Consider a one-parameter real analytic family of metrics $g_t$ on a compact manifold $M$ converging to a metric $g$ in $C^k$-norm, for some $k$. It is known that the Laplace spectrum of $g_t$ will converge to the Laplace spectrum of $g$. However, is it true that $\lambda_n(t)$ will converge to $\lambda_n$, where $0 \leq \lambda_1(t) \leq \lambda_2(t) \leq \cdots$ denotes the Laplace spectrum of $g_t$, and $0 \leq \lambda_1 \leq \lambda_2 \leq \cdots$ denotes the Laplace spectrum of $g$? Thanks!

Addendum: I just found the following paper. Regarding the theorem on page 1, I wish to ask more generally: the theorem says that if the operators $A(t)$ are $C^M$, where $M$ could be $\omega, \infty$ or Holder class, then the eigenvalues of $A(t)$ can be parametrized (under certain conditions) to be $C^M$. But does that mean the following: if the operators $A(t)$ are positive, self-adjoint and have discrete spectrum, and for each $t$, we have $0 \leq \lambda_1(t) \leq \lambda_2(t) \leq....$ as the spectrum of $A(t)$, then are $\lambda_i(t)$ parametrized to be $C^M$?

Edit: I am new to MO and MSE and unfortunately was not aware of the practices. I have edited the MSE question requesting people to post their answers here instead.

Further edit: As commented below, the existence of the limit metric is not that important. All I want to know is, if we rearrange the eigenvalues at each step according to increasing order, will the rearranged functions still be $C^M$?

cross each other. In the ordering given by the OP, two otherwise smooth trajectories would develop kinks. $\endgroup$