# Mariş's Theorem

During a literature search (to answer my MathOverflow question concerning symmetries of "ground states" in variational problems, I came across a very nice theorem due to Mihai Mariş. The theorem itself is, more than anything else, a statement about the geometry of Euclidean spaces. I will give a rather elementary write-up of (a special case of) the theorem here. (The proof presented here can equally well be applied to get the full strength of the theorem as presented in Maris's paper; I give just the special case for clarity of the discussion.)

Theorem    [Mariş; rough version]
Fix a $C^1(\mathbf{R})$ function $V$, and fixed a dimension $d \geq 2$. Let us study the constrained minimization problem where the functional to minimize is the energy functional $\int_{\mathbf{R}^d} |\nabla f|^2 ~\mathrm{d}x$, and the constraint is that $\int V(f) ~\mathrm{d}x$ is required to be a fixed constant.
Then a minimizer, should it exist, must be radially symmetric about some point $x_0$.
Remark
For motivation and historical context, see the MathOverflow question referenced above.

The proof of Mariş's theorem boils down to mostly a statement about the geometry of the Euclidean spaces, using the following properties of minimizers:

1. Under suitable assumptions on the growth rate of $V$ near $\infty$, by using Lagrange multipliers we see that minimizers satisfy a semilinear elliptic partial differential equations, and by elliptic regularity the minimizers are always $C^{1,\alpha}$.
2. Given any minimizer, if we act on it with a symmetry of $\mathbf{R}^d$ we get another minimizer.
3. Given $f$ a minimizer: let $\Pi$ be a hyperplane such that the integral of $V(f)$ on each of the two halves partitioned by $\Pi$ are equal, then the integral of $|\nabla f|^2$ on each of the two halves are also equal. (For if not, replace $f$ on the larger half by its mirror image from the smaller half. This function is still in $W^{1,2}$ (it is continuous along $\Pi$), and this new function would have lower energy than $f$.)

These properties allow us to reduce Mariş's theorem to a statement about sets of $C^1$ functions on $\mathbf{R}^d$ satisfying certain closure properties with respect to a certain functional. We begin by first setting up the notation and formalizing the three properties above in a convenient-to-use way for the geometric argument.

## Notations

Throughout we let $\mathcal{Y}$ be a set of functions defined over $\mathbf{R}^d$ for some $d \geq 2$; this $\mathcal{Y}$ will be the domain of our functionals. We will denote by $\mathcal{X}$ a subset of $\mathcal{Y}$ to be described below.

By $S\mathbf{R}^d$ I denote the tangent sphere bundle of Euclidean $d$-space.1 Notice that we can freely identify elements of $S\mathbf{R}^d$ with oriented hyperplanes in $\mathbf{R}^d$; given a position $x$ and a unit vector $v$ based at $x$, we can associate to it the hyperplane through $x$ orthogonal to $v$, with positive orientation given by the direction of $v$. This identification is of course not-unique: given another $y$ on the same hyperplane, $(y,v)$ defines the same hyperplane with the same orientation as $(x,v)$.

By $\mathfrak{g}$ we denote a function $\mathcal{Y}\times S\mathbf{R}^d \to \mathbf{R}$; we put its first argument in square brackets and its second argument in parentheses. We will refer to $\mathfrak{g}$ as the "constraint function(al)".

By $\Gamma$ we denote the set of all isometries of $\mathbf{R}^d$, which induces also a mapping of $S\mathbf{R}^d$ to itself. We will note a few special elements:

• For $(x,v)\in S\mathbf{R}^d$ we let $r_{x,v}: y\mapsto y - [ 2(y-x)\cdot v] v$ be the reflection over the corresponding hyperplane.
• Similarly, if $V = \{v_1,v_2,\ldots,v_k\}$ is a orthonormal set of vectors, we let $r_{x,V} = r_{x,v_1}r_{x,v_2}\cdots r_{x,v_k}$. Notice that the operations on the right hand side commute because the $v_i$'s are mutually orthogonal.
• And lastly, we use the short hand $r_x: y\mapsto 2x - y$ the complete reflection about the point $x$, which is equivalent to $r_{x,V}$ when $V$ is a set of orthonormal basis vectors.
• For $v,w\in \mathbb{S}^{d-1}$ such that $v\perp w$, and for $\theta\in \mathbf{R}$, we let $\rho^\theta_{x,v,w}$ denote the rotation in the plane spanned by $\{v,w\}$ through $x$ by angle $\theta$, with direction of the rotation determined by the ordering $v,w$ (as opposed to $w,v$). Analytically we can write it as sending $y \mapsto y + (y\cdot v)[ (\cos \theta - 1)v + \sin\theta w] + (y\cdot w)[-\sin\theta v + (\cos\theta -1)w].$

