Symmetric Convex/Concave Functions: A Property of Their Extremizers
Abstract
This note discusses a fundamental property of symmetric convex (concave) functions: if a symmetric convex function attains a minimum, then there exists a minimizer lying on the diagonal; the analogous statement holds for symmetric concave functions. We present a rigorous treatment for the 2-dimensional case, an $n$-dimensional generalization, important corollaries, and remarks.
The core idea of the proof is: symmetry generates a set of symmetric extremizers, and convexity (via Jensen’s inequality) implies that their arithmetic mean — which necessarily lies on the diagonal — is also an extremizer. Differentiability is not required.
2-Dimensional Case
Theorem 1 (Minimum)
Let $F: \mathbb{R}^2 \to \mathbb{R}$ satisfy:
- Symmetry: $F(x, y) = F(y, x)$ for all $x, y \in \mathbb{R}$;
- Convexity: $F$ is convex;
- Existence of an extremum: $F$ has a local minimum on $\mathbb{R}^2$ (for a convex function, any local minimum is global).
Then there exists $t \in \mathbb{R}$ such that $(t, t)$ is also a (global) minimizer of $F$.
Proof
Let $(a, b)$ be a global minimizer of $F$, and denote $m = F(a, b) = \min_{(x,y) \in \mathbb{R}^2} F(x, y)$.
Step 1: Symmetry yields another minimizer.
By symmetry, $F(b, a) = F(a, b) = m$, so $(b, a)$ is also a global minimizer.
Step 2: Convexity controls the value at the midpoint.
Since $F$ is convex, Jensen’s inequality states that for any $\mathbf{u}, \mathbf{v} \in \mathbb{R}^2$ and $\lambda \in [0, 1]$,
Take $\mathbf{u} = (a, b)$, $\mathbf{v} = (b, a)$, $\lambda = \frac{1}{2}$. The midpoint
\[\frac{1}{2}(a, b) + \frac{1}{2}(b, a) = \left( \frac{a + b}{2},\; \frac{a + b}{2} \right)\]lies on the line $y = x$. Substituting into the inequality gives
\[F\!\left( \frac{a + b}{2},\; \frac{a + b}{2} \right) \le \frac{1}{2} F(a, b) + \frac{1}{2} F(b, a).\]Step 3: The minimum forces equality.
Since $F(a, b) = F(b, a) = m$, the right-hand side is $\frac{1}{2} m + \frac{1}{2} m = m$, hence
But $m$ is the global minimum, so no point can have value less than $m$. Therefore
\[F\!\left( \frac{a + b}{2},\; \frac{a + b}{2} \right) = m.\]Step 4: Conclusion.
The point $\displaystyle \left( \frac{a + b}{2},\; \frac{a + b}{2} \right)$ lies on the diagonal and attains the value $m$; setting $t = \dfrac{a + b}{2}$ completes the proof. $\square$
Theorem 1’ (Maximum)
If “convex” is replaced by “concave” and “minimum” by “maximum” in Theorem 1, the conclusion still holds: there exists $t \in \mathbb{R}$ such that $(t, t)$ is a (global) maximizer of $F$.
Proof
Let $(a, b)$ be a global maximizer of the concave function $F$, with maximum value $M = F(a, b)$. By symmetry, $(b, a)$ is also a maximizer.
For a concave function, Jensen’s inequality reverses:
\[F(\lambda \mathbf{u} + (1 - \lambda) \mathbf{v}) \ge \lambda F(\mathbf{u}) + (1 - \lambda) F(\mathbf{v}).\]Taking $\mathbf{u} = (a, b)$, $\mathbf{v} = (b, a)$, $\lambda = \frac{1}{2}$ yields
\[F\!\left( \frac{a + b}{2},\; \frac{a + b}{2} \right) \ge \frac{1}{2} M + \frac{1}{2} M = M.\]Since $M$ is the global maximum, equality must hold, and the midpoint is also a maximizer. $\square$
$n$-Dimensional Generalization
Let $F: \mathbb{R}^n \to \mathbb{R}$, and denote $x = (x_1, x_2, \dots, x_n)$.
$F$ is called symmetric if for every permutation $\sigma \in S_n$ (the symmetric group of degree $n$),
\[F(x_1, x_2, \dots, x_n) = F(x_{\sigma(1)}, x_{\sigma(2)}, \dots, x_{\sigma(n)}).\]Theorem 2 (Minimum)
Let $F: \mathbb{R}^n \to \mathbb{R}$ satisfy:
- Symmetry: as defined above;
- Convexity: $F$ is convex;
- Existence of an extremum: $F$ has a local minimum on $\mathbb{R}^n$.
Then there exists $t \in \mathbb{R}$ such that $(t, t, \dots, t) \in \mathbb{R}^n$ is also a (global) minimizer of $F$.
Proof Sketch
Let $\mathbf{x}^{(0)} = (x_1^{(0)}, x_2^{(0)}, \dots, x_n^{(0)})$ be a global minimizer of $F$ with value $m$. Consider all points obtained by permuting the coordinates:
\[\mathbf{x}_\sigma = (x_{\sigma(1)}^{(0)}, x_{\sigma(2)}^{(0)}, \dots, x_{\sigma(n)}^{(0)}), \quad \sigma \in S_n.\]By symmetry, each $\mathbf{x}_\sigma$ is a minimizer with value $m$. The set of minimizers of a convex function is convex, so their arithmetic mean
\[\frac{1}{n!} \sum_{\sigma \in S_n} \mathbf{x}_\sigma\]is also a minimizer. It is easy to verify that all coordinates of this mean are equal to the arithmetic mean of the original coordinates:
\[\left( \frac{1}{n} \sum_{i=1}^n x_i^{(0)},\; \frac{1}{n} \sum_{i=1}^n x_i^{(0)},\; \dots,\; \frac{1}{n} \sum_{i=1}^n x_i^{(0)} \right).\]Thus $t = \frac{1}{n} \sum_{i=1}^n x_i^{(0)}$ gives a diagonal minimizer $(t, t, \dots, t)$. $\square$
Theorem 2’ (Maximum)
If “convex” is replaced by “concave” and “minimum” by “maximum” in Theorem 2, the conclusion holds analogously.
