There is no need to consider a Riemannian cover here so I’m not sure what role that plays. As such, I will drop the tildes from the metrics.

The key idea is the classical result that it is possible to characterize the $k$-th eigenvalue as the following min-max problem.

$$\lambda_k = \inf_{V \subset C^2(M), \, dim(V) = k} \max_{f \in V \backslash \{0\}} \frac{\int_M |\nabla f|^2 \,dVol} {\int_M f^2\, dVol}$$

Intuitively, the infimum is realized when $V$ is the subspace spanned by the first $k$-eigenfunctions and then the maximization picks out the largest value of the Raleigh quotient. For simplicity denote the integral quotient in the previous equation as $R(f)$.

To prove the inequalities in your question, first consider a single function $f \in C^2(M)$. We then have that the Raleigh quotient satisfies $$ \frac{\int_M \frac{1}{b^2} |\nabla f|_g^2 \, \frac{1}{b^n} dVol_g} {\int_M f^2\, \frac{1}{a^n} dVol_g} \leq \frac{\int_M |\nabla f|_{g_0}^2 \,dVol_{g_0}} {\int_M f^2\, dVol_{g_0}}, $$ which follows from bounds on the metric and the way that the volume form and gradient scale with the metric. Note that the left hand side is simply $\frac{a^n}{b^{n+2}}$ times the Rayleigh quotient with respect to $g$.

To obtain the first inequality, we then consider the space $V$ spanned by the first $k$ eigenfunctions with respect to $g_0$. From the previous inequality, we have that for every $f \in V$, $\frac{a^n}{b^{n+2}} R_g(f) \leq R_{g_0}(f)$. For the function $f_k \in V$ which maximizes the quantity $$\frac{\int_M |\nabla f_k|_g^2 \, dVol_g} {\int_M f_k^2\, dVol_g},$$ we must have that $$\frac{a^n}{b^{n+2}} R_g(f_k) \leq R_{g_0}(f_k) \leq \max_{f \in V} R_{g_0}(f) = \lambda_k(g),$$ where the final inequality holds by the definition of $V$. To finish the proof, we note that $V$ might not be the subspace which realizes the infimum of the Rayleigh quotient with respect to $g$, so we need to minimize the left hand side over all subspaces. However, doing so makes the Rayleigh quotient smaller, so we find the desired inequality $$\frac{a^n}{b^{n+2}} \lambda_k(g_0) \leq \lambda_k(g).$$

For the other inequality, we simply switch the roles of $g$ and $g_0$ and argue in the same way.

There is no need to consider a Riemannian cover here so I’m not sure what role that plays. As such, I will drop the tildes from the metrics.

The key idea is the classical result that it is possible to characterize the $k$-th eigenvalue as the following min-max problem.

$$\lambda_k = \inf_{V \subset C^2(M), \, dim(V) = k} \max_{f \in V \backslash \{0\}} \frac{\int_M |\nabla f|^2 \,dVol} {\int_M f^2\, dVol}$$

Intuitively, the infimum is realized when $V$ is the subspace spanned by the first $k$-eigenfunctions and then the maximization picks out the largest value of the Raleigh quotient. For simplicity denote the integral quotient in the previous equation as $R(f)$.

To prove the inequalities in your question, first consider a single function $f \in C^2(M)$. We then have that the Raleigh quotient satisfies $$ \frac{\int_M \frac{1}{b^2} |\nabla f|_g^2 \, \frac{1}{b^n} dVol_g} {\int_M f^2\, \frac{1}{a^n} dVol_g} \leq \frac{\int_M |\nabla f|_{g_0}^2 \,dVol_{g_0}} {\int_M f^2\, dVol_{g_0}}, $$ which follows from bounds on the metric and the way that the volume form and gradient scale with the metric. Note that the left hand side is simply $\frac{a^n}{b^{n+2}}$ times the Rayleigh quotient with respect to $g$.

