About a theorem of Wiener on the Bessel-Kingman Hypergroup
AAbout a theorem of Wiener on the
Bessel-Kingman Hypergroup
Lukas Innig [email protected]
March 25, 2019
A theorem of Wiener on the circle group was strengthened and extended byFournier in [2] to locally compact abelian groups and extended further to theBessel-Kingman hypergroup with parameter α = / by Bloom/Fournier/Leinertin [1]. We further extend this theorem to Bessel-Kingman hypergroups withparameter α > / . In this paper we will prove a theorem of
Fournier ( 4The Bessel-Kingman Hypergroup-pro.4) on the Bessel-Kingman Hypergroup with parameter α ≥ . We will use the proofin [1] for the case α = as a guideline, which will be altered where necessary. 4TheBessel-Kingman Hypergrouppro.4 is based upon a theorem of Wiener but treats a moregeneral case. Following [3] and [1] we define:
Definition 1:
The Bessel-Kingman Hypergroup with parameter α is defined as K = ( R + , ˚ α ), where ε x ˚ α ε y ( f ) = Z x + y | x − y | K α ( x, y, z ) f ( z ) z α +1 dz with K α ( x, y, z ) = Γ( α + 1)Γ( / )Γ( α + / )2 α − (cid:2)(cid:0) z − ( x − y ) (cid:1) (cid:0) ( x + y ) − z (cid:1)(cid:3) α − / ( xyz ) α and identity involution ( x − = x ). The Haar measure ω α ( dz ) is of the form ω α ( dz ) = z α +1 dz . The characters are given by χ λ ( x ) := j α ( λx ) , x ∈ R + where j α denotes themodified Bessel function of order α : j α ( x ) := ∞ X k =0 ( − k Γ( α + 1)2 k k !Γ( α + k + 1) x k , ∀ x ∈ R . a r X i v : . [ m a t h . C A ] F e b ne has χ ≡
1. Furthermore K is a Pontryagin hypergroup. In fact K ∼ = K ∧ , wherethe isomorphism is given by λ χ λ . We note that ( R + , ˚ α ) is commutative because K ( x, y, z ) = K ( y, x, z ) and R x + y | x − y | · · · = R y + x | y − x | · · · .As a convention, we denote • I n := [ n − , n ), • C Γ := Γ( α +1)Γ( / )Γ( α + / )2 α − , • ω n := ω α ( I n ).Furthermore let α ≥ / . The case − / < α < / will not be treated. Definition 2:
For a hypergroup ( R + , ˚ ) with Haar measure ω , the discrete amalgam norm is given by k f k p,q := ∞ X n =1 ω n (cid:18) ω n Z I n | f | p dω (cid:19) q / p ! / q . In the case p or q equal to ∞ we set by convention k f k ∞ ,q := ∞ X n =1 ω n sup x ∈ I n | f ( x ) | q ! / q , k f k p, ∞ := sup n ∈ N (cid:18) ω n Z I n | f | p dω (cid:19) / p and k f k ∞ , ∞ := sup n ∈ N sup x ∈ I n | f ( x ) | ! = k f k ∞ . The function spaces { f measurable | k f k p,q < ∞} will be denoted as ( L p , ‘ q )( R + , ˚ ). Forthese spaces the following properties hold: k f k p ,q ≤ k f k p ,q , if p ≤ p . k f k p,q ≤ C k f k p,q , if q ≥ q . particularly, it holds for p ≤ p and q ≥ q ( L p , ‘ q )( R + , ˚ ) ⊂ ( L p , ‘ q )( R + , ˚ )and ( L p , ‘ q )( R + , ˚ ) ⊂ L p ( R + , ˚ ) ∩ L q ( R + , ˚ ) for p ≥ q,L p ( R + , ˚ ) ∪ L q ( R + , ˚ ) ⊂ ( L p , ‘ q )( R + , ˚ ) for p ≤ q. efinition 3: Because we are situated on a hypergroup, we can also form amalgam spaces by shiftingthe unit interval I using the left-translation τ y defined as τ y f ( x ) = f ( y ˚ α x ) = Z x + y | x − y | K α ( x, y, z ) f ( z ) z α +1 dz . For the Bessel-Kingman hypergroup ( R + , ˚ α ) the continuous ( p, ∞ ) -amalgam norm isgiven by sup y ∈ R + (cid:18)Z | f | p τ y [0 , dω (cid:19) / p . Now we are able to state our more general version of the theorem of Fournier [2,Theorem 3.1]. The proof of this theorem is the main part of this paper. As said beforeit is closely related to a theorem of Wiener which implies that an integrable functionwith non-negative Fourier transform on the unit circle which is square integrable on aunit neighborhood is already square integrable on the whole circle. In [2] this theoremwas extended to R and to LCA-Groups. Theorem 4 (Theorem of Fournier):
For f ∈ L ( R + , ˚ α ) with ˆ f ≥ the following statements are equivalent:i) f is square integrable on a neighborhood of .ii) ˆ f ∈ ( L , ‘ )( R + , ˚ α ) .iii) f ∈ ( L , ‘ ∞ )( R + , ˚ α ) . To prove this we follow [1]. So we have to check the following properties of ( R + , ˚ α ) with α ≥ / . Equivalence of the discrete and the continuous amalgam norms.
