Algebra $A$ and its Gelfand spectrum 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:


*

*commutativity

*associativity

*distributivity

*scalar multiplication property 

*the norm of the product is less than or equal to the product of the norms,


The first four properties are easy to prove:


*

*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}

*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}

*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 $(g(x)+h(x))f(x)=g(x)f(x)+h(x)f(x)$

*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}

*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.
 A: 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.
