On page 15 of Spectral Graph Theory by Fan Chung, http://www.math.ucsd.edu/~fan/research/cb/ch1.pdf, before eq (1.14) is the step,
$\displaystyle || \sum_{i\neq 0} (1-\lambda_i)^s a_i \phi_i T^{\frac{1}{2}} || \leq (1-\lambda')^s\frac{\max_x \sqrt{d_x}}{\min_y \sqrt{d_y}}$.
Popping out the eigenvalue term is obvious and I've seen this inequality proven different ways, eg. http://www.ccs.neu.edu/home/rraj/Courses/7880/S12/Lectures/lec7_RandomWalks2.pdf, but would like to know why the max/min part should be obvious directly from the line above. Maybe I'm tired, but it just isn't jumping out at me.
To clarify, this relates to bounding the distance between the distribution resulting from a random walk and the stationary distribution.
Thanks.
Maybe she is estimating as: $$ || \sum_i a_i \phi_i T^{\frac{1}{2}} || \leq || T^{\frac{1}{2}} || \cdot || \sum_{i \neq 0} a_i \phi_i || \\ \leq || T^{\frac{1}{2}} || \cdot || \sum_{i} a_i \phi_i || \leq || T^{\frac{1}{2}} || \cdot || f T^{-\frac{1}{2}} || \\ \leq || T^{\frac{1}{2}} || \cdot || T^{-\frac{1}{2} } || \cdot || f|| \leq \frac{\max_x \sqrt{d_x}}{ \min_y \sqrt{d_y} }$$