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Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Update: Surprisingly this problem seems harder than it looks. So far the best estimates on $\mu_1(\Omega)$ that I found here: $$\mu_1(\Omega)\geq \frac{\pi^2}{D^2} + \frac{1}{2}$$ where $D=\operatorname{diameter}(\Omega).$ Thus if $\mu_1(\Omega)<2$ then $D> \frac{\pi}{\sqrt{3/2}}.$ This also gives a lower bound on the perimeter of $\Omega$ since $P(\Omega)\geq 2 D \geq \frac{2\sqrt{2}}{\sqrt{3}} \pi.$ On the other hand, it is also know that $\mu_1(\Omega)\leq \mu_1(B)$ where $B$ is a geodesic ball of the same volume as $\Omega.$

Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Update: Surprisingly this problem seems harder than it looks.

• So far the best estimates on $\mu_1(\Omega)$ that I found here: $$\mu_1(\Omega)\geq \frac{\pi^2}{D^2} + \frac{1}{2}$$ where $D=\operatorname{diameter}(\Omega).$ Thus if $\mu_1(\Omega)<2$ then $D> \frac{\pi}{\sqrt{3/2}}.$ This also gives a lower bound on the perimeter of $\Omega$ since $P(\Omega)\geq 2 D \geq \frac{2\sqrt{2}}{\sqrt{3}} \pi.$
• On the other hand, it is also known that $\mu_1(\Omega)\leq \mu_1(B)$ where $B$ is a geodesic ball of the same volume as $\Omega.$

Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Update: Surprisingly this problem seems harder than it looks. So far the best estimates on $\mu_1(\Omega)$ that I had found here the following: $$\mu_1(\Omega)\geq \frac{\pi^2}{D^2} + \frac{1}{2}$$ where $D=\operatorname{diameter}(\Omega).$ Thus if $\mu_1(\Omega)<2$ then $D> \frac{\pi}{\sqrt{3/2}}.$ This also gives a lower bound on the perimeter of $\Omega$ since $P(\Omega)\geq 2 D \geq \frac{2\sqrt{2}}{\sqrt{3}} \pi.$ On the other hand, it is also know that $\mu_1(\Omega)\leq \mu_1(B)$ where $B$ is a geodesic ball of the same volume as $\Omega.$

Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Update: Surprisingly this problem seems harder than it looks. So far the best estimates on $\mu_1(\Omega)$ that I found here: $$\mu_1(\Omega)\geq \frac{\pi^2}{D^2} + \frac{1}{2}$$ where $D=\operatorname{diameter}(\Omega).$ Thus if $\mu_1(\Omega)<2$ then $D> \frac{\pi}{\sqrt{3/2}}.$ This also gives a lower bound on the perimeter of $\Omega$ since $P(\Omega)\geq 2 D \geq \frac{2\sqrt{2}}{\sqrt{3}} \pi.$ On the other hand, it is also know that $\mu_1(\Omega)\leq \mu_1(B)$ where $B$ is a geodesic ball of the same volume as $\Omega.$

Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Question: Let $\Omega\subset \mathbb{S}^2_+$ be any geodesically convex subset of the hemisphere $\mathbb{S}^2_+$. Then is it true that $\mu_1(\Omega)\geq \mu_1(\mathbb{S}^2_+)$ where $\mu_1$ is the first (non-trivial) Neumann eigenvalue of the Laplacian?

The above statement clearly fails when $\mathbb{S}^2_{+}$ is replaced by a rectangle in the plane and $\Omega$ is a thin strip around the diagonal of the rectangle. Still, I am wondering if such a construction generalizes to the hemisphere as well.

Update: Surprisingly this problem seems harder than it looks. So far the best estimates on $\mu_1(\Omega)$ that I had found here the following: $$\mu_1(\Omega)\geq \frac{\pi^2}{D^2} + \frac{1}{2}$$ where $D=\operatorname{diameter}(\Omega).$ Thus if $\mu_1(\Omega)<2$ then $D> \frac{\pi}{\sqrt{3/2}}.$ This also gives a lower bound on the perimeter of $\Omega$ since $P(\Omega)\geq 2 D \geq \frac{2\sqrt{2}}{\sqrt{3}} \pi.$ On the other hand, it is also know that $\mu_1(\Omega)\leq \mu_1(B)$ where $B$ is a geodesic ball of the same volume as $\Omega.$