- Email: [email protected]
Corrigendum to “On sharp constants for dual Segal-Bargmann Lp spaces” [J. Math. Anal. Appl. 424 (2) (2015) 1198–1222]
J. Math. Anal. Appl. 475 (2019) 1992–1995
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Corrigendum
Corrigendum to “On sharp constants for dual Segal-Bargmann Lp spaces” [J. Math. Anal. Appl. 424 (2) (2015) 1198–1222] William E. Gryc a,∗ , Todd Kemp b,1 a b
Muhlenberg College, United States of America University of California, San Diego, United States of America
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 28 December 2018 Available online 15 March 2019 Submitted by R.M. Aron
In our previous article, a particular function was stated to give an equality in a particular inequality, proving that inequality was sharp. In this corrigendum it is shown that the function given in the original paper does not have the stated property, but prove that another function does have the desired property, thus preserving the overall result from the original paper. © 2019 Elsevier Inc. All rights reserved.
There is an error in the article titled On sharp constants for dual Segal-Bargmann Lp spaces that appeared in the Journal of Mathematical Analysis and Applications on pages 1198–1222 in issue 424 in 2015 (doi 10.1016/j.jmaa.2014.11.063) which was authored by the same authors of this document. In that paper, the following statement appeared near the bottom on page 1206 in section 2.2: n and f ∈ (Pαn )−1 h such that there is equality in . . . it suffices to show that there exists some h ∈ Hp,α (2.12). That is
hp,α = f p,α . Cpn Let f (z) =
pα p/n 2π
(2.14)
α αw,z e− 2 |z| ; then Pαn f ≡ 1 ≡ hα ). 0 (the z = 0 case of the function hz (w) = e α
2
A straightforward computation shows that h = hα 0 and f satisfy (2.14). The function f in the above snippet (highlighted here) does not quite have the stated properties. As shown below, with this function, Pαn f is constant but not equal to 1, and (2.14) is not satisfied in this case. DOI of original article: https://doi.org/10.1016/j.jmaa.2014.11.063.
* Corresponding author. 1
E-mail addresses: [email protected] (W.E. Gryc), [email protected] (T. Kemp). Supported in part by NSF CAREER Award DMS-1254807 and NSF Grant DMS-1800733.
https://doi.org/10.1016/j.jmaa.2019.03.003 0022-247X/© 2019 Elsevier Inc. All rights reserved.
W.E. Gryc, T. Kemp / J. Math. Anal. Appl. 475 (2019) 1992–1995
1993
2−p However, if one replaces the definition of f (z) above with f (z) = (p/[2(p − 1)])n exp(− α2 p−1 |z|2 ), then the rest of the statement follows. That is, the following snippet (with the change highlighted) is correct: n . . . it suffices to show that there exists some h ∈ Hp,α and f ∈ (Pαn )−1 h such that there is equality in (2.12). That is
hp,α = f p,α . Cpn α 2−p
(2.14)
α Let f (z) = (p/[2(p − 1)])n e− 2 p−1 |z| ; then Pαn f ≡ 1 ≡ hα 0 (the z = 0 case of the function hz (w) = 2
eαw,z ). A straightforward computation shows that h = hα 0 and f satisfy (2.14). Below the “straightforward computation” is performed to demonstrate the correctness of the new statement, as well as the incorrectness of the first statement. Throughout what follows, the following lemma will be useful to compute Gaussian integrals. It is labeled as Lemma 2.1 in the original paper: Lemma 0.1. Let A be a complex symmetric matrix, v a vector in Rk , and let (·, ·) denote the standard inner product on Rk . Define the function f (x) = exp(−(x, Ax) + 2(v, x)). Then f ∈ L1 (Rk ) if and only if (A) is positive definite, and in this case,
e−(x,Ax)+2(v,x) dx =
Rk
π k/2 det(A)
e(v,A
−1
v)
.
(0.1)
Next some notation and definitions that were used in the original paper are restated below. The operator Pαn is defined as (Pαn f )(z)
eαz,w f (w)γαn (dw),
= Cn
where ·, · is the standard sesquilinear inner product on Cn and γαn (dw) =
α n π
e−α|w| dw. 2
Also, the constant Cpn is defined as Cpn :=
2 p1/p p 1p
n
where 1/p + 1/p = 1.
Finally, the norm · p,α is defined as ⎛
f p,α = ⎝
Cn
For this corrigendum, define g˜(z) and g(z) as
⎞1/p n |f (z)|p γαp/2 (dz)⎠
.