Important Corollaries
Corollary 1 (Uniqueness)
If a symmetric convex function $F$ has a unique minimizer, then that point must lie on the diagonal $(t, t, \dots, t)$. Similarly, if a symmetric concave function has a unique maximizer, it must also lie on the diagonal.
Corollary 2 (Strictly Convex/Concave Case)
If $F$ is a strictly convex symmetric function and has a minimizer, then the minimizer is unique and lies on the diagonal. The analogous statement holds for strictly concave symmetric functions.
Corollary 3 (Constrained Optimization)
Let $C \subseteq \mathbb{R}^n$ be a symmetric convex set (i.e., if $x \in C$, then $\sigma(x) \in C$ for every permutation $\sigma \in S_n$), and let $F: C \to \mathbb{R}$ be a symmetric convex function. If $F$ attains a minimum on $C$, then there exists $t \in \mathbb{R}$ such that $(t, t, \dots, t) \in C$ and this point is a minimizer of $F$ on $C$.
Worked Example
Problem
Find the minimum of
\[f(x, y, z) = x^2 + y^2 + z^2 + xy + yz + zx - 2(x + y + z)\]over $\mathbb{R}^3$.
Solution Using the Theorem
1. Verify the hypotheses of the theorem
- Symmetry: Swapping any two variables leaves the expression unchanged, so $f$ is symmetric.
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Convexity: $f$ is a quadratic form plus a linear term. Its Hessian matrix is
\[H = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}.\]Compute the leading principal minors: \(\Delta_1 = 2 > 0,\quad \Delta_2 = \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 3 > 0,\quad \Delta_3 = \det(H) = 4 > 0,\) so $H$ is positive definite and $f$ is strictly convex.
- Existence of a minimum: Since the quadratic form is positive definite, $|(x, y, z)| \to \infty$ implies $f \to +\infty$. By continuity, $f$ attains a global minimum.
By the Diagonal Minimizer Theorem for Symmetric Convex Functions (Theorem 2), there exists a minimizer on the diagonal $x = y = z$.
2. Reduce to a one-dimensional problem
Set $x = y = z = t$:
\[f(t, t, t) = 3t^2 + 3t^2 - 6t = 6t^2 - 6t = 6\left(t - \frac12\right)^2 - \frac32.\]The minimum is $-\dfrac32$, attained at $t = \dfrac12$.
Hence the global minimum is $-\dfrac32$, achieved at $\left(\dfrac12, \dfrac12, \dfrac12\right)$.
3. Verification
Directly compute the partial derivative:
\[\frac{\partial f}{\partial x} = 2x + y + z - 2 = 0,\]and by symmetry the other two equations are identical, giving $x = y = z = \dfrac12$, confirming the result.
Summary
Without the theorem, one would need to solve a $3 \times 3$ linear system. Using the theorem, we only need to verify symmetry and convexity, then reduce the problem to minimizing a single-variable quadratic function — a significant simplification.
Remarks
Differentiability Is Not Required
The theorem does not require $F$ to be differentiable. Only convexity (or concavity) and the existence of an extremum are needed. The proof relies solely on Jensen’s inequality and the definition of an extremum — no gradients, Hessians, or other analytic tools are necessary.
Necessity of Convexity
Without convexity (or concavity), the conclusion generally fails. Two counterexamples are given below.
Counterexample 1:
\[F(x, y) = -[(x - y)^2 - 1]^2 - [e^{x+y} - (x - y)^2]^2\]is a differentiable symmetric function with two strict local maxima at $(0.5, -0.5)$ and $(-0.5, 0.5)$, yet has no critical point on the diagonal $y = x$.
Counterexample 2:
\[F(x, y) = \bigl((x-1)^2 + (y+1)^2\bigr) \bigl((x+1)^2 + (y-1)^2\bigr)\]is also differentiable and symmetric, with minimizers at $(1, -1)$ and $(-1, 1)$, but has no extremum (not even a stationary point) on the diagonal $y = x$.
Common Misconception
The theorem only guarantees the existence of an extremizer on the diagonal. It does not assert that all extremizers lie on the diagonal.
Related Literature
The result presented here does not typically appear as a named theorem in standard textbooks, but related ideas can be found in the following areas:
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Majorization Theory and Schur-Convex Functions: Symmetric convex functions are closely related to Schur-convex functions. A classical result states that if $F$ is a symmetric convex function, then $F$ is Schur-convex, and its minimum is attained at a point with equal components (i.e., on the diagonal). See Marshall, Olkin & Arnold (2011) Inequalities: Theory of Majorization and Its Applications.
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Convex Optimization Under Group Action: More generally, if a convex function $F$ is invariant under the action of a group $G$ and the feasible set is also $G$-invariant, then the set of minimizers is $G$-invariant. By convexity, the $G$-orbit average yields a $G$-invariant minimizer. The symmetric group considered in this note is a special case of this framework.
This theorem elegantly combines the “dimension reduction” effect of symmetry with the “convexity of the solution set” from convexity — a concise yet powerful tool in optimization, game theory, and the calculus of variations.
