Reducing dimensions by considering ex-ante instead of ex-post symmetry

Question

Let $\mathbf x = (x_1,\ldots,x_n) \in \mathbb R^n$. Consider the following problem: \begin{align} \max_{\mathbf x}f(\mathbf x). \end{align} Presume that $\mathbf{\bar x} = \underbrace{(\bar x,\ldots,\bar x)}_{n \text{ times}}$ is the unique maximizer of $f$, i.e. \begin{align} \mathbf{\bar x} = \arg\max_{\mathbf x}f(\mathbf x). \end{align} Let $F(x) = f(x,\ldots,x)$. I was wondering if the following statement is true: \begin{align} \bar x = \arg\max_{x \in \mathbb R}F(x). \end{align}

Answer 1

Short answer: Yes!

Proposition. Let $n\in\mathbb N, f: \mathbb R^n \rightarrow \mathbb R$ and $\mathbf{\bar x} = (\bar x, \ldots, \bar x)\in\mathbb R^n$ such that $f(\mathbf{\bar x}) > f(\mathbf y)$ for all $\mathbf{\bar x} \neq \mathbf y$ in $\mathbb R^n$. Let $F: \mathbb R \rightarrow \mathbb R: x \mapsto f(x,\ldots,x)$. Then $F(\bar x) > F(y)$ for all $\bar x \neq y \in \mathbb R$.

Proof. Since we have $f(\mathbf{\bar x}) > f(\mathbf y)$ for all $\mathbf{\bar x} \neq \mathbf y\in \mathbb R^n$, we have $f(\mathbf{\bar x}) > f(\mathbf y)$ in particular for all $\mathbf y = (y, \ldots, y) \in \mathbb R^n$ (where $y\neq \bar x)$. By definition it follows that for all $\mathbf y = (y, \ldots, y) \in \mathbb R^n$ ($y\neq \bar x$): $\; F(y) := f(y,\ldots, y) = f(\mathbf y) < f(\mathbf {\bar x}) = f(\bar x,\ldots, \bar x) = F(\bar x)$ and thus $\bar x$ is a unique maximizer of $F$. $\quad\square$