1994
W.E. Gryc, T. Kemp / J. Math. Anal. Appl. 475 (2019) 1992–1995
pα p/n
e− 2 |z| , 2π n 2 α 2−p p g(z) := e− 2 p−1 |z| . 2(p − 1)
g˜(z) :=
α
2
The assertion of this corrigendum is that when f (z) is set equal to g˜(z) in the quote above, then the quote is incorrect, but when f (z) is set equal to g(z) in the quote above, then the quote is correct. To that end, we first prove a lemma that will allow us to prove the preceding statements. Lemma 0.2. Let c be any real number greater than −1. Define the function fc : Cn → R as fc (z) := e−αc|z| . 2
Then Pαn fc is a constant function and 1 , (c + 1)n 1 = , (2c + 1)n/p
(Pαn fc )(z) =
(0.2)
fc p,α
(0.3)
and Pαn (fc )p,α /fc p,α is maximized when c =
2−p 2(p−1) .
Proof. We first compute (Pαn fc )(z). To that end, define the matrix J2n in block form as
J2n
0 = In
−In , 0
T where Ik is the k × k identity matrix. Note that J2n = −J2n . For z ∈ Cn , define the associated real vector T r(z) ∈ R2n as r(z) := [ (z) (z) ] . Also, recall that ·, · is the standard sesquilinear inner product on Cn and (·, ·) is the standard bilinear inner product on Rk . Note that
z, w = (r(z), (I2n + iJ2n )r(w)).
(0.4)
Also note that a a a −b a −b (I2n − iJ2n ) , (I2n − iJ2n ) = −i , −i b b b a b a =
n j=1
(a2j
+
b2j )
n n + 2i (bj aj − aj bj ) − (a2j + b2j ) j=1
j=1
= 0. Identifying w ∈ Cn with x = r(w) ∈ R2n , Lemma 0.1 and Equation (0.4) yield α n 2 2 n (Pα fc )(z) = eαz,w e−αc|w| e−α|w| dw π Cn α n α e2(x, 2 (I2n −iJ2n )r(z))−(x,α(c+1)I2n x) dx = π R2n
= =
α n π
α 1 α πn e( 2 (I2n −iJ2n )r(z), α(c+1) 2 (I2n −iJ2n )r(z)) (α(c + 1))n
1 , (c + 1)n
(0.5)
W.E. Gryc, T. Kemp / J. Math. Anal. Appl. 475 (2019) 1992–1995
1995
where the last equality is justified by Equation (0.5). This proves Equation (0.2). To compute fc p,α , again applying Lemma 0.1 yields: αp n 2 2 fc pp,α = |e−αcx |p e−(αp/2)x dx 2π R2n
=
=
αp n 2π
e−(x,αp(c+1/2)I2n x) dx
R2n
αp n 2π
1 πn = . n n (αp) (c + 1/2) (2c + 1)n
Taking pth roots yields Equation (0.3). Finally, note that d dc
Thus, c =
2−p p−1
=
1 p−1
Pαn (fc )p,α fc p,α
1/n =
d (2c + 1)1/p dc (c + 1)
=
(2c + 1)1/p−1 (2(c + 1) − p(2c + 1)) p(c + 1)2
=
(2c + 1)1/p−1 (−2(p − 1)c + (2 − p)). p(c + 1)2
− 1 is the global maximum of
Pαn (fc )p,α fc p,α
for c ∈ (−1, ∞), completing the proof. 2
Using the notation of Lemma 0.2, note that g(z) =
p 2(p − 1)
n e
−α 2
2−p 2 p−1 |z|
=
2−p 1+ 2(p − 1)
n f
2−p 2(p−1)
(z).
Thus, by Lemma 0.2, Pαn (g)(z) ≡ 1, and 2−p n Pαn (f 2−p )p,α ( p−1 + 1)1/p Pαn (g)p,α 2(p−1) = = 2−p gp,α f 2−p p,α ( 2(p−1) + 1) 2(p−1) 1 1/p n p 1/p n ) ( p−1 ) 2( p−1 = = p p ( 2(p−1) ) p1/p (p−1) n n 2(p )1/p 2 = = = Cpn . p1/p p p1/p p 1/p Thus, if h = Pαn (g) and f = g, then Equation (2.14) holds. Lastly, note that by Lemma 0.2 Pαn (˜ g )(z) =
pα p/n 2π
Pαn (f1/2 )(z) =
pα p/n 2π
1 , (3/2)n
(0.6)
which is a constant function but is not generally equal to 1. Also, unless p = 3/2 (in which case 1/2 = (2 − p)/(2(p − 1))), we have Pαn (f1/2 )p,α Pαn (f(2−p)/(2(p−1)) )p,α Pαn (˜ g )p,α = < = Cpn , ˜ g p,α f1/2 p,α f(2−p)/(2(p−1)) p,α so that Equation (2.14) does not generally hold for h = Pαn (˜ g ) and f = g˜. This completes the correction to the article.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Corrigendum
Corrigendum to “On sharp constants for dual Segal-Bargmann Lp spaces” [J. Math. Anal. Appl. 424 (2) (2015) 1198–1222] William E. Gryc a,∗ , Todd Kemp b,1 a b
Muhlenberg College, United States of America University of California, San Diego, United States of America
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 28 December 2018 Available online 15 March 2019 Submitted by R.M. Aron
In our previous article, a particular function was stated to give an equality in a particular inequality, proving that inequality was sharp. In this corrigendum it is shown that the function given in the original paper does not have the stated property, but prove that another function does have the desired property, thus preserving the overall result from the original paper. © 2019 Elsevier Inc. All rights reserved.
