The spectrum of the Hardy Banach algebra $(H^1(mathbbT),+,*,||_1)$.

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Let $mathbbT$ be the $1$-torus and define: $$H^1(mathbbT):= forall n<0, hatf(n)=0,$$
where if $fin L^1(mathbbT)$ we have denoted by $hatf$ the Fourier transform of $f$.

By the linearity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a subspace of $L^1(mathbbT)$.

By $forall f,g in L^1(mathbbT), widehatf*g=hatfhatg$ and by Young inequality for convolution, it is clear that $left(H^1(mathbbT),+,*,||_1right)$ is a commutative normed algebra.

By the continuity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a closed subspace of $left(L^1(mathbbT),||_1right),$ and so $left(H^1(mathbbT),+,*,||_1right)$ is a commutative Banach algebra.

Then I start wondering how the spectrum (i.e. the set of non null multiplicative linear functional) of the commutative Banach algebra $left(H^1(mathbbT),+,*,||_1right)$ looks like...

Clearly, every element of the spectrum of the commutative Banach algebra $(L^1(mathbbT),+,*,||_1)$ is also an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right)$, provided that this element does not vanish on the whole $H^1(mathbbT)$. Being the spectrum of $left(L^1(mathbbT),+,*,||_1right)$ formed by the elements $$varphi_n: L^1(mathbbT)rightarrowmathbbC, fmapsto hatf(n)$$
for some $n in mathbbZ$, and being clear that, for all integers $n$, the multiplicative functional $varphi_n$ does not vanish identically on $H^1(mathbbT)$ if and only if $n$ is non-negative, we found that $forall nge0, varphi_n$ is an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right).$

So the question: are there any other elements of the spectrum out there?

spectral-theory banach-algebras hardy-spaces

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edited Jul 30 at 14:49

asked Jul 30 at 9:41

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Let $mathbbT$ be the $1$-torus and define: $$H^1(mathbbT):= forall n<0, hatf(n)=0,$$
where if $fin L^1(mathbbT)$ we have denoted by $hatf$ the Fourier transform of $f$.

By the linearity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a subspace of $L^1(mathbbT)$.

By $forall f,g in L^1(mathbbT), widehatf*g=hatfhatg$ and by Young inequality for convolution, it is clear that $left(H^1(mathbbT),+,*,||_1right)$ is a commutative normed algebra.

By the continuity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a closed subspace of $left(L^1(mathbbT),||_1right),$ and so $left(H^1(mathbbT),+,*,||_1right)$ is a commutative Banach algebra.

Then I start wondering how the spectrum (i.e. the set of non null multiplicative linear functional) of the commutative Banach algebra $left(H^1(mathbbT),+,*,||_1right)$ looks like...

Clearly, every element of the spectrum of the commutative Banach algebra $(L^1(mathbbT),+,*,||_1)$ is also an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right)$, provided that this element does not vanish on the whole $H^1(mathbbT)$. Being the spectrum of $left(L^1(mathbbT),+,*,||_1right)$ formed by the elements $$varphi_n: L^1(mathbbT)rightarrowmathbbC, fmapsto hatf(n)$$
for some $n in mathbbZ$, and being clear that, for all integers $n$, the multiplicative functional $varphi_n$ does not vanish identically on $H^1(mathbbT)$ if and only if $n$ is non-negative, we found that $forall nge0, varphi_n$ is an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right).$

So the question: are there any other elements of the spectrum out there?

spectral-theory banach-algebras hardy-spaces

share|cite|improve this question

edited Jul 30 at 14:49

asked Jul 30 at 9:41

Bob

1,517522

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Let $mathbbT$ be the $1$-torus and define: $$H^1(mathbbT):= forall n<0, hatf(n)=0,$$
where if $fin L^1(mathbbT)$ we have denoted by $hatf$ the Fourier transform of $f$.

By the linearity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a subspace of $L^1(mathbbT)$.

By $forall f,g in L^1(mathbbT), widehatf*g=hatfhatg$ and by Young inequality for convolution, it is clear that $left(H^1(mathbbT),+,*,||_1right)$ is a commutative normed algebra.

By the continuity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a closed subspace of $left(L^1(mathbbT),||_1right),$ and so $left(H^1(mathbbT),+,*,||_1right)$ is a commutative Banach algebra.

Then I start wondering how the spectrum (i.e. the set of non null multiplicative linear functional) of the commutative Banach algebra $left(H^1(mathbbT),+,*,||_1right)$ looks like...

Clearly, every element of the spectrum of the commutative Banach algebra $(L^1(mathbbT),+,*,||_1)$ is also an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right)$, provided that this element does not vanish on the whole $H^1(mathbbT)$. Being the spectrum of $left(L^1(mathbbT),+,*,||_1right)$ formed by the elements $$varphi_n: L^1(mathbbT)rightarrowmathbbC, fmapsto hatf(n)$$
for some $n in mathbbZ$, and being clear that, for all integers $n$, the multiplicative functional $varphi_n$ does not vanish identically on $H^1(mathbbT)$ if and only if $n$ is non-negative, we found that $forall nge0, varphi_n$ is an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right).$

So the question: are there any other elements of the spectrum out there?

spectral-theory banach-algebras hardy-spaces

share|cite|improve this question

edited Jul 30 at 14:49

asked Jul 30 at 9:41

Bob

1,517522

Let $mathbbT$ be the $1$-torus and define: $$H^1(mathbbT):= forall n<0, hatf(n)=0,$$
where if $fin L^1(mathbbT)$ we have denoted by $hatf$ the Fourier transform of $f$.

