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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 ...
Markus's user avatar
  • 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 \...
Pierre Dubois's user avatar
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-...
Transcendental's user avatar
3 votes
0 answers
136 views

Unitary operators with the same inner product as vectors

Suppose we have a set of real unit vectors $v_1,\ldots,v_m \in \mathbb{R}^n$. We can always find a set of unitary operators $U_1,\ldots,U_m$ acting on $\mathbb{C}^N$ (for $N$ that is possibly much ...
user3826143's user avatar
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=\...
Manish Kumar's user avatar
2 votes
1 answer
406 views

Mean squared absolute value of inner product of unit vectors

Given a nonempty finite subset $S$ of the unit sphere of $d$-dimensional complex Hilbert space, let $\lambda(S) = \frac{1}{\lvert S \rvert^2} \sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2$ be ...
Alec Edgington's user avatar
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 ...
S-F's user avatar
  • 63
1 vote
1 answer
90 views

A linear algebra question for semi-Euclidean norm

Let us consider Minkowski inner product on $\mathbb R^{1+n}$, defined by $$ \langle v,w \rangle = -v_0w_0+\sum_{j=1}^n v_j w_j\quad \,\forall\, v,w \in \mathbb R^{1+n}.$$ We say that a vector $v$ is ...
Ali's user avatar
  • 4,145
1 vote
1 answer
142 views

An inner product and projection property in RKHS

I'm reading an article regarding the condition for the existence of a solution to a constrained Pick interpolation problem, and the author wrote the following equality - for background, we have the ...
GBA's user avatar
  • 167
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 ...
Saj_Eda's user avatar
  • 395
1 vote
0 answers
94 views

Invariance signature in infinite dimension

Let $V$ be an infinite dimensional vector space and suppose we have a smooth family $\{g_t\}_{t\ge 0}$ of symmetric bi-linear forms such that: $g_0$ is positive-definite $g_t$ is non-degenerate for ...
John117's user avatar
  • 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) ...
epsilon's user avatar
  • 43
1 vote
0 answers
517 views

Complex conjugate and unitary complex conjugate

Definition: Let V be complex finite dimensional inner product space Complex Conjugate: $J$ is called complex conjugate on V iff $(i) J$ is antilinear $(ii) J^2 =I$ Definition: Anti-unitary Complex ...
Sushil's user avatar
  • 121
-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 (...
hookah's user avatar
  • 1,096