There is an error in the article titled On sharp constants for dual Segal-Bargmann Lp spaces that appeared in the Journal of Mathematical Analysis and Applications on pages 1198–1222 in issue 424 in 2015 (doi 10.1016/j.jmaa.2014.11.063) which was authored by the same authors of this document. In that paper, the following statement appeared near the bottom on page 1206 in section 2.2: n and f ∈ (Pαn )−1 h such that there is equality in . . . it suffices to show that there exists some h ∈ Hp,α (2.12). That is
hp,α = f p,α . Cpn Let f (z) =
pα p/n 2π
(2.14)
α αw,z e− 2 |z| ; then Pαn f ≡ 1 ≡ hα ). 0 (the z = 0 case of the function hz (w) = e α
2
A straightforward computation shows that h = hα 0 and f satisfy (2.14). The function f in the above snippet (highlighted here) does not quite have the stated properties. As shown below, with this function, Pαn f is constant but not equal to 1, and (2.14) is not satisfied in this case. DOI of original article: https://doi.org/10.1016/j.jmaa.2014.11.063.
* Corresponding author. 1
E-mail addresses: [email protected] (W.E. Gryc), [email protected] (T. Kemp). Supported in part by NSF CAREER Award DMS-1254807 and NSF Grant DMS-1800733.
https://doi.org/10.1016/j.jmaa.2019.03.003 0022-247X/© 2019 Elsevier Inc. All rights reserved.
W.E. Gryc, T. Kemp / J. Math. Anal. Appl. 475 (2019) 1992–1995
1993
2−p However, if one replaces the definition of f (z) above with f (z) = (p/[2(p − 1)])n exp(− α2 p−1 |z|2 ), then the rest of the statement follows. That is, the following snippet (with the change highlighted) is correct: n . . . it suffices to show that there exists some h ∈ Hp,α and f ∈ (Pαn )−1 h such that there is equality in (2.12). That is
hp,α = f p,α . Cpn α 2−p
(2.14)
α Let f (z) = (p/[2(p − 1)])n e− 2 p−1 |z| ; then Pαn f ≡ 1 ≡ hα 0 (the z = 0 case of the function hz (w) = 2
eαw,z ). A straightforward computation shows that h = hα 0 and f satisfy (2.14). Below the “straightforward computation” is performed to demonstrate the correctness of the new statement, as well as the incorrectness of the first statement. Throughout what follows, the following lemma will be useful to compute Gaussian integrals. It is labeled as Lemma 2.1 in the original paper: Lemma 0.1. Let A be a complex symmetric matrix, v a vector in Rk , and let (·, ·) denote the standard inner product on Rk . Define the function f (x) = exp(−(x, Ax) + 2(v, x)). Then f ∈ L1 (Rk ) if and only if (A) is positive definite, and in this case,
e−(x,Ax)+2(v,x) dx =
Rk
π k/2 det(A)
e(v,A
−1
v)
.
(0.1)
Next some notation and definitions that were used in the original paper are restated below. The operator Pαn is defined as (Pαn f )(z)
eαz,w f (w)γαn (dw),
= Cn
where ·, · is the standard sesquilinear inner product on Cn and γαn (dw) =
α n π
e−α|w| dw. 2
Also, the constant Cpn is defined as Cpn :=
2 p1/p p 1p
n
where 1/p + 1/p = 1.
Finally, the norm · p,α is defined as ⎛
f p,α = ⎝
Cn
For this corrigendum, define g˜(z) and g(z) as
⎞1/p n |f (z)|p γαp/2 (dz)⎠
.
