I really need the help of this forum on this one. My supervisor and I are in disagreement about how a complex version of Mercer's theorem should look like. Since we couldn't find any sources for the complex case, we are both a bit puzzled about how the other one arrived at their conclusion (hence a literature reference would also be crucial and appreciated). Let $\mathcal{M}$ be a nice enough manifold. His version is:
Let $$Tf(\textbf{y}) := \int_{\mathcal{M}} K(\textbf{x},\textbf{y}) f(\textbf{x}) \, \mathrm{d} \mathbf{x}$$ Suppose $K$ is a continuous symmetric (or hermitian) non-negative definite kernel. Then there is an orthonormal basis $\varphi_n(\mathbf x)$ of $L_2(\mathcal M)$ consisting of eigenfunctions of $TK$ such that the corresponding sequence of eigenvalues $\lambda_1\ge \lambda_2\ge \ldots>0$ is nonnegative. The eigenfunctions corresponding to non-zero eigenvalues are continuous on $\mathcal M$ and $K$ has the representation $$ K(x,y)=\sum_{n=1}^{\infty }\lambda_{n}\,\varphi_{n}(x)\, \overline{\varphi_{n}(y)} $$
where the convergence is absolute and uniform with \begin{equation*} \lambda_n \varphi_n(y) =\int_{\mathcal M} K(x,y) \varphi_n(x) \mathrm d x \, . \end{equation*}
My version is:
Let $$Tf(\textbf{y}) := \int_{\mathcal{M}} K(\textbf{x},\textbf{y}) f(\textbf{x}) \, \mathrm{d} \mathbf{x}$$ Suppose $K$ is a continuous symmetric (or hermitian) non-negative definite kernel. Then there is an orthonormal basis $\varphi_n(\mathbf x)$ of $L_2(\mathcal M)$ consisting of eigenfunctions of $TK$ such that the corresponding sequence of eigenvalues $\lambda_1\ge \lambda_2\ge \ldots>0$ is nonnegative. The eigenfunctions corresponding to non-zero eigenvalues are continuous on $\mathcal M$ and $K$ has the representation $$ K(x,y)=\sum_{n=1}^{\infty }\lambda_{n}\,\color{red}{\overline{\varphi_{n}(x)}\, \varphi_{n}(y)} $$
where the convergence is absolute and uniform with \begin{equation*} \lambda_n \varphi_n(y) =\int_{\mathcal M} \color{red}{K(x,y) \varphi_n(x)} \mathrm d x \, . \end{equation*}
$\color{red}{\text{Expressions that were changed in response to the comments (conjugated to fit the proposed solution).}}$
He explains his solution as follows: Since the literature does not tell us explicitly about the complex case, we must assume that the complex case follows by almost exact analogy.
I explain my solution as follows: By assuming correctness of the stated theorem, we should be able to plug in a kernel $K$ with finite representation (i.e. finitely many non-zero eigenvalues) and show that the second equality for $\lambda_n \varphi_n(\mathbf{y})$ holds. I can even support this with numerical evidence from an algorithm that rests upon Mercer's theorem for this case -- but my supervisor wouldn't budge and I feel a bit icky trying to get behind his reasoning (barring obvious counterexamples)...
Thank you for any input on the matter!
Since there are too much comments, posting this as an answer. I think its just about change of basis. $T,f,K(x,y)$ written in basis $\phi_n$ is your formula and $T,f,K(x,y)$ written in basis $\phi^*_n$ is your advisors formula. Having said that both formulas are equally correct under the constraint $T\phi_n = \lambda_n \phi_n$ and $T\phi_n^* = \lambda_n \phi_n^*$.
I am assuming orthonormal basis $\{\phi_n\}$ are orthonormal w.r.t $L_2$ inner product and not inner product w.r.t kernel. If what your advisor said is correct, you will have the two equations i mentioned: $T\phi_n = \lambda_n \phi_n$ and $T\phi_n^* = \lambda_n \phi_n^*$. May be these 2 equations are your advisor's motivation.
These 2 conditions imply $T_{K-K^*}$ is a $0$ operator. This implies $K = K^*$. This implies $K$ is real. May be this can be your argument.