<s>[INST] Determine the existence of a function Does there exist a function $f:\mathbb{R}^n \to \mathbb{R}$ such that for all $x = (x_1,...,x_n) \in \mathbb{R}^n$:


*

*$f(A) = f(x)$ for every permutation A of $\{x_1,...,x_n\}$.

*$f(x + (a,...,a)) = f(x) + a$ for every $a \in \mathbb{R}$.

*$f(ax) = af(x)$ for every $a \in \mathbb{R}$.

*There exist $y = (y_1,...,y_n) \in \mathbb{R}^n$ such that $f(y) > max\{y_1,...,y_n\}$


?
I have managed to prove that for $n = 1$ and $n = 2$ such a function doesn't exist, but I don't know to prove it for $n>2$.
 [/INST] Let
\begin{align*}
m(x) &:= \min\{x_k \colon\ k=1,\dots,n\}, \\
M(x) &:= \min\{x_k \colon\ k=1,\dots,n\}, \\
h(x) &:= -\frac{n}{2}(m(x) + M(x)) + \sum_{k=1}^n x_k, \\
f(x) &:= \frac{m(x) + M(x)}{2} + h(x).
\end{align*}
We have already determined that $\frac{m(x) + M(x)}{2}$ satisfies $1-3$. Let us now check the behaviour of $h(x)$.
Notice that $h(x) = h(p(x))$ for any permutation $p$ of $x$ because neither $n$, $m(x)$, $M(x)$, nor sum of the elements of $x$ depend on the ordering of the elements.
As for what happens when we add $(a,\dots,a)$,
\begin{align*}
h(x+(a,\dots,a)) &= -\frac{n}{2}(m(x) + a + M(x) + a) + \sum_{k=1}^n (x_k + a), \\
&= -na - \frac{n}{2}(m(x) + M(x)) + na + \sum_{k=1}^n x_k = h(x).
\end{align*}
Now, for $a \ge 0$, we have:
\begin{align*}
h(ax) &= -\frac{n}{2}(m(ax) + M(ax)) + \sum_{k=1}^n ax_k, \\
&= -\frac{an}{2}(m(x) + M(x)) + a\sum_{k=1}^n x_k = ah(x).
\end{align*}
If we prove the same for $a = -1$, this will work in general. Notice that
$$m(-x) = -M(x), \quad M(-x) = -m(x).$$
So,
\begin{align*}
h(-x) &= -\frac{n}{2}(m(-x) + M(-x)) + \sum_{k=1}^n (-x_k), \\
&= -\frac{n}{2}(-M(x) - m(x)) - \sum_{k=1}^n x_k \\
&= - \left(-\frac{n}{2}(m(x) + M(x)) + \sum_{k=1}^n x_k\right) = -h(x).
\end{align*}
Hence, $h(ax) = ah(x)$ for any $a \in \mathbb{R}$.
Using all of this, it is easy to verify $1-3$ for $f$:


*

*Neither $m(x)$, $M(x)$, nor $h(x)$ change with a permutation.

*$f(x+(a,\dots,a)) = \frac{m(x)+a + M(x)+a}{2} + h(x+(a,\dots,a)) = a + \frac{m(x) + M(x)}{2} + h(x) = f(x) + a.$

*$f(ax) = \frac{m(ax) + M(ax)}{2} + h(ax) = a\frac{m(x) + M(x)}{2} + ah(x) = af(x)$.
Now, notice that for $n = 1$ and $n = 2$, $h(x) = 0$, so we did nothing to those cases. But, for $n > 2$, this is no longer true. Let $n = 5$ and let $r > 2$ be any real number bigger than $2$. We construct $y$:
$$y := (1, r-1, r-1, r-1, r-1).$$
Now,
\begin{align*}
f(y) &= \frac{m(y) + M(y)}{2} + h(y) = \frac{m(y) + M(y)}{2} - \frac{n}{2}(m(y) + M(y)) + \sum_{k=1}^n y_k \\
&= \frac{1 + (r-1)}{2} - \frac{5}{2}(1 + (r-1)) + (1+(r-1)+(r-1)+(r-1)+(r-1)) \\
&= \frac{r}{2} - \frac{5r}{2} + 4r - 3 = 2r - 3 = r + (r-3) > r - 1 = M(x).
\end{align*}
My guess is that we could probably construct an appropriate $y$ for $n=3,4$, but the above example with $n=5$ seemed neater. Obviously, $n>5$ can be constructed the same way as the above example.
Also, notice that
$$f(x) := \frac{m(x) + M(x)}{2} + rh(x)$$
also satisfies $1-3$ for any $r \in \mathbb{R}$, which gives us pretty much freedom in constructing examples for $n=3,4$.
</s>