1994
W.E. Gryc, T. Kemp / J. Math. Anal. Appl. 475 (2019) 1992–1995
pα p/n
e− 2 |z| , 2π n 2 α 2−p p g(z) := e− 2 p−1 |z| . 2(p − 1)
g˜(z) :=
α
2
The assertion of this corrigendum is that when f (z) is set equal to g˜(z) in the quote above, then the quote is incorrect, but when f (z) is set equal to g(z) in the quote above, then the quote is correct. To that end, we first prove a lemma that will allow us to prove the preceding statements. Lemma 0.2. Let c be any real number greater than −1. Define the function fc : Cn → R as fc (z) := e−αc|z| . 2
Then Pαn fc is a constant function and 1 , (c + 1)n 1 = , (2c + 1)n/p
(Pαn fc )(z) =
(0.2)
fc p,α
(0.3)
and Pαn (fc )p,α /fc p,α is maximized when c =
2−p 2(p−1) .
Proof. We first compute (Pαn fc )(z). To that end, define the matrix J2n in block form as
J2n
0 = In
−In , 0
T where Ik is the k × k identity matrix. Note that J2n = −J2n . For z ∈ Cn , define the associated real vector T r(z) ∈ R2n as r(z) := [ (z) (z) ] . Also, recall that ·, · is the standard sesquilinear inner product on Cn and (·, ·) is the standard bilinear inner product on Rk . Note that
z, w = (r(z), (I2n + iJ2n )r(w)).
(0.4)
Also note that a a a −b a −b (I2n − iJ2n ) , (I2n − iJ2n ) = −i , −i b b b a b a =
n j=1
(a2j
+
b2j )
n n + 2i (bj aj − aj bj ) − (a2j + b2j ) j=1
j=1
= 0. Identifying w ∈ Cn with x = r(w) ∈ R2n , Lemma 0.1 and Equation (0.4) yield α n 2 2 n (Pα fc )(z) = eαz,w e−αc|w| e−α|w| dw π Cn α n α e2(x, 2 (I2n −iJ2n )r(z))−(x,α(c+1)I2n x) dx = π R2n
= =
α n π
α 1 α πn e( 2 (I2n −iJ2n )r(z), α(c+1) 2 (I2n −iJ2n )r(z)) (α(c + 1))n
1 , (c + 1)n
(0.5)
W.E. Gryc, T. Kemp / J. Math. Anal. Appl. 475 (2019) 1992–1995
1995
where the last equality is justified by Equation (0.5). This proves Equation (0.2). To compute fc p,α , again applying Lemma 0.1 yields: αp n 2 2 fc pp,α = |e−αcx |p e−(αp/2)x dx 2π R2n
=
=
αp n 2π
e−(x,αp(c+1/2)I2n x) dx
R2n
αp n 2π
1 πn = . n n (αp) (c + 1/2) (2c + 1)n
Taking pth roots yields Equation (0.3). Finally, note that d dc
Thus, c =
2−p p−1
=
1 p−1
Pαn (fc )p,α fc p,α
1/n =
d (2c + 1)1/p dc (c + 1)
=
(2c + 1)1/p−1 (2(c + 1) − p(2c + 1)) p(c + 1)2
=
(2c + 1)1/p−1 (−2(p − 1)c + (2 − p)). p(c + 1)2
− 1 is the global maximum of
Pαn (fc )p,α fc p,α
for c ∈ (−1, ∞), completing the proof. 2
Using the notation of Lemma 0.2, note that g(z) =
p 2(p − 1)
n e
−α 2
2−p 2 p−1 |z|
=
2−p 1+ 2(p − 1)
n f
2−p 2(p−1)
(z).
Thus, by Lemma 0.2, Pαn (g)(z) ≡ 1, and 2−p n Pαn (f 2−p )p,α ( p−1 + 1)1/p Pαn (g)p,α 2(p−1) = = 2−p gp,α f 2−p p,α ( 2(p−1) + 1) 2(p−1) 1 1/p n p 1/p n ) ( p−1 ) 2( p−1 = = p p ( 2(p−1) ) p1/p (p−1) n n 2(p )1/p 2 = = = Cpn . p1/p p p1/p p 1/p Thus, if h = Pαn (g) and f = g, then Equation (2.14) holds. Lastly, note that by Lemma 0.2 Pαn (˜ g )(z) =
pα p/n 2π
Pαn (f1/2 )(z) =
pα p/n 2π
1 , (3/2)n
(0.6)
which is a constant function but is not generally equal to 1. Also, unless p = 3/2 (in which case 1/2 = (2 − p)/(2(p − 1))), we have Pαn (f1/2 )p,α Pαn (f(2−p)/(2(p−1)) )p,α Pαn (˜ g )p,α = < = Cpn , ˜ g p,α f1/2 p,α f(2−p)/(2(p−1)) p,α so that Equation (2.14) does not generally hold for h = Pαn (˜ g ) and f = g˜. This completes the correction to the article.