For each $(x,v)\in S\mathbf{R}^d$ we define three "projection" maps on $\mathbf{R}^d$:

• $\pi_{x,v} : y\mapsto y - (y-x)\cdot v$,
• $\pi^+_{x,v}$ which acts as the identity if $(y-x) \cdot v \geq 0$ and as $r_{x,v}$ otherwise,
• $\pi^-_{x,v}$ which acts as the identity if $(y-x)\cdot v \leq 0$ and as $r_{x,v}$ otherwise.

Note that $\pi_{x,v}$ is the projection map to the hyperplane defined by $(x,v)$, and $\pi^+_{x,v}$ is the projection map to the positive half space; elements in the negative half space is mapped to their image under reflection across the corresponding hyperplane to $(x,v)$. Also observe that $\pi^{-}_{x,v} = \pi^+_{x,-v}$.

We extend the above definition to sets $V$ of orthonormal vectors:

• $\pi_{x,V} = \pi_{x,v_1}\pi_{x,v_2} \cdots \pi_{x,v_k}$ and
• $\pi_{x,V}^\perp = 1 - \pi_{x,V}$

are the orthogonal projections to the subspace orthogonal to $V$ and the subspace spanned by $V$ respectively (the subscript denotes the directions to project away, not the ones to keep).

## Geometry background

We note a few facts concerning our definitions above.

1. $r_{x,v} \pi^{\pm}_{x,v} = \pi^{\mp}_{x,v}$ and $\pi^{\pm}_{x,v} r_{x,v} = \pi^{\pm}_{x,v}$.
2. If $v\perp w$, we have $r_{x,v}\pi^{\pm}_{x,w} = \pi^{\pm}_{x,w} r_{x,v}$, and $\pi^-_{x,v}\pi^+_{x,w} = \pi^+_{x,w}\pi^-_{x,v}$.
3. In general, given $v\neq w$, let $u \perp v$ be in the span of $\{v,w\}$, then there exists $\theta$ such that $r_{x,v}r_{x,w} = \rho^\theta_{x,v,u}$. Let $h$ be a continuous function on the span of $\{v,w\}$ invariant under this rotation, we have three cases
• $\theta / \pi$ is irrational. Then orbit of $v$ under $\left( \rho^\theta_{x,v,u}\right)^k$ is dense on the unit circle. Hence $h$ must be radially symmetric on this plane.
• $\theta / \pi = p/q$ where $p$ is odd. Then $r_{x,v}$ and $r_{x,w}$ generate a dihedral group of an even polygon, and $r_{x,\{v,u\}}$ is a symmetry of $h$.
• $\theta/\pi = p/q$ where $p$ is even and $q$ is odd. Then $r_{x,v}$ and $r_{x,w}$ generate a dihedral group of an odd polygon, and hence the two reflections are conjugate by a reflection. (To see this, let $k$ be the order of the element $r_{x,v}r_{x,w}$, by assumption $k$ is odd. Hence $\left(r_{x,v}r_{x,w}\right)^kr_{x,v} = r_{x,v}$. Letting $u'$ be a vector such that $r_{x,u'} = \left(r_{x,v}r_{x,w}\right)^{\frac{k-1}{2}}r_{x,w}$, the LHS can be written as $r_{x,u'}r_{x,w} r_{x,u'}$. Observe that $r_{x,u'}$ is a symmetry of $h$ which swaps the spans of $v$ and $w$.)

## Hypotheses

We will state our theorems and lemmas in terms of the following hypotheses imposed on the set $\mathcal{X}$ and the constraint functional $\mathfrak{g}$.

