)_V)$ and $(\mathcal{W}, (\cdot,\cdot)_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ \langle v \...
- 643
4
votes
1
answer
384
views
A Hilbert-space completion of a Hilbert $ C^{*} $-module over a separable $ C^{*} $-algebra
Let $ B $ be a separable $ C^{*} $-algebra and $ \mathcal{E} $ a Hilbert $ B $-module. We know that $ B $ has a faithful state $ \phi $. Using $ \phi $, we can construct a $ \mathbb{C} $-valued pre-...
- 1,396
3
votes
0
answers
1k
views
Inner Product on tensor product of Hilbert spaces is unique?
Given two Hilbert Spaces $H$ and $K$, a natural inner product on $H\otimes K$(= vector space tensor product of $H$ and $K$) is given by
$\hspace{.5in}\langle h_1\otimes k_1,h_2\otimes k_2\rangle=\...
- 165
2
votes
0
answers
318
views
What are alternative or equivalent definitions of a positive-definite function on a group?
The standard definition of a positive-definite function on a group goes as follows:
Let $\varphi : G \rightarrow L(H)$, where $G$ is a group (with an involution) and $H$ a Hilbert space. $L(H)$ is the ...
- 63
1
vote
1
answer
273
views
Adjoint of an operator-valued linear operator
I have come across a linear bounded operator $B:K\to \mathcal{L}(U,Z)$ where $K$, $U$, and $Z$ are separable Hilbert spaces. I need a reference (any source) to find out about:
The adjoint of such an ...
- 395
1
vote
0
answers
83
views
What is lost after RKHS embedding of the L1 space?
We know that every distribution or $L^1$ function $f$ over space $\mathcal{X}$ (e.g., $R^d$) can be embedded to an RKHS $\mathcal{H}$ with a $1$-bounded kernel $\mathcal{K}$ (e.g., the RBF kernel) ...
- 43
-1
votes
1
answer
323
views
Expressing the sum of two squared inner products more compactly: is it possible to lift the dimension? [closed]
Let $v_1,v_2\in\mathbb{R}^d$ be two fixed vectors, and $\langle \cdot,\cdot\rangle_{\mathbb{R}^d}$ be the usual Euclidean inner product in $\mathbb{R}^d$.
My question is as follows. Is there an (...
- 1,096