To obtain the first inequality, we then consider the space $V$ spanned by the first $k$ eigenfunctions with respect to $g_0$. From the previous inequality, we have that for every $f \in V$, $\frac{a^n}{b^{n+2}} R_g(f) \leq R_{g_0}(f)$. For the function $f_k \in V$ which maximizes the quantity $$\frac{\int_M |\nabla f_k|_g^2 \, dVol_g} {\int_M f_k^2\, dVol_g},$$ we must have that $$\frac{a^n}{b^{n+2}} R_g(f_k) \leq R_{g_0}(f_k) \leq \max_{f \in V} R_{g_0}(f) = \lambda_k(g_0),$$ where the final inequality holds by the definition of $V$. To finish the proof, we note that $V$ might not be the subspace which realizes the infimum of the Rayleigh quotient with respect to $g$, so we need to minimize the left hand side over all subspaces. However, doing so makes the Rayleigh quotient smaller, so we find the desired inequality $$\frac{a^n}{b^{n+2}} \lambda_k(g) \leq \lambda_k(g_0).$$

For the other inequality, we simply switch the roles of $g$ and $g_0$ and argue in the same way.

It is possible to characterize the $k$-th eigenvalue as the following min-max problem.

$$\lambda_k = \inf_{V \subset C^2(M), \, dim(V) = k} \max_{f \in V \backslash \{0\}} \frac{\int_M |\nabla f|^2 \,dVol} {\int_M f^2\, dVol}$$

Intuitively, $V$ is the subspace spanned by the first $k$-eigenfunctions and the maximization picks out the largest value of the Raleigh quotient.

To prove the inequalities in your question, fix a substance $V$ and bound the numerator and denominator in terms of $a$ and $b$. The power of $n$ appears because this is how the volume form scales under rescaling and the extra power of $2$ comes from the scaling of the gradient term.

To obtain the first inequality, we consider the space $V$ spanned by the first $k$ eigenfunctions with respect to $g$. Then we can bound the Raleigh quotient with respect to $\tilde g$ and then pass to the infimum. For the second inequality, we instead use the space spanned by the first $k$ eigenfunctions of $\tilde g$.

Also, there is no need to consider a Riemannian cover here so I’m not sure what role that plays.

There is no need to consider a Riemannian cover here so I’m not sure what role that plays. As such, I will drop the tildes from the metrics.

The key idea is the classical result that it is possible to characterize the $k$-th eigenvalue as the following min-max problem.

$$\lambda_k = \inf_{V \subset C^2(M), \, dim(V) = k} \max_{f \in V \backslash \{0\}} \frac{\int_M |\nabla f|^2 \,dVol} {\int_M f^2\, dVol}$$

Intuitively, the infimum is realized when $V$ is the subspace spanned by the first $k$-eigenfunctions and then the maximization picks out the largest value of the Raleigh quotient. For simplicity denote the integral quotient in the previous equation as $R(f)$.

To prove the inequalities in your question, first consider a single function $f \in C^2(M)$. We then have that the Raleigh quotient satisfies $$ \frac{\int_M \frac{1}{b^2} |\nabla f|_g^2 \, \frac{1}{b^n} dVol_g} {\int_M f^2\, \frac{1}{a^n} dVol_g} \leq \frac{\int_M |\nabla f|_{g_0}^2 \,dVol_{g_0}} {\int_M f^2\, dVol_{g_0}}, $$ which follows from bounds on the metric and the way that the volume form and gradient scale with the metric. Note that the left hand side is simply $\frac{a^n}{b^{n+2}}$ times the Rayleigh quotient with respect to $g$.

To obtain the first inequality, we then consider the space $V$ spanned by the first $k$ eigenfunctions with respect to $g_0$. From the previous inequality, we have that for every $f \in V$, $\frac{a^n}{b^{n+2}} R_g(f) \leq R_{g_0}(f)$. For the function $f_k \in V$ which maximizes the quantity $$\frac{\int_M |\nabla f_k|_g^2 \, dVol_g} {\int_M f_k^2\, dVol_g},$$ we must have that $$\frac{a^n}{b^{n+2}} R_g(f_k) \leq R_{g_0}(f_k) \leq \max_{f \in V} R_{g_0}(f) = \lambda_k(g),$$ where the final inequality holds by the definition of $V$. To finish the proof, we note that $V$ might not be the subspace which realizes the infimum of the Rayleigh quotient with respect to $g$, so we need to minimize the left hand side over all subspaces. However, doing so makes the Rayleigh quotient smaller, so we find the desired inequality $$\frac{a^n}{b^{n+2}} \lambda_k(g_0) \leq \lambda_k(g).$$

For the other inequality, we simply switch the roles of $g$ and $g_0$ and argue in the same way.