By the linearity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a subspace of $L^1(mathbbT)$.

By $forall f,g in L^1(mathbbT), widehatf*g=hatfhatg$ and by Young inequality for convolution, it is clear that $left(H^1(mathbbT),+,*,||_1right)$ is a commutative normed algebra.

By the continuity of the Fourier transform, it is clear that $H^1(mathbbT)$ is a closed subspace of $left(L^1(mathbbT),||_1right),$ and so $left(H^1(mathbbT),+,*,||_1right)$ is a commutative Banach algebra.

Then I start wondering how the spectrum (i.e. the set of non null multiplicative linear functional) of the commutative Banach algebra $left(H^1(mathbbT),+,*,||_1right)$ looks like...

Clearly, every element of the spectrum of the commutative Banach algebra $(L^1(mathbbT),+,*,||_1)$ is also an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right)$, provided that this element does not vanish on the whole $H^1(mathbbT)$. Being the spectrum of $left(L^1(mathbbT),+,*,||_1right)$ formed by the elements $$varphi_n: L^1(mathbbT)rightarrowmathbbC, fmapsto hatf(n)$$
for some $n in mathbbZ$, and being clear that, for all integers $n$, the multiplicative functional $varphi_n$ does not vanish identically on $H^1(mathbbT)$ if and only if $n$ is non-negative, we found that $forall nge0, varphi_n$ is an element of the spectrum of $left(H^1(mathbbT),+,*,||_1right).$

So the question: are there any other elements of the spectrum out there?

spectral-theory banach-algebras hardy-spaces

share|cite|improve this question

edited Jul 30 at 14:49

asked Jul 30 at 9:41

Bob

1,517522

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 14:49

edited Jul 30 at 14:49

edited Jul 30 at 14:49

asked Jul 30 at 9:41

Bob

1,517522

asked Jul 30 at 9:41

Bob

1,517522

asked Jul 30 at 9:41

Bob

1,517522

1,517522

add a comment | 

add a comment | 

1 Answer
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There are none. Let $phi$ be such an element. Since the linear span of $z^n:ninmathbbN $ is dense in $H^1$, there exists $n$ such that $phi(z^n)ne 0$. Since $z^n*z^n = z^n$, it follows that $phi(z^n)^2 = phi(z^n)$, so $phi(z^n)=1$. Then for any $fin H^1$ we have
$$phi(f) = phi(f)phi(z^n) = phi(f*z^n) =phi(hat f(n) z^n) = hat f(n)$$

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answered Jul 30 at 15:14

user357151

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1 Answer
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1 Answer
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up vote
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accepted

There are none. Let $phi$ be such an element. Since the linear span of $z^n:ninmathbbN $ is dense in $H^1$, there exists $n$ such that $phi(z^n)ne 0$. Since $z^n*z^n = z^n$, it follows that $phi(z^n)^2 = phi(z^n)$, so $phi(z^n)=1$. Then for any $fin H^1$ we have
$$phi(f) = phi(f)phi(z^n) = phi(f*z^n) =phi(hat f(n) z^n) = hat f(n)$$

share|cite|improve this answer

answered Jul 30 at 15:14

user357151

13.8k31140

add a comment | 

up vote
1
down vote



accepted

There are none. Let $phi$ be such an element. Since the linear span of $z^n:ninmathbbN $ is dense in $H^1$, there exists $n$ such that $phi(z^n)ne 0$. Since $z^n*z^n = z^n$, it follows that $phi(z^n)^2 = phi(z^n)$, so $phi(z^n)=1$. Then for any $fin H^1$ we have
$$phi(f) = phi(f)phi(z^n) = phi(f*z^n) =phi(hat f(n) z^n) = hat f(n)$$

share|cite|improve this answer

answered Jul 30 at 15:14

user357151

13.8k31140

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

There are none. Let $phi$ be such an element. Since the linear span of $z^n:ninmathbbN $ is dense in $H^1$, there exists $n$ such that $phi(z^n)ne 0$. Since $z^n*z^n = z^n$, it follows that $phi(z^n)^2 = phi(z^n)$, so $phi(z^n)=1$. Then for any $fin H^1$ we have
$$phi(f) = phi(f)phi(z^n) = phi(f*z^n) =phi(hat f(n) z^n) = hat f(n)$$

share|cite|improve this answer

answered Jul 30 at 15:14

user357151

13.8k31140

There are none. Let $phi$ be such an element. Since the linear span of $z^n:ninmathbbN $ is dense in $H^1$, there exists $n$ such that $phi(z^n)ne 0$. Since $z^n*z^n = z^n$, it follows that $phi(z^n)^2 = phi(z^n)$, so $phi(z^n)=1$. Then for any $fin H^1$ we have
$$phi(f) = phi(f)phi(z^n) = phi(f*z^n) =phi(hat f(n) z^n) = hat f(n)$$

share|cite|improve this answer

answered Jul 30 at 15:14

user357151

13.8k31140

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 15:14

user357151

13.8k31140

answered Jul 30 at 15:14

user357151

13.8k31140

answered Jul 30 at 15:14

user357151

13.8k31140

13.8k31140

add a comment | 

add a comment | 

 

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