We show thatthe norms defined in 2The Bessel-Kingman Hypergrouppro.2 and 3The Bessel-KingmanHypergrouppro.3 are equivalent.
Uniform boundedness of the translation operator on ( L p , ‘ q )( R + , ˚ α ) . We showthat k τ y f k p,q ≤ C · k f k p,q , ∀ y ∈ R + with a constant C independent of y . For this we will first prove the uniform boundednesson ( L ∞ , ‘ ). Then by invoking duality and interpolation arguments we get the generalcase. 3 roperties of the convolution. For the convolution a generalized version of theYoung inequality on amalgams must hold:
Theorem 5 (Young inequality on amalgams):
For f ∈ ( L p , ‘ q ) and f ∈ ( L p , ‘ q ) , where (cid:18) p , q (cid:19) = (cid:18) p , q (cid:19) + (cid:18) p , q (cid:19) − (1 , we have f ˚ f ∈ ( L p , ‘ q ) and k f ˚ f k p,q ≤ C · k f k p ,q · k f k p ,q . Properties of the Fourier transformation.
For the Fourier transform of a function f we need a more generalized form of the Hausdorff-Young theorem: Theorem 6 (Hausdorff-Young theorem on amalgams):
For f ∈ ( L p , ‘ q )( R + , ˚ α ) with ≤ p, q ≤ it holds, that ˆ f ∈ ( L q , ‘ p )( R + , ˚ α ) . Here too, we will consider the special cases ( p, q ) ∈ { (1 , , (1 , , (2 , , (2 , } . Theresult then follows with interpolation arguments. Proposition 7:
We have k f k p, ∞ ≤ C sup y ∈ R + (cid:18)Z | f | p τ y [0 , dω α (cid:19) / p . (1.20) Proof:
Let us concentrate on the term τ y [0 , : τ y [0 , ( x ) = C Γ ( xy ) α Z x + y | x − y | [0 , ( z ) z h(cid:16) z − ( x − y ) (cid:17) (cid:16) ( x + y ) − z (cid:17)i α − / dz. Like in [1], it will be sufficient to look at the supremum in (1.20) taken only over all y of the form y = n + / , ≤ n ∈ Z . We therefore consider τ n + / [0 , ( x ) for x ∈ I n +1 .The domain of the integration is[ | x − y | , x + y ] ∩ [0 ,
1] = [ | x − n − / | , x + n + / ] ∩ [0 , | x − n − / | , ⊃ [ / ,
1] for x ∈ I n +1 .
