# The spectrum of the Hardy Banach algebra $(H^1(\mathbb{T}),+,*,\|\|_1)$.

Let $\mathbb{T}$ be the $1$-torus and define: $$H^1(\mathbb{T}):=\{f\in L^1(\mathbb{T})\ | \ \forall n<0, \hat{f}(n)=0\},$$ where if $f\in L^1(\mathbb{T})$ we have denoted by $\hat{f}$ the Fourier transform of $f$.

By the linearity of the Fourier transform, it is clear that $H^1(\mathbb{T})$ is a subspace of $L^1(\mathbb{T})$.

By $\forall f,g \in L^1(\mathbb{T}), \widehat{f*g}=\hat{f}\hat{g}$ and by Young inequality for convolution, it is clear that $\left(H^1(\mathbb{T}),+,*,\|\|_1\right)$ is a commutative normed algebra.

By the continuity of the Fourier transform, it is clear that $H^1(\mathbb{T})$ is a closed subspace of $\left(L^1(\mathbb{T}),\|\|_1\right),$ and so $\left(H^1(\mathbb{T}),+,*,\|\|_1\right)$ 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 $\left(H^1(\mathbb{T}),+,*,\|\|_1\right)$ looks like...

Clearly, every element of the spectrum of the commutative Banach algebra $(L^1(\mathbb{T}),+,*,\|\|_1)$ is also an element of the spectrum of $\left(H^1(\mathbb{T}),+,*,\|\|_1\right)$, provided that this element does not vanish on the whole $H^1(\mathbb{T})$. Being the spectrum of $\left(L^1(\mathbb{T}),+,*,\|\|_1\right)$ formed by the elements $$\varphi_n: L^1(\mathbb{T})\rightarrow\mathbb{C}, f\mapsto \hat{f}(n)$$ for some $n \in \mathbb{Z}$, and being clear that, for all integers $n$, the multiplicative functional $\varphi_n$ does not vanish identically on $H^1(\mathbb{T})$ if and only if $n$ is non-negative, we found that $\forall n\ge0, \varphi_n$ is an element of the spectrum of $\left(H^1(\mathbb{T}),+,*,\|\|_1\right).$

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

## 1 Answer

There are none. Let $\phi$ be such an element. Since the linear span of $\{z^n:n\in\mathbb{N} \}$ 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 $f\in H^1$ we have $$\phi(f) = \phi(f)\phi(z^n) = \phi(f*z^n) =\phi(\hat f(n) z^n) = \hat f(n)$$