Hypothesis    [on the constraint]
We assume that

1. For fixed $f \in \mathcal{Y}$ and $x\in\mathbf{R}^d$, the function $\mathfrak{g}[f](x,\cdot):\mathbb{S}^{d-1} \to \mathbf{R}$ is continuous.
2. The function $\mathfrak{g}$ is odd under reflection: $\mathfrak{g}[f](x,v) = - \mathfrak{g}[f](x,-v)$.
3. The function $\mathfrak{g}$ only depends on the hyperplane: if $(x,v)$ and $(y,v)$ define the same hyperplane, then $\mathfrak{g}[f](x,v) = \mathfrak{g}[f](y,v)$.
4. For every $f, (x, v)$, there exists a $\lambda \in \mathbf{R}$ such that $\mathfrak{g}[f](x-\lambda v, v) = 0$.
5. If $f\in \mathcal{Y}$ is such that $f = f \circ r_{x,v}$ (so that $f$ is symmetric across the corresponding hyperplane), then $\mathfrak{g}[f](x,v) = 0$.
6. $\mathfrak{g}$ is symmetric under Euclidean group of motions: if $\gamma \in \Gamma$ is an isometry, then for any $f\in \mathcal{Y}$, we have that
• $f \circ \gamma \in \mathcal{Y}$
• $\mathfrak{g}[f\circ \gamma](\gamma(x,v)) = \mathfrak{g}[f](x,v)$
Hypothesis    [on the solutions]

We assume the following closure conditions on $\mathcal{X}\subset \mathcal{Y}$:

1. If $f\in \mathcal{X}$ and $(x,v)\in S\mathbf{R}^d$ is such that $\mathfrak{g}[f](x,v) = 0$, then $f\circ \pi^{\pm}_{x,v} \in \mathcal{X}$.
2. If $f\in \mathcal{X}$ and $\gamma \in \Gamma$, then $f\circ \gamma \in \mathcal{X}$.

Additionally, we assume that $\mathcal{X} \subset C^1(\mathbf{R}^d)$.

We our hypotheses stated, we can now state the main geometric result.

Theorem    [Mariş; geometric]
Fix $d \geq 2$. Let the constraint function $\mathfrak{g}$ and the solution set $\mathcal{X}$ satisfy the two sets of hypotheses above. Then for any element $f\in \mathcal{X}$ there exists a point $x_0\in\mathbf{R}^d$ such that $f$ is radially symmetric about $x_0$.

We can reduce Theorem 1 to this theorem by setting $\mathcal{Y} = W^{1,2}$, letting $\mathcal{X}$ be the set of minimizers, and setting the functional $\mathfrak{g}[f](x,v) := \int_{\pi^+_{x,v}\mathbf{R}^d} V(f) ~\mathrm{d}x - \int_{\pi^-_{x,v}\mathbf{R}^d} V(f) ~\mathrm{d}x,$ the difference between the integral of $V(f)$ between the two half spaces demarcated by $(x,v)$.

That the items of Hypothesis 1 is satisfied follows mainly from the properties of Lebesgue integration. That the items of Hypothesis 2 is satisfied follow from the discussion in the introductory paragraphs of this post.

Remark

The dimensional restriction is necessary. In dimension 1, if you let $\mathfrak{g}[f](x,v) = v\cdot f'$, then we can allow $\mathcal{X} = \mathcal{Y}$ to be the space of all continuously differentiable functions with compact support, and the theorem is clearly false in that case.

One should think of the theorem as providing a very stringent constraint on how many elements there can be in $\mathcal{X}$.

## Intermediate lemmas and proof of the theorem

We will use, without proof, the Borsuk-Ulam Theorem

Theorem    [Borsuk-Ulam]
A continuous odd map $\phi:\mathbb{S}^{d_1} \to \mathbf{R}^{d_2}$ where $d_1 \geq d_2 \geq 1$ admits a vector $v$ such that $\phi(v) = 0$.

Now some technical lemmas.

Lemma
Let $f,x$ be fixed. If $V$ is a set of orthonormal vectors such that $\mathfrak{g}[f](x,v) = 0$ for all $v\perp V$, we have that $f(y) = f(z)$ whenever $\pi_{x,V}^\perp y = \pi_{x,V}^\perp z$ and $|\pi_{x,V} y| = |\pi_{x,V} z|$. In this case we say $f$ is radially symmetric about $(x,V)$.
Proof:
Since $f$ is continuously differentiable, it suffices to show that all "angular derivatives" vanish, i.e. that for every $v\perp V$, $v\cdot \nabla f(y) = 0$ when $y\neq x, (y-x)\cdot v = 0$. By hypothesis since $\mathfrak{g}[f](x,v) = 0$, $f\circ \pi^+_{x,v} \in \mathcal{X}$ and so is $C^1$. Hence for $y = \pi_{x,v} y$ we have that $0 = v\cdot \nabla (f\circ \pi^+_{x,v})(y) = v\cdot \nabla f(y)$ as needed.