4y using x α = x · ( x ) α − / together with | x − n − / | ≤ / we get C Γ ( n + / ) α x α Z | x − n − / | z h(cid:16) z − ( x − n − / ) (cid:17) (cid:16) ( x + n + / ) − z (cid:17)i α − / dz ≥ C Γ ( n + / ) α x Z / z " (cid:18) z − (cid:19) ( x + 2 x ( n + / ) + ( n + / ) − z x !| {z } =( ∗ ) α − / dz. ( ∗ ) = 1 + 2( n + / ) x + ( n + / ) − z x and 1 x are decreasing in x .Therefore it follows that τ n + / [0 , ( x ) ≥ τ n + / [0 , ( n + 1) ∀ x ∈ I n +1 . Further wewill show that τ n + / [0 , ( n + 1) ≥ C · n α +1 .τ n + / [0 , ( n + 1)= C Γ (( n + 1)( n + / )) α · Z / z "(cid:18) z − (cid:19) (cid:18) n + 32 (cid:19) − z ! α − / dz (sort by powers of n )= C Γ (( n + 1)( n + / )) α · Z / z (cid:20) n (4 z −
1) + n (cid:18) z − (cid:19) + (cid:18) z − z − (cid:19)(cid:21) α − / dz (place n outside the brackets)= C Γ · n α − (( n + 1)( n + / )) α · Z / z (cid:20) (4 z −
1) + n − (cid:18) z − (cid:19) + n − (cid:18) z − z − (cid:19)(cid:21) α − / dz ≥ C Γ · n α − (( n + 1)( n + / )) α Z / z (4 z − α − / dz | {z } constant = C · n α − (( n + 1)( n + / )) α ≥ C · n α − (( n + n )( n + n )) α = C · n α − (4 n ) α = C · n α +1 . Above we used that both f ( z ) := (cid:16) z − (cid:17) and f ( z ) := (cid:16) z − z − (cid:17) are greaterthan 0 on the interval [ / , .5 1.0 1.5 2.0 z (cid:45) (cid:45) (cid:45) f (cid:72) z (cid:76) Figure 1: f ( z ) f ( z ) ≥ ⇔ z − ≥ ⇔ z ≥ ⇔ | z | ≥ f ( z ) ≥ ⇔ z − z − ≥ ⇔ − (cid:18) z − (cid:19) + 1 ≥ ⇔ ≥ | z − |⇔ ≥ | z | ≥ . Further we have ω n +1 = Z n +1 n z α +1 dz ≥ n α +1 . So τ n + / [0 , ( n + 1) ≥ C · n α +1 ≥ C · ω n +1 for n ≥ y ∈ [1 , ∞ ) (cid:18)Z | f | p τ y [0 , dω α (cid:19) / p ≥ sup n ≥ Z n + / n − / | f | p τ n + / [0 , dω α ! / p ≥ sup n ≥ Z I n +1 | f | p τ n + / [0 , dω α ! / p ≥ sup n ≥ τ n + / [0 , ( n + 1) Z I n +1 | f | p dω α ! / p ≥ sup n ≥ C ω n +1 Z I n +1 | f | p dω α ! / p , x, y ∈ I it obviously holds that sup y ∈ I τ y [0 , ( x ) ≥ τ [0 , ( x ) = 1. Sosup y ∈ R + (cid:18)Z | f | p τ y [0 , dω α (cid:19) / p ≥ C · k f k p, ∞ . (cid:3) Proposition 8:
It holds that k f k p, ∞ ≥ C sup y ∈ R + (cid:18)Z | f | τ y [0 , dω α (cid:19) / p . Proof: i) Let y ∈ [0 , τ y [0 , ( x ) = 1 , if x + y ≤ , ≤ , if | x − y | ≤ x + y ≥ , = 0 , if | x − y | ≥ . Note that τ y [0 , ( x ) = [0 , ( x ˚ y ) = ε x ˚ ε y ( [0 , ). Therefore if supp( ε x ˚ ε y ) =[ | x − y | , x + y ] ⊂ [0 , ε x ˚ ε y ( [0 , ) = 1 because ε x ˚ ε y ∈ M ( K ).It follows now that τ y [0 , ≤ [0 , and therefore Z | f | p τ y [0 , dω α ≤ Z | f | p [0 , dω α + Z | f | p [1 , dω α ≤ Z ω [0 , | f | p dω α + 2 α +1 Z ω [1 , | f | p dω α because ω n = R nn − z α +1 dz ≤ n α +1 , hence 1 ≤ n α +1 ω n .So (cid:18)Z | f | p τ y [0 , dω α (cid:19) / p ≤ (cid:18)Z I ω | f | p dω α (cid:19) / p + 2 α +1 p (cid:18)Z I ω | f | p dω α (cid:19) / p ≤ (1 + 2 α +1 p ) k f k p, ∞ . .ii) If y ∈ [1 ,
