Prove or disprove: Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ be twice continuously partially differentiable such that $Hf(x) = 0$ for all $x \in \mathbb{R}^n$. Then there exist $A \in \mathbb{R}^{1 \times n}$ and $b \in \mathbb{R}$ such that $f(x) = A \cdot x + b$ for all $x \in \mathbb{R}^n$.
Given that $Hf(x) = 0$ for all $x \in \mathbb{R}^n$, it implies that the Hessian matrix of $f$ at any point $x$ is zero. The Hessian matrix $Hf(x)$ is a symmetric matrix whose entries are the second partial derivatives of $f$.
Since all entries of the Hessian matrix are zero, it implies that all second partial derivatives of $f$ are zero. This implies that all first-order partial derivatives of $f$ must be constant.
Let's assume that the partial derivatives of $f$ with respect to the first variable $x_1, x_2, \ldots, x_n$ are constant. Let $c_i$ be the constant value of the partial derivative with respect to $x_i$ for $i = 1, 2, \ldots, n$. Then we can write $f(x)$ as:
$ f(x) = c_1 x_1 + c_2 x_2 + \ldots + c_n x_n + b $
Here, $b$ is a constant representing the constant part of $f(x)$.
If we set $A = [c_1, c_2, \ldots, c_n] \in \mathbb{R}^{1 \times n}$, we obtain:
$ f(x) = A \cdot x + b $ Thus, we have shown that $f(x)$ can be written in the form $f(x) = A \cdot x + b$, where $A \in \mathbb{R}^{1 \times n}$ and $b \in \mathbb{R}$.
Therefore, the claim is proved. Is my proof correct?