Random walk and isoperimetric constant I assume that a result of the following kind is known, and I would really appreciate a reference for it... Or at least, some hints as to where to start looking.

Theorem(?): Let $\varepsilon>0$ and Let $\Gamma=(V,E)$ be a finite regular non-bipartite graph of degree $d$, with $V=\{v_1,\dots, v_n\}$ and with isoperimetric constant $\delta>0$. Suppose we start at $v_1$ and take a random walk on $\Gamma$ for $t$ steps. Then, provided $t>f(\varepsilon, \delta, n, d)$, we have
  $$ \frac{1}{n}-\varepsilon \leq \mathbb{P}(\textrm{we are at vertex }v_i) \leq \frac{1}{n}+\varepsilon$$
  for every $i=1,\dots, n$.

Obviously, the proper version of this theorem would give an explicit form for the function $f(\varepsilon, \delta, n, d)$. I'm figuring such a function would need to depend on all of these four variables, although I'm not sure about $d$.
I'm also interested in the case where the random walk is "lazy"; indeed, I'm generally interested in any results of this type, although the precise form given above is my primary focus.
Any hints would be appreciated. Thanks in advance!
 A: Given the proper keywords, a quick search gives 
https://ocw.mit.edu/courses/mathematics/18-409-topics-in-theoretical-computer-science-an-algorithmists-toolkit-fall-2009/lecture-notes/MIT18_409F09_scribe4.pdf
as one of many many references.
Assume $\mu_t$ is the measure at the $t^\text{th}$-step of your random walk and $\pi$ is the uniform distribution (which is the stationary one in this case).
Then $\| \mu_t -\pi \|_{\ell^2} \leq 2 \rho^t$ where $\rho$ is the spectral radius (see below).
To get a bound on the $\ell^1$-norm, one can add a factor of $\sqrt{n}$ due to the distortion of the $\ell^2$ and $\ell^1$ norm in a $n$-dimensional vector space  $\| \mu_t -\pi \|_{\ell^1} \leq  \sqrt{n} 2 \rho^t$
In the lazy case, this estimate is worse (because the corresponding $\rho$ is bigger), but it works also on bipartite graphs.
If $\Delta$  is the Laplacian on your graph (normalised so that its norm is 2, i.e. $\Delta f(x) = f(x) - \tfrac{1}{d} \sum_{y \sim x} f(x)$), the the isoperimetric constant gives you a bound on the smallest non-zero eigenvalue, call it $\lambda$. There is some debate as to who you should name for this inequality (and whether one should write $\lambda_1$ and $\lambda_2$). But it depends on $d$, $\delta$ and that's it: $\lambda \geq \delta^2 / 2d^2$ (or $\lambda \geq 1 - \sqrt{ 1 - \delta^2/d^2}$ which is better but less known).
The spectral radius $\rho$ above is just $1-\lambda$ for the random walk, and $1-\lambda/2$ for the lazy walk.
Here is an outline of the proof. Your random walk is encoded as a matrix operator. This is $P = 1-\Delta$. Since $\Delta$ has spectrum in $[0,2]$ so $P$ has spectrum in $[-1,1]$. The (usual) lazy walk is given by $L = I - \tfrac{1}{2} \Delta$.
The estimate on the first positive eigenvalue of $\Delta$ (thanks to the isoperimetric constant), tells you that orthogonal to the constant function, the spectrum of $P$ is inside $[-1,1-\lambda]$ (and the spectrum of $L$ is inside $[0,1-\lambda/2]$.
The only "tricky" part for $P$, is that if the graph is not bipartite, you can prove there is nothing to worry about on the negative part of the spectrum, that is the spectrum is inside $[-1+\lambda,1-\lambda]$.
Now you know that if a vector lies orthogonal to the eigenspace for the eignevalue 1, then $\|Pv\|_2 \leq c \|v\|_2$ where $c$ is the next biggest eignevalue. Applying this to $v =  s - \pi$ (where $s$ is some starting distribution), you get: $\| P(s-\pi) \|_2 \leq \rho \|s - \pi\|_2$.
This is because $s - \pi$ is orthogonal to constant functions (i.e. it sums to 0). After another step of the random walk its
$$
\| P^2(s-\pi) \|_2 \leq \rho  \| P(s-\pi) \|_2 \leq \rho^2 \|s - \pi\|_2
$$
Since the norm of $\|s - \pi\|_2 \leq 2$ (in your case its $1+1/\sqrt{n}$) one gets the estimate:
$$
\| P^t(s -\pi) \|_{\ell^2} \leq 2 \rho^t
$$
Now use that $P^t s = \mu_t$ and $P^t \pi = \pi$ (because $P\pi =\pi$) to get:
$$
\| \mu_t -\pi\|_{\ell^2} = \| P^t( s -\pi) \|_{\ell^2} \leq 2 \rho^t
$$
which is the estimate claimed above.
A: Some further intuitions: Let $A$ be the normalized adjacency matrix of $G$, let $A_L$ be the normalized adjacency matrix of $G$ w a self-loop at each vertex; so $A_L =\frac{d}{d+1}A +\frac{1}{d+1}I$ where $I$ is the identity matrix. Let $n$ be the number of vertices and let us assume that $G$ is $d$-regular (and so $G$ with a self-loop added at each vertex is $d+1$-regular.) Next let $\chi$ be the stochastic vector that represents the initial position of the random walk; if the walk starts at a vertex $v$ with probability $p_v$ then $\chi_v = p_v$. If the walk definitely starts at a vertex $v$ then $\chi_v = 1$ and $\chi_u$ is 0 for all vertices $u \not = v$. 
Case 1 the non-bipartite case. Then write $\chi = \frac{1}{n}{\bf{1}} + z$, where ${\bf{1}}$ is the vector that is 1 everywhere, and $z$ is orthonormal to ${\bf{1}}$ such that $||z||_2 \le 1$. Then after $t$ steps the probability distribution of the walk after $t$ steps is represented by the vector $\frac{1}{n}{\bf{1}} + A^t z$ and the probability distribution of the lazy walk after $t$ steps is represented by the vector $\frac{1}{n}{\bf{1}} + A_L^t z$. Now let $\lambda$ be the 2nd largest eigenvalue of $A$ and $\lambda_L$ the 2nd-larges eignevalue of $A_L$. The key take-aways are the following:


*

*if $G$ is an expander, then $\lambda$ is bounded away from 1, and in fact so is $\lambda_L= \frac{d \lambda}{d+1} + \frac{1}{d+1}$, and 

*$||A^tz||_2 \le \lambda^{2t}||z||_2$ and $||A_L^tz||_2 \le \lambda_L^{2t}||z||_2$. 


So 1. and 2. together imply that the probability distribution of where the walk is after $t$ steps lazy case or not, converges exponentially fast to the uniform distribution.
Case 2 the bipartite case. Then for 2. to hold for $z$, the vector $\chi$ as in 1. has to be written $\chi = \frac{1}{n} {\bf{1}} + \frac{[p_s-p_T]}{n}y +z$, where $y$ is the vector such that $y_v$ is 1 if $v$ is in one side $S$ of the bipartition of $G$ and $-1$ if $v$ is in the other side $T$ of the bipartition of $G$, and $p_S$ is the probability of the random walk starting at a vertex in $S$ and $p_T$ the probability of the random walk starting at a vertex $T$. Then the probability of the random walk after $t$ steps (not lazy) is represented by the vector $A^t\chi = \frac{1}{n} {\bf{1}} + (-1)^t \frac{[p_S-p_T]}{n}y +A^tz$. And the probability of the lazy random walk after $t$ steps is represented by the vector $A_L^t\chi =$ $\frac{1}{n} {\bf{1}} + (-1)^t(\frac{d}{d+1})^t \frac{[p_S-p_T]}{n}y +A_L^t z$.
So for Case 2 a non-lazy walk that starts at one side $S$ of $G$ with probability $p$ and at $T$ with probability $1-p$ and for $t$ large, if $t$ is even you will be at a vertex $S$ with probability $p$ and at a vertex in $T$ with probability $1-p$, and given you are at $S$ ($T$) you will be uniformly distributed amongst a vertex in $S$ $(T)$; and if $t$ is odd you will be at a vertex $S$ with probability $1-p$ and at a vertex in $T$ with probability $p$, and given you are at $S$ ($T$) you will be uniformly distributed amongst a vertex in $S$ $(T)$. So for $t$ large and even $A^t\chi$ is close to the vector $\frac{2}{n} x(p,(1-p))$, where $x(p,(1-p))_v$ is $p$ if $v$ is in $S$ and $1-p$ if $v$ is in $T$. So