Algebra $A$ and its Gelfand spectrum

Question

Let $A$ be the set of all function $f$ on $\mathbb{R}$ of the form $$ f(x)=d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt,\qquad\quad x\in\mathbb{R}, $$ where $d\in\mathbb{C}$ and $k\in L_1([0,\infty])$. The norm on $A$ is defined by

$$ \|f\|:=|d|+\int\limits_0^\infty |k(t)|dt. $$

I want to show that $A$ is a commutative Banach algebra and find its Gelfand spectrum

We need the following properties in order to show that $A$ is a commutative Banach algebra:

1. commutativity
2. associativity
3. distributivity
4. scalar multiplication property
5. the norm of the product is less than or equal to the product of the norms,

The first four properties are easy to prove:

1. Let $f,g\in A$, then \begin{align} f(x)g(x)&=\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)\\ &=\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\\ &=g(x)f(x). \end{align}
2. Let $f,g,h\in A$, then \begin{align} (f(x)g(x))h(x)&=\left(\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)\right)\left(d_2+\int\limits_{0}^{\infty}e^{ixt}k_2(t)dt\right)\\ &=\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)\left(d_2+\int\limits_{0}^{\infty}e^{ixt}k_2(t)dt\right)\right)\\ &=f(x)(g(x)h(x)) \end{align}

3. Let $f,g,h\in A$, then \begin{align} f(x)(g(x)+h(x))&=\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)+\left(d_2+\int\limits_{0}^{\infty}e^{ixt}k_2(t)dt\right)\right)\\ &=f(x)g(x)+f(x)h(x) \end{align} This implies that $(g(x)+h(x))f(x)=g(x)f(x)+h(x)f(x)$

4. Let $f,g\in A$ and $\alpha\in\mathbb{R}$, then \begin{align} \alpha (f(x)g(x))&=\alpha\left(\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)\right)\\ &=(\alpha f(x))g(x)=f(x)(\alpha g(x)). \end{align}

5. Let $f,g\in A$, then \begin{align} f(x)g(x)&=\left(d+\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(d_1+\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right)\\ &=dd_1+d\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt+d_1\int\limits_{0}^{\infty}e^{ixt}k(t)dt+\left(\int\limits_{0}^{\infty}e^{ixt}k(t)dt\right)\left(\int\limits_{0}^{\infty}e^{ixt}k_1(t)dt\right) \end{align} Here I get stuck, I cannot show that $\|fg\|\leq\|f\|\|g\|$, any hints?

Secondly, I have some difficulties mastering the Gelfand spectrum concept. How can I find Gelfand spectrum of $A$?

Any hints are appreciated.

Answer 1

The first thing to show is that the decomposition is unique. That is, if $f$ is continuous on $\mathbb{R}$ has such a representation, then $d$ and $k$ are unique ($k$ is unique as an element of $L^1[0,\infty)$.) Equivalently, if $f=d+\int_{0}^{\infty}e^{ixt}k(t)dt$ is the $0$ function on $\mathbb{R}$, then $d=0$ and $k=0$ as an element of $L^1[0,\infty)$.

Once you have that, you need to show that $A$ is closed under addition, scalar multiplication, and multiplication. To show that it is closed under function multiplication, assume $f_1,f_2$ are two such functions, and write $$ f_1f_2 = d_1d_2+d_1f_2+d_2f_1+\int_{0}^{\infty}e^{ixt}k_1(t)dt\int_{0}^{\infty}e^{ixt}k_2(t)dt \\ = d_1d_2+\int_{0}^{\infty}\left(d_1k_2(t)+d_2k_1(t)+\int_{0}^{t}k_1(t-s)k_2(s)ds\right)e^{ixt}dt. $$ Show that the expression in parentheses is an $L^1[0,\infty)$ function. Showing that $A$ is closed under scalar multiplication and function addition is not difficult.

All properties of operations then following from the ordinary operations for complex functions on $\mathbb{R}$, including (1),(2),(3),(4). Property (5) requires the convolution property: $$ \left\|\int_{0}^{t}k_1(t-s)k_2(s)ds\right\|_{L^1} \le \|k_1\|_{L^1}\|k_2\|_{L^1}. $$ This gives the required norm identity: \begin{align} \|f_1f_2\|_{A} & = |d_1||d_2|+\left\|d_1k_2(t)+d_2k_1(t)+\int_{0}^{t}k_1(t-s)k_2(s)d\right\|_{L^1} \\ & \le |d_1||d_2|+|d_1|\|k_2\|_{L^1}+|d_2|\|k_1\|_{L^1}+\|k_1\|_{L^1}\|k_2\|_{L^2} \\ & = (|d_1|+\|k_1\|_{L^1})(|d_2|+\|k_2\|_{L^1}) \\ & = \|f_1\|_A\|f_2\|_A \end{align}

Concerning the Spectrum ...
The functions $f \in A$ have holomorphic extensions to the upper half-plane, and the point evaluations in the upper half plane have the form $$ E_z(f) = d+\int_{0}^{\infty}e^{izt}k(t)dt,\;\;\;\Im z \ge 0. $$ All of these are evaluations $E_z$ are in the Gelfand spectrum, and none of these evaluations can be $0$ for an invertible element. The function $\tilde{f}(z)=E_z(f)$ is holomorphic in the open upper half plane, has radial limits at $\infty$ as is continuous on the closed upper half plane.