Lemma
Let $f\in\mathcal{X}$ and $(x,v)\in S\mathbf{R}^d$, such that $f$ is radially symmetric about $(x,v)$. Then there exists $\lambda\in\mathbf{R}$ such that $f$ is radially symmetric about $(x-\lambda v,\{\})$.
Proof:

By our hypothesis, there exists some $\lambda$ such that $\mathfrak{g}[f](x-\lambda v,v) = 0$. Denote by $z = x-\lambda v$ for convenience.

Consider the function $f^+ = f \circ \pi^+_{z,v}$, which by definition is symmetric under the reflection $r_{z,v}$. Clearly $f^+$ is still radially symmetric about $(z,v)$. So now let $w$ be an arbitrary direction $\neq v$. There exists a unique direction $u\perp v$ such that $w$ is in the span of $\{u,v\}$. Since $u\perp v$ and $f^+$ is radially symmetric about $(z,v)$, we must also have $f^+$ being symmetric under the reflection $r_{z,u}$. This implies that $f^+ = f^+\circ \rho^{\pi}_{z,u,v}$ using that $\rho^\pi_{z,u,v} = r_{z,\{u,v\}}$. By our hypothesis then $\mathfrak{g}[f^+](z,w) = \mathfrak{g}[f^+](z,-w)$ using that the isometry $\rho^{\pi}_{z,u,v}(z,w) = (z,-w)$. This implies therefore $\mathfrak{g}[f^+](z,w) = 0.$ Since $w\neq v$ is arbitrary, this implies that $\mathfrak{g}[f^+](z,w) = 0$ for any $w$. Combined with the fact that $z$ is chosen so that $\mathfrak{g}[f^+](z,v) = 0$, we have, by the previous Lemma, that $f^+$ must be radially symmetric about $(z,\{\})$.

The same argument also shows that $f^-$ is radially symmetric. As $f^+$ and $f^-$ agree along the hyperplane defined by $(x,v)$, we conclude that $f = f^+ = f^-$ everywhere and is radially symmetric.

Corollary    [2 dimensions]
If $d = 2$, and $f\in \mathcal{X}$, then there exists $x$ such that $f$ is radially symmetric about $(x,\{\})$.
Proof:

Choose $y$ arbitrarily. By our hypotheses $\mathfrak{g}[f](y,\cdot)$ is an odd continuous function from the circle to the reals. Hence it must vanish for some direction $u$. Let $v$ be orthogonal to $u$. Then we can choose $x = y + \lambda v$ such that $\mathfrak{g}[f](x,v) = \mathfrak{g}[f](x,u) = 0$.

Now, $f^{++}:= f \circ \pi^+_{x,u} \pi^+_{x,v}$ is invariant under rotation by angle $\pi$, and by construction belongs to $\mathcal{X}$. Hence the same argument as the previous Lemma shows that $f^{++}$ is radially symmetric. Similarly we can construction $f^{+-}, f^{-+}, f^{--}$ and conclude they are all radially symmetric about $x$. The overlaps along the $\{u,v\}$ axes then allow us to conclude that $f = f^{++} = f^{--}= f^{+-} = f^{-+}$ and is radially symmetric.

Lemma
Let $O$ be a set of orthonormal basis vectors. If $f\in\mathcal{X}$ is radially symmetric about $(x,V)$ and $(x,W)$ where $V,W\subseteq O$, and $V^\perp \cap W^\perp \neq \emptyset$, then $f$ is radially symmetric about $V\cap W$.
Proof:
We can write $O = (V\cap W) \cup (V^\perp \cap W^\perp) \cup (V\cap W^\perp) \cup (W\cap V^\perp)$, and hence every point $y$ in $\mathbf{R}^d$ can be written as a linear combination of $p+q+r+s$ each belonging to a subspace spanned by one of the four sets. The two radial symmetries imply that $f(p,q,r,s) = f(p,\sqrt{q^2+r^2},0,s) = f(p,\sqrt{q^2 + r^2 + s^2},0,0)$. This demonstrates the radial symmetry.

With this set up, we can prove the theorem by induction on dimension.

