By Walsh J.L.

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**Example text**

H+ Hence ei(x1 ,y1 ) dν B (y1 ) , g(x1 ) = HB+ so that g¯ ∈ F(HB+ ) and we may therefore compute the outer integral in the proposition and we get ∼ i e2 |x1 |2B + g(x1 )dx1 = HB+ e − 2i |y1 |2B + dν B+ (y1 ) HB+ e = −1 − 2i |B+ y1 |2B + dν(y1 ) H+ −1 e− 2 (y1 ,B+ i = y1 ) dν(y1 ) . 32) of dν(y1 ) we get this equal to −1 e− 2 (y1 ,B+ i y1 ) −1 i e 2 (y2 ,B− y2 ) dµ(y1 , y2 ) H+ ×H− e− 2 (y,B i = H −1 y) e− 2 i dµ(y) = D (y,y) dµ(y) . 4 The Fresnel Integral Relative to a Non-singular Quadratic Form 47 which by deﬁnition is the left hand side of the last equality in the proposition.

19) must hold for all x in HB . ) we have ∼ e 2 |x|B f (x)dx = i e− 2 |x|B dµB (x) 2 HB 2 i HB e− 2 |B i = −1 x|2B dµ(x) . 17) proves the identity in the proposition. 2. , for y ∈ D(B), i i e 2 (x+y,B(x+y)) f (x + y)dx = H e 2 (x,Bx) f (x)dx . 1 1 1 The square root |(1/2πi)B| 2 is given by the formula 1 |(1/2πi)B| 2 = (1/2π)n/2 B 1 2 π e−i 4 signB , 1 where (2π)n/2 is the positive root, B 2 is the positive root of the absolute value of the determinant of B and sign B is the signature of the form B.

E. 9) for all x ∈ D, and we shall assume that the form (x, y) is a continuous symmetric bilinear form on D. 10) where D∗ is the dual space of D. e. 9), a continuous linear mapping on D. 9) that ||x||∗ ≤ a|x|, where ||x||∗ is the norm in D∗ . 10) are continuous. In the general case is not uniquely given by B. However if B is non degenerate in the stronger sense that it has a bounded continuous inverse B −1 on H, then it follows easily that D = H, since D contains the range of B, and that (x, y) = (x, B −1 y).