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A bounded continuous normalized function that is not of positive type
Statement
On the additive topological group , let This function is real-valued, even, continuous, bounded by , and normalized by , but it is not of positive type. In the positive-type matrix for , , , the coefficient vector has a negative quadratic form.
Facts & Assumptions
Positive type requires the matrix to be positive semidefinite for every finite list and every complex coefficient vector (Continuous positive-type functions and normalization).
The real exponential is continuous and strictly increasing (The exponential function is strictly increasing).
For every real , , , and (The exponential is positive and satisfies ).
For every real , ( for every real , hence ).
Reciprocation reverses strict inequalities between positive reals (Inverses of positives are positive, and reciprocation reverses order).
A composite of continuous real functions is continuous (A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).
The square of every nonzero real number is positive (Squares of nonzero elements are positive).
Integer powers are defined by finite repeated multiplication (Integer powers ).
A topological group has continuous multiplication and inversion (Topological group: multiplication and inversion are continuous).
The positive reals are closed under addition (Ordered field).
Proof
Given: The additive real group with its usual topology and the function .
Proof technique: direct.
Addition on is continuous because , and inversion preserves distances; hence the usual additive group satisfies [A11]. The maps and are polynomials and are continuous by [A6]. Composing with the continuous exponential by [A2] and [A7] proves that is continuous.
Write . If then ; otherwise [A8] applied to gives . Also by finite multiplication, so is even. By [A3] and strict monotonicity in [A2], for all , and . Thus is real-valued, bounded by in modulus and normalized at the identity.
Put and . The matrix on the listed points is , since is even. For , direct multiplication gives .
Applying [A4] at gives , and at gives . By [A10] and [A12], . By [A3], ; if then , while if then [A5] gives . Hence , and step 2.1 yields . Thus is not positive semidefinite by [A1], so is not of positive type.
Depends on
- Continuous positive-type functions and normalization
- Topological group: multiplication and inversion are continuous
- Ordered field
- The real exponential function and the number $e$ by a power series
- The exponential function is strictly increasing
- The exponential is positive and satisfies $\exp(-x)=1/\exp(x)$
- $1+x\le\exp(x)$ for every real $x$, hence $(1-p)^m\le\exp(-mp)$
- Inverses of positives are positive, and reciprocation reverses order
- The multiplicative identity is positive
- Integer powers $a^m$
- Squares of nonzero elements are positive
- Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function
- A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs
Used by
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Sources
- Bekka, de la Harpe and Valette, Kazhdan's Property (T), Definition C.4.1 and Proposition C.4.2, Appendix C, printed pp. 373–374 (standard reference, not scraped)
- Emmanuel Kowalski, An Introduction to the Representation Theory of Groups, §3.4 (standard reference, not scraped)