Proposition    [Induction hypothesis]
If $d \geq 3$, and $f\in \mathcal{X}$. For every $x$ there is a $d-2$ dimensional subspace such that $f$ is radially symmetric about $(x,V)$.
Proof:
Since $d \geq 3$, using the Borsuk-Ulam theorem we can choose $v$ such that $\mathfrak{g}[f](x,v) = 0$. Let $f^\pm = f\circ \pi^\pm_{x,v}$. Applying the Borsuk-Ulam theorem to the continuous functions $\mathfrak{g}[f^\pm](x,\cdot):\mathbb{S}^{d-1}\cap v^\perp$ we get two vectors $w^\pm$ for which $\mathfrak{g}[f^\pm](x,w^\pm) = 0$. Writing $K^\pm = \{v,w^\pm\}^\perp$, the argument in the proof of Lemma 9 suffices to show that $f^\pm$ are radially symmetric about $(x,K^\pm)$ respectively. We split into several cases depending on how $w^\pm$ compare. Let $K = K^+\cap K^-$.

1. $w^+ = \pm w^-$. Then $K^+ = K^- = K$. Hence $f^\pm$ are two functions radially symmetric about $(x,K)$, and such that they agree on $\pi^\perp_{x,K\cup\{w^+\}}$, so they agree wholly and $f$ is radially symmetric about $(x,K)$.
2. Suppose $r_{x,w^+}r_{x,w^-}$ generate the circle in the plane spanned by $w^\pm$. Then from the definition of radial symmetry we see that $f^\pm$ are both radially symmetric around $(x,K\cup \{v\})$, and hence by Lemma 11 radially symmetric around $(x,K)$. Since they agree on $\pi_{x,v}$, we get $f^\pm = f$ is also radially symmetric..
3. Suppose that $r_{x,w^+}r_{x,w^-}$ generate the dihedral group for an even polygon. Letting $u,u'$ be an orthonormal basis for the span of $w^\pm$, we have then $r_{x,\{u,u'\}}$ is a symmetry of $f^\pm$. Let $u''$ be in the span of $u,u'$, we have that using the invariance property of $\mathfrak{g}$ under isometries, $\mathfrak{g}[f](x,-u'') = \mathfrak{g}[f\circ r_{x,\{u,u'\}}](r_{x,\{u,u'\}}(x,u'')) = \mathfrak{g}[f](x,u'')$. This implies, by Lemma 8, that both $f^\pm$ are radially symmetric around $(x,K\cup\{v\})$. This implies via Lemma 11 that $f = f^\pm$ are radially symmetric around $(x,K)$.
4. Lastly, we have the odd polygon dihedral case. By assumption $f^\pm$ agree on $\pi_{x,v}$, and by radial symmetry is completely determined by the values there. By the discussion previously on geometry, we have that there exists some $u'$ such that $f^\pm |_{\pi_{x,v}(\mathbf{R}^d)}$ is symmetric under $r_{x,u'}$. A direct computation shows that thus $f^\pm \circ r_{x,u'} \circ r_{x,v} = f^\mp$. This implies that $f \circ r_{x,\{u',v\}} = f$. So for any $u$ that is a linear combination of $u',v$ we have that $\mathfrak{g}[f](x,u) = \mathfrak{g}[f](x,-u) = 0$ (again using the action of $\mathfrak{g}$ under isometry). Hence $f$ is radially symmetric about $(x,\{u',v\}^\perp)$.

Proposition    [Induction step]
If $f\in\mathcal{X}$ is radially symmetric about $(x,V)$ where $V$ spans a subspace of dimension at least 2, then there exists $w\in V$ such that $f$ is radially symmetric about $(x,V\cap\{w\}^\perp)$.
Proof:
Using Borsuk-Ulam theorem again, there exists $w\in V$ such that $\mathfrak{g}[f](x,w) = 0$. We can then use the same method as the proof of Lemma 9 to conclude that $f$ is radially symmetric about $V\cap \{w\}^\perp$.

After the induction brings $V$ down to only 1 dimensional, the theorem follows by applying Lemma 9.

1. If you are not familiar with the tangent sphere bundle, just think of it as the Cartesian product $\mathbf{R}^d \times \mathbb{S}^{d-1}$, or ordered pairs $(x,v)$ where $x$ is a point in $\mathbf{R}^d$ and $v$ is a unit vector. ^
##### Willie WY Wong
###### Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.