Q:

The choice of the eigenfunction of Laplacian

$M$ is a closed Riemannian manifold and $\lambda_1>0$ is the first nontrivial eigenvalue of $\Delta$. Can we find a eigenfunction $f$ of $\lambda_1$ such that $\mathop {\sup }\limits_M f - \mathop {\inf }\limits_M f = 2$ and $\mathop {\inf }\limits_M f \geq-1$?

A:

Yes. Let $f$ be a eigenfunction of $\Delta$ with respect to $\lambda_1$. By multiplying a constant, we can assume that $\sup f - \inf f = 2$ (It can be found as $f$ is nonconstant). If $\inf f \geq -1$, we are done. If $\inf f < -1$, then $\sup f < 1$, and $h=-f$ satisfies $\inf h \geq -1$.

