Let $\odot$ represents the point-wise product and $L$ is the Laplacian matrix (which is symmetric positive semidefinite matrix) and it can be diagonalized as $L = \Phi \Lambda\Phi^*$ and $\Phi = [\varphi_1, \ldots, \varphi_N]$ and $\Phi \Phi^* = \Phi^*\Phi = I$ and $\Lambda = Diag(\lambda_1, \ldots, \lambda_N)$ is the eigenvalue matrix such that $\lambda_i \geq 0$ for all $i$.
In this paper (page 3), the authors generalized a similarity notion on compact manifolds to the graph setting claiming that
$$ \alpha (\varphi_k,\varphi_\ell) \in [0,1]. $$ where $\varphi_k$ and $\varphi_\ell$ are the $k$-th and $\ell$-th eigenvectors of graph Laplacian matrix, respectively.
I am curious to write a rigorous mathematical reasoning for this. My derivations are as follows.
$$\alpha(\varphi_k,\varphi_\ell) = \frac{\|e^{-tL} (\varphi_k \odot \varphi_{\ell})\|}{\|\varphi_k \odot \varphi_{\ell}\|} \leq \frac{\|\Phi e^{-t\Lambda} \Phi^{*}\| \|\varphi_k \odot \varphi_{\ell}\|}{\|\varphi_k \odot \varphi_{\ell}\|} \leq \|e^{-t\Lambda}\| = e^{-t\lambda_{\min}}\leq 1, $$ where $\lambda_{\min}$ is the largest eigenvalue of $L$.
I just get confused about that. Would you please elaborate on this case?
Note: Based on the nice comment of @MaoWao, it is clear that the problem is about the "negative sign" in the definition of this metric.
Additional note: This question from the first concerning the graph setting (NOT the manifold - as the graph Laplcian matrix and its diagonalization are used and an example also provided for that). Moreover, the paper is actually about generalizing that concept from manifold to the graph setting addressing that equation when doing this generalization. Then the answer on manifold is not helpful and it is off-topic.
I have not read the paper carefully enough to understand what everything means, but this property might just follow by Cauchy-Schwarz. Because $p(t,x,\cdot)$ and $p(t,\cdot,y)$ are probability distributions, we have (using Fubini towards the end) \begin{align} \|e^{t\Delta}f\|_{L^2}^2 &= \int_M|e^{t\Delta}f(x)|^2\,dx = \int_M\left|\int_Mp(t,x,y)f(y)\,dy\right|^2\,dx\\ &\le \int_M\left(\int_Mp(t,x,y)\,dy\right)\left(\int_Mp(t,x,y)|f(y)|^2\,dy\right)\,dx\\ &= \int_M\int_Mp(t,x,y)|f(y)|^2\,dy\,dx\\ &= \int_M|f(y)|^2\int_Mp(t,x,y)\,dx\,dy\\ &=\|f\|_{L^2}^2. \end{align}