Is this a correct definition of the affine dual space? Since I have no good references, I recall some definitions.
Affine space Let $V$ be a real vector space. An affine space over $V$ is a set $A$ equipped with a map $\alpha\colon A\times V\to A$ satisfying the following conditions
A1) $\alpha(x,0)=x$ for any $x\in A$.
A2 $\alpha(\alpha(x,u),v)=\alpha(x,u+v)$ for any $x\in A$ and $u,v\in A$.
A3) For any $x,y\in A$ there exists a unique $u\in V$ such that $y=\alpha(x,u)$. We write $u=y-x$.
Affine functional Let $(A,\alpha)$ be an affine space over a vector space $V$. A function $f\colon A\to \mathbb R$ is called an affine functional if there exists a linear functional $\varphi_f\colon V\to \mathbb R$ such that
$$ f(x)-f(x_0)=\varphi_f(x-x_0)$$
whenever $x,x_0\in A$.
Here comes the question.

Is the set of all affine functionals an affine space?

Trying to prove it, for any affine functional $f$ on $A$ and linear functional $\psi$ on $V$ I have defined
$$
\alpha^\ast(f,\psi)(x)=f(x)+\psi(x-x_0)=f(x_0)+(\varphi_f+\psi)(x-x_0)\,,
$$
where $x_0\in A$ is fixed and $x$ ranges in $A$.
I've proved that $\alpha^\ast(f,\psi)$ is an affine functional and $\alpha^\ast$ satisfies A1) and A2). I'm struggling to prove the validity of the third hypothesis.
$$g(x)=g(x_0)+\varphi_g(x-x_0)= g(x_0)+[f(x)-f(x_0)-\varphi_f(x-x_0)]+\varphi_g(x-x_0)=[g(x_0)-f(x_0)]+f(x)+(\varphi_f-\varphi_g)(x-x_0)=[g(x_0)+f(x_0)]+\alpha^\ast(f,\varphi_f-\varphi_g)$$
The presence of a constant doesn't allow me to deduce A3). Have I chosen an incorrect definition of $\alpha^\ast$? Or is the statement false?
I appreciate any help you can provide.
UPDATE
With the following result, we can deduce the answer to my question. As Hanas, I denote with $Aff(\mathcal A,\mathcal B)$ the set of affine maps between two affine spaces $\mathcal A$ and $\mathcal B$.

Let $V,W$ be vector spaces and let $(\mathcal{A},\alpha,V), (\mathcal{B},\beta,W)$ be affine spaces.

*

*$Aff(\mathcal{A}, W)$ - seeing $W$ with its natural affine structure - is a real vector space with respect to the operations defined for any $\psi,\chi\in Aff(\mathcal{A}, W)$, $\lambda\in \mathbb R$ by
\begin{align*}
    \psi+\chi\colon x\in \mathcal{A}&\mapsto\psi(x)+\chi(x)\in W\,,\\
    \lambda \psi\colon x&\mapsto \lambda \psi(x)\in W\,.
\end{align*}

*$(Aff(\mathcal{A},\mathcal{B}),\omega_{\mathcal{A},\mathcal{B}},Aff(\mathcal{A},W))$ is an affine space where $\omega_{\mathcal{A},\mathcal{B}}$ is defined for any $f\in Aff(\mathcal{A},\mathcal{B})$ and $\psi\in Aff(\mathcal{A},W)$ by
$$
\omega_{\mathcal{A},\mathcal{B}}(f,\psi)\colon x\in \mathcal{A}\mapsto \beta(f(x),\psi(x))\in \mathcal{B}\,.
$$

*If $\dim V=n$ and $\dim W=m$, then $\dim Aff(\mathcal{A},W)=m(n+1)$.


 A: At first I'd like to ask why you use one of the most difficult definitions of affinity. Do you have to, are you working on something or could you change it?
Now to your question: You have to think what the vector space $V$ is, in which the set of all affine functionals is supposed to be an affine space or to with it is affine. Because you take the second argument $\psi$ of $\alpha^*$ as a linear function, it seems you want to proof that the set of affine functions , which I will call $Aff$, is a affine space of the set of linear functions $Lin$, but $Aff$ is not included in $Lin$, so it can't be a affine space of it, or you want to proof, that $Aff$ is affine to $Lin$, which cant be the case, because "$Aff$ has more free variables". It is however included in the set of all functionals to $\mathbb R$. So you could prove it there.
To be more precise: I'm not totally sure whether you want to proof, that $Aff$ is affine to a vector space or whether it is an affine subset of something. I say 99% the first. Then you would need a vector space with a translation, that generates the affine set from the vector space. For this you can't use $Lin$. as simple counter example use linear functions over $\mathbb R$, which are $f(x)=ax$ und affine functions, namely $f(x)=ax+b$. Then $Aff$ has one more degree of freedom, so their dimension is one more then the dimension of $Lin$, so there can't exist the bijection from (A3) of $\alpha$. I like the second way more: affine subspace point of view. Find a big vector space, in which $A$ is contained and then $A$ is an affine subspace, if $A-a$ for every (or equivalent any) $a\in A$ is a vector space (I hope this general statement works). But then $Aff$ is in the vector space of all mappings from ... (from where are we even mapping) to $\mathbb R$ and we see that it is a vector space in there. So, as a vector space, it also is an affine space, because it is affine to itself. So, you could use $\alpha =+$.
I hope this helpes.
