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A bounded continuous normalized function that is not of positive type

Statement

On the additive topological group G=(R,+), let f(t)=exp⁡(−t4). This function is real-valued, even, continuous, bounded by ∣f(t)∣≤1, and normalized by f(0)=1, but it is not of positive type. In the positive-type matrix for g1=0, g2=12, g3=1, the coefficient vector (1,−2,1) has a negative quadratic form.

Facts & Assumptions

[A1]

Positive type requires the matrix (φ(gi−1gj))i,j to be positive semidefinite for every finite list and every complex coefficient vector (Continuous positive-type functions and normalization).

[A2]

The real exponential is continuous and strictly increasing (The exponential function is strictly increasing).

[A3]

For every real x, exp⁡(x)>0, exp⁡(−x)=1/exp⁡(x), and exp⁡(0)=1 (The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x)).

[A5]

Reciprocation reverses strict inequalities between positive reals (Inverses of positives are positive, and reciprocation reverses order).

[A8]

The square of every nonzero real number is positive (Squares of nonzero elements are positive).

[A9]

Integer powers are defined by finite repeated multiplication (Integer powers am).

[A11]

A topological group has continuous multiplication and inversion (Topological group: multiplication and inversion are continuous).

[A12]

The positive reals are closed under addition (Ordered field).

Proof

Given: The additive real group with its usual topology and the function f(t)=exp⁡(−t4).

Proof technique: direct.

1.1A2A6A7A11algebra

Addition on R is continuous because ∣(x+y)−(x0+y0)∣≤∣x−x0∣+∣y−y0∣, and inversion x↦−x preserves distances; hence the usual additive group satisfies [A11]. The maps t↦t4 and t↦−t4 are polynomials and are continuous by [A6]. Composing with the continuous exponential by [A2] and [A7] proves that f is continuous.

1.2A2A3A8A9

Write t4=(t2)2. If t2=0 then t4=0; otherwise [A8] applied to t2 gives t4>0. Also (−t)4=t4 by finite multiplication, so f is even. By [A3] and strict monotonicity in [A2], 0<f(t)=exp⁡(−t4)≤exp⁡(0)=1 for all t, and f(0)=1. Thus f is real-valued, bounded by 1 in modulus and normalized at the identity.

2.1A1step 1.2algebra

Put a=exp⁡(−1/16) and b=exp⁡(−1). The matrix on the listed points is M=(1aba1aba1), since f is even. For c=(1,−2,1), direct multiplication gives c∗Mc=6−8a+2b.

3.1A1A3A4A5A10A12step 2.1∎

Applying [A4] at x=−1/16 gives a≥15/16, and at x=1 gives 2≤exp⁡(1). By [A10] and [A12], 2=1+1>0. By [A3], b=1/exp⁡(1)>0; if exp⁡(1)=2 then b=1/2, while if 2<exp⁡(1) then [A5] gives b<1/2. Hence b≤1/2, and step 2.1 yields c∗Mc≤6−8(15/16)+2(1/2)=−1/2<0. Thus M is not positive semidefinite by [A1], so f is not of positive type.

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