2) then it follows, analogously to the calculation above, that τ y [0 , ≤ [0 , . Z | f | p τ y [0 , dω α ≤ Z | f | p [0 , dω α + Z | f | p [1 , dω α + Z | f | p [2 , dω α ≤ Z ω [0 , | f | p dω α + 2 α +1 Z ω [1 , | f | p dω α + 3 α +1 Z ω [2 , | f | p dω α , (cid:18)Z | f | p τ y [0 , dω α (cid:19) / p ≤ (cid:18)Z I ω | f | p dω α (cid:19) / p + 2 α +1 p (cid:18)Z I ω | f | p dω α (cid:19) / p + 3 α +1 p (cid:18)Z I ω | f | p dω α (cid:19) / p ≤ (1 + 2 α +1 p + 3 α +1 p ) k f k p, ∞ . iii) Let k ≥
2. We consider y ∈ I k +1 = [ k, k + 1] ⇒ supp τ y [0 , = [ y − , y + 1] ⊂ [ k − , k + 2].Now it follows τ y [0 , ( x ) = C Γ x α y α Z | x − y | z h ( z − ( x − y ) )(( x + y ) − z ) i α − / dz = C Γ xy Z | x − y | z h ( z − ( x − y ) ) | {z } ≤ ( x + y ) − z x y !| {z } ≤ ( x + y )2 − ( x − y )2 x y = xy i α − / dz ≤ C Γ xy Z | x − y | z (cid:18) xy (cid:19) α − / dz ≤ C Γ · α − / ( xy ) α + / . The last inequality is due to 1 − | x − y | ≤ k − ≤ x < k + 2 and k ≤ y < k + 1, we get ≤ C Γ α − / ( k − α + / k α + / ≤ C Γ α − / (cid:16) kk − (cid:17) α + / k α + / k α + / ≤ C Γ α − / α + / k α + / k α + / ≤ C Γ α − / α + / k α +1 = C · k α +1 . We note that ω k +2 = Z k +2 k +1 z α +1 dz ≤ ( k + 2) α +1 ≤ α +1 k α +1 . Hence we get τ y [0 , ( x ) ≤ C ω k +2 ≤ C ω k +1 ≤ C ω k for y ∈ I k , k ≥ . Z I j | f | p τ y [0 , dω α ≤ C · Z I j ω j | f | p dω α ( j = k, k + 1 , k + 2) ⇒ (cid:18)Z | f | p τ y [0 , dω α (cid:19) / p ≤ C / p · k +2 X j = k Z I j ω j | f | p dω α ! / p ≤ · C / p · k f k p, ∞ (cid:3) Choosing C as the maximum of the constants in i) - iii) concludes the proof. ( L p , ‘ q )( R + , ˚ α ) Proposition 9 (The case ( L ∞ , ‘ )( R + , ˚ α )): For f ∈ ( L ∞ , ‘ )( R , ˚ α ) and y ∈ R + it holds that k τ y f k ∞ , ≤ C · k f k ∞ , with a constant C independent of y . Proof:
Like in [1] it is enough to consider only functions f n := I n = [ n − ,n ) . Thatmeans we show only that k τ y f n k ∞ , ≤ C · k f n k ∞ , = C · ω n with C independent of n and y . For the sake of completeness, we repeat the proof forthe correctness of our constraint.Let c n := k P n f k ∞ ( P n denotes the restriction to the interval I n ) and let g = P n c n f n . k τ y f k ∞ , ≤ k τ y g k ∞ , ≤ X n c n k τ y f n k ∞ , ≤ X n c n C k f n k ∞ , = C k f k ∞ , . Fix y and n . We denote a k ∈ Z + as exceptional if k = 1 or if there exists x ∈ I k so that | x − y | or x + y lies in I n . The set of all exceptional indices will be denoted as E . Anindex k ∈ Z + which is not exceptional will be called generic . G := Z + \ E .For k ∈ G the intersection of [ | x − y | , x + y ] and I n is either empty or all of I n for all x ∈ I k . Then τ y f n either vanishes on all of I k or is in the form of τ y f n ( x ) = C Γ ( xy ) α Z nn − z h(cid:16) z − ( x − y ) (cid:17) (cid:16) ( x + y ) − z (cid:17)i α − / dz, for x ∈ I k .
