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Let $(M^n,g)$ be a closed Riemannian manifold. Define $$\lambda(g)=\inf\{\mathcal{F}(g,f),\;0<f\in C^{\infty}(M),\; \int_Mf^2\;d\nu=1\},$$ where $$\mathcal{F}(g,f)=\int_M\left(|\nabla f|^2+ af^2\log f + bRf^2\right)\;d\nu,$$ and $R$ is the scalar curvature of manifold.

Can one be confident that there is a $0<f_0\in C^{\infty}(M)$ with $\int_Mf_0^2\;d\nu=1$, such that $\lambda(g)=\mathcal{F}(g,f_0)$?

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    $\begingroup$ Looks like a job for the direct method. However, the functional is not convex in $f$ due to the term $f^2\log(f)$, so there is some work to do. Another problem may be the "open" constraint $f>0$. Both problems would not be there if there would be $f\log(f)$ instead of $f^2\log(f)$. Since I do not work on manifolds, I have no idea how to formulate the respective Banach spaces (should be similar to some Sobolev or Orlicz-Sobolev spaces…). $\endgroup$
    – Dirk
    Jan 10, 2019 at 12:58
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    $\begingroup$ Does your manifold have finite volume? Otherwise if you choose $f<\epsilon$ everywhere we have $\int f^2 log(f)< log(\epsilon) \int f^2= log(\epsilon)$ and $F(g,f)$ can get arbitrarily small. $\endgroup$
    – user100927
    Jan 14, 2019 at 12:57
  • $\begingroup$ Yes, the volume is finite. $\endgroup$
    – user162551
    Jan 14, 2019 at 21:03

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