Background:

Let $\mathbb{S}^2_+$ be the hemisphere. Then we know that for $f:\mathbb{S}^2_+\to \mathbb{R}$ satisfying (when written in coordinates) $\int_{0}^{2\pi}\int_{-\pi/2}^{\pi/2}f(r,\theta)\sin(r)dr d\theta = 0$ we have $$\int_{0}^{2\pi}\int_{-\pi/2}^{\pi/2} |f|^2 \sin(r)drd\theta \leq \frac{1}{2}\int_{0}^{2\pi}\int_{-\pi/2}^{\pi/2} |\nabla f|^2 \sin(r)drd\theta.$$

Here the best constant $1/2$ can be computed by simply recalling the first eigenvalue on the hemisphere (which is equal to two) along with fact that the poincare constant is $\frac{1}{\lambda_1(\mathbb{S}^2_+)}.$

Question:

I am interested in understanding the case when the test functions in the above inequality are of the form $f(r,\theta)=f(r)$. In particular if we work with $\int_{-\pi/2}^{\pi/2} f(r)\sin(r)dr=0$ then what is the best constant $C>0$ in the inequality

$$\int_{-\pi/2}^{\pi/2} |f|^2 \sin(r)dr \leq C \int_{-\pi/2}^{\pi/2} |\partial_r f|^2 \sin(r)dr?$$

Background:

Let $\mathbb{S}^2_+$ be the hemisphere. Then we know that for $f:\mathbb{S}^2_+\to \mathbb{R}$ satisfying (when written in coordinates) $\int_{0}^{2\pi}\int_{0}^{\pi/2}f(r,\theta)\sin(r)dr d\theta = 0$ we have $$\int_{0}^{2\pi}\int_{0}^{\pi/2} |f|^2 \sin(r)drd\theta \leq \frac{1}{2}\int_{0}^{2\pi}\int_{0}^{\pi/2} |\nabla f|^2 \sin(r)drd\theta.$$

Here the best constant $1/2$ can be computed by simply recalling the first eigenvalue on the hemisphere (which is equal to two) along with fact that the poincare constant is $\frac{1}{\lambda_1(\mathbb{S}^2_+)}.$

Question:

I am interested in understanding the case when the test functions in the above inequality are of the form $f(r,\theta)=f(r)$. In particular if we work with $\int_{0}^{\pi/2} f(r)\sin(r)dr=0$ then what is the best constant $C>0$ in the inequality

$$\int_{0}^{\pi/2} |f|^2 \sin(r)dr \leq C \int_{0}^{\pi/2} |\partial_r f|^2 \sin(r)dr?$$