9o it is easy to see that we can claim the following statements about x ∈ { τ y f n > } if x ∈ I k and k generic. | x − y | < n − < n ≤ x + y ⇒ x − y < n − y − x < n − n ≤ x + y ⇒ x < n + y − y − n + 1 < x and n − y ≤ x. Taken together it holds that supp τ y f n ∩ ( S k ∈ G I k ) ⊂ [max { , y − n + 1 , n − y } , n + y − = 0) of the lower bound yields y = n − / . Thus we candistinguish two cases:i) y ≤ n − / ⇒ n − y ≤ x < n + y − y ≥ n − / ⇒ y − n + 1 < x < n + y − ω k ≤ k α +1 and x α ≤ k − α ≤ ( kk − ) α k α ≤ C · k α we get for x ∈ I k , k ≥ ω n X k ∈ G ω k k P k ( τ y f n ) k ∞ = C Γ · X k ∈ G ω k ω n sup x ∈ I k Z nn − z ( xy ) α h(cid:16) z − ( x − y ) (cid:17) (cid:16) ( x + y ) − z (cid:17)i α − / dz ≤ C · X k ∈ G k α +1 ω n y α k α sup x ∈ I k Z nn − z h(cid:16) z − ( x − y ) (cid:17) (cid:16) ( x + y ) − z (cid:17)i α − / dz ≤ C · X k ∈ G kω n y α sup x ∈ I k n Z n − z (cid:2) ( z − x + y )( z + x − y )( x + y − z )( x + y + z ) | {z } =: (cid:126) (cid:3) α − / dz. ( ∗ )We consider n and y as in the first case and substitute, according to the sign, z by n resp. n − x by n − y resp. n + y −
1. It follows that( ∗ ) ≤ C · X k ∈ G kω n y α sup x ∈ I k n (cid:2) ( n − n + y + y )( n + n + y − − y )( n + y − − n + 1 + y )( n + y − y + n ) (cid:3) α − / ≤ C · X k ∈ G kω n y α n [2 y (2 n − y (2 n + 2 n − α − / (because 2 y ≤ n − ≤ C · X k ∈ G kny α − n α − n α +1 y α = C · X k ∈ G kny ≤ C · X k ∈ G y ≤ C .
10n the last row it was used that k ≤ n + y − ≤ n and that the sum consists of fewerthan 2 y terms: n + y − − ( n − y ) = 2 y − y and n be as in the second case. Then we can estimate (cid:126) analogously. (cid:126) ≤ (cid:2) ( n − y + n − y )( n + n + y − − y )( n + y − y − n + 1)( n + y − y + n ) (cid:3) α − / ≤ [(2 n − n − y (2 y + 2 y )] α − / (because 2 n − ≤ y ) ≤ C · n α − y α − ⇒ ( ∗ ) ≤ C · X k ∈ G kn α y α − y α n α +1 = C · X k ∈ G kny ≤ C · X k ∈ G n ≤ C . Similar to the first case we used here that k ≤ n + y − ≤ y and that the sum consistsof fewer than n + y − − ( y − n + 1) = 2 n − k is now an exceptional index, we have to estimate ω k k P k ( τ y f n ) k ∞ . For exceptionalindices we have by definition either x + y ∈ I n for x ∈ I k , i.e. y + I k intersects the interval I n , or I n − y intersects I k . There are at most two such indices k ∈ E . The other casesof exceptional indices derive from the cases where I n + y or y − I n intersect the interval I k , or where k = 1. Each of these cases, with the exception of k = 1, yields at most 2exceptional indices. Thus there are at most 7. By looking closer one can see that thereare in fact only 5 of them.If k ≤ n then one has, with the use of τ y f n ( x ) = Z x + y | x − y | f n ( x ) K ( x, y, z ) ω α ( dz ) ≤ Z x + y | x − y | K ( x, y, z ) ω α ( dz ) = 1and ω n = Z n n − z α +1 dz = 12 α + 2 (cid:16) (3 n ) α +2 − (3 n − α +2 (cid:17) = 3 α +2 α + 2 (cid:18) n α +2 − ( n −
13 ) α +2 (cid:19) ≤ α +2 α + 2 (cid:16) n α +2 − ( n − α +2 (cid:17) = 3 α +2 Z nn − z α +1 dz = 3 α +2 ω n , the following estimate: ω k k P k ( τ y f n ) k ∞ ≤ ω k ≤ ω n ≤ α +2 ω n = 3 α +2 k f n k ∞ , . (1.70)11f k is exceptional and k > n , then one of the intervals y ± I n must intersect I k . Thesmallest value for y must therefore satisfy y + n = k −
1. That implies that y + 13 k > k − ⇒ y > k − > k, (1.71)because k >
3. Particularly we have y > x for all x ∈ I k in these cases. Now we canfind an upper bound for the remaining exceptional indices. τ y f n ( x ) = C Γ ( xy ) α Z n | x − y | z h (cid:16) z − ( x − y ) (cid:17)| {z } ≤ n − ( x − y ) (cid:16) ( x + y ) − z (cid:17)| {z } ≤ ( x + y ) −| x − y | =4 xy i α − / dz ≤ C Γ ( xy ) α ( n − | x − y | ) | {z } ≤ · n · h(cid:16) n − ( x − y ) (cid:17) xy i α − / ≤ C Γ α − / ( xy ) α + / n h n − ( x − y ) | {z } ≤ n − ( n − =2 n − i α − / ≤ C Γ α − / ( x ∗ x ) α + / n (2 n − | {z } ≤ n α − / ≤ C Γ α − / α + / x α +1 α − / n α + / ≤ C n α + ( k − α +1 . Hence we can estimate the remaining terms of the norm. ω k k P k ( τ y f n ) k ∞ ≤ C · k α +1 n α + / ( k − α +1 ≤ C · (cid:18) (cid:19) α +1 · n α + / ≤ C · ω n = C · k f n k ∞ , (cid:3) Using duality and complex interpolation as in [1], the boundedness of translation on( L ∞ , ‘ )( R + , ˚ α ) can be extended to the general amalgam spaces ( L p , ‘ q )( R + , ˚ α ). ( L p , ‘ q )( R + , ˚ α ) We have yet to prove the following theorem:
Theorem 10 (Hausdorff-Young theorem for amalgams):
For f ∈ ( L p , ‘ q )( R + , ˚ α ) with ≤ p, q ≤ holds, that ˆ f ∈ ( L q , ‘ p )( R + , ˚ α ) . The special case p = q = 2 is already known (see Theorem 2.2.22 in [3]). We willconsider the other extreme cases p, q ∈ { , } first. Therefore we will have to show that12 d I k ∞ , < ∞ . d I ( λ ) = Z R + I ( x ) χ λ ( x ) dω α ( x )= Z ∞ X k =0 ( − k Γ( α + 1)2 k k !Γ( α + k + 1) ( λx ) k dω α ( x )= ∞ X k =0 ( − k Γ( α + 1) λ k k k !Γ( α + k + 1) Z x k x α +1 dx = ∞ X k =0 ( − k Γ( α + 1) λ k k k !Γ( α + k + 1) 12 k + 2 α + 2= ∞ X k =0 ( − k Γ( α + 1) λ k k +1 k !Γ( α + k + 2)= Γ( α + 1)2 α λ α +1 ∞ X k =0 ( − k λ k + α +1 k + α +1 k !Γ( α + k + 2)= Γ( α + 1)2 α λ α +1 J α +1 ( λ ) . Here J α +1 denotes the Bessel function of degree α + 1. According to 9.2.1 in [4] it holdsfor λ → ∞ that J α +1 ( λ ) ≈ r πλ cos( λ − ( α + 1) π − π ) . Altogether for large enough λ we have (cid:12)(cid:12)(cid:12)d I ( λ ) (cid:12)(cid:12)(cid:12) ≈ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ( α + 1)2 α λ α +1 r πλ cos( λ − ( α + 1) π − π ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:47) C · λ − α − / . Now we can show that k d I k ∞ ,q < ∞ ⇔ q > α +1 α + / . Particularly for q ≥ k d I k ∞ ,q = ∞ X k =0 ω k sup λ ∈ I k | d I ( λ ) | q ! / q ≤ C · C + ∞ X k = N α k α +1 k q ( − α − / ) / q = C + C · ∞ X k = N α k α +1 − q ( α + / ) / q , where N α and C be chosen such that (cid:12)(cid:12)(cid:12)d I ( λ ) (cid:12)(cid:12)(cid:12) ≤ C · λ − α − / for λ ≥ N α . The seriesconverges iff 2 α + 1 − q ( α + / ) < − ⇔ q > α +1 α + / . One can easily see that d I does not lie in ( L p , ‘ q )( R + , ˚ α ) if q ≤ α +1 α + / and for all p . Now we proceed again as in131]. Let g = ω · I ˚ α [0 , and g n = ω · I ˚ α [ n − ,n +1] for n ≥
2. We show that g n ( x ) = 1 ∀ x ∈ I n . For this let x ∈ I . g ( x ) = 1 ω Z R + I ( y ) τ x [0 , ( y ) dω α ( y )= 1 ω Z Z x + y | x − y | [0 , ( z ) K ( x, y, z ) dω α ( z ) dω α ( y )= 1 ω Z Z [ | x − y | ,x + y ] ∩ [0 , K ( x, y, z ) dω α ( z ) dω α ( y )= 1 ω Z dω α ( y ) = 1 , because for x, y ∈ [0 , ⇒ [ | x − y | , x + y ] ⊂ [0 , ⇒ [ | x − y | , x + y ] ∩ [0 ,
2] = [ | x − y | , x + y ]and thus the inner integral yields the value 1. Entirely analog it holds for x ∈ I n , that g n ( x ) = 1 ω Z R + I ( y ) τ x [ n − ,n +1] ( y ) dω α ( y )= 1 ω Z Z [ | x − y | ,x + y ] ∩ [ n − ,n +1] K ( x, y, z ) dω α ( z ) dω α ( y )= 1 ω Z dω α ( y ) = 1 , because for x ∈ [ n − , n ] and y ∈ [0 , ⇒ [ | x − y | , x + y ] ⊂ [ n − , n + 1] ⇒ [ | x − y | , x + y ] ∩ [ n − , n + 1] = [ | x − y | , x + y ].The rest of the proof runs exactly as in [1], with 3 replaced by ω and ˚ / replacedby ˚ α . Note that Young’s inequality can be obtained as in [1] as well. (cid:3) Equivalence on compact/discrete hypergroups
As an addendum, if K is a compact or discrete Hypergroup, let us note the equivalenceof the continuous amalgam norm (here denoted as k·k ∗ p, ∞ ) with the discrete amalgamnorm.We first describe the discrete case. The discrete amalgam norm on a discrete hypergroupis defined as one would expect as k f k p,q = X k ∈ K ω ( { k } ) (cid:18) ω ( { k } ) | f ( k ) | p ω ( { k } ) (cid:19) q / p / q = X k ∈ K ω ( { k } ) | f ( k ) | q / q = k f k q . For the equivalence of the norms we compute k f k ∗ p, ∞ = sup n ∈ K X k ∈ K τ n { } ( k ) | f ( k ) | p ω ( { k } ) / p = sup n ∈ K X k ∈ K { } ( n ˚ k ) | f ( k ) | p ω ( { k } ) / p so, using 0 ∈ supp ε n ˚ ε k ⇔ k = n ¯, we get= sup n ∈ K (cid:16) ε n ¯ ˚ ε n ( { } ) | f ( n ) | p ω ( { n } ) (cid:17) / p and finally, with Theorem 1.3.26 in [3]: ω ( { n } ) = ε n ¯ ˚ ε n ( { } ) − , we have= sup n ∈ K | f ( n ) | = k f k ∞ . On the other hand, in the compact case, the discrete amalgam norm shrinks to ω ( K ) (cid:18) ω ( K ) Z K | f | p dω (cid:19) q / p ! / q = (cid:18)Z K | f | p dω (cid:19) / p = k f k p . The equivalence now follows: k f k ∗ p, ∞ = sup y ∈ K (cid:18)Z τ y K | f | p dω (cid:19) / p = sup y ∈ K (cid:18)Z Z K K d ( ε y ˚ ε x ) | f | p ( x ) ω ( dx ) (cid:19) / p = sup y ∈ K (cid:18)Z | f | p dω (cid:19) / p = (cid:18)Z | f | p dω (cid:19) / p = k f k p . So, in both cases, we obtained not only equivalence, but equality of norms.15 eferences [1] Walter R. Bloom, John F. Fournier und Michael Leinert,
Wiener’s theorem on hy-pergroups (2014), preprint.[2] Fournier, John J. F.,
Local and global properties of functions and their Fourier trans-forms , Tˆohoku Math. J. 49 (1997), 115 - 131.[3] Walter R. Bloom, Herbert Heyer,
Harmonic Analysis of Probability Measures onHypergroups , de Gruyter Studies in Mathematics 20, de Gruyter, Berlin/New York(1995).[4] Milton Abramowitz, Irene A Stegun,
Handbook of mathematical functions with for-mulas, graphs, and mathematical tables , Washington, U.S. Govt. Print. Off. (1964).[5] Michael Leinert,
On a Theorem of Wiener , Manuscripta Math. 110, 1–12 (2003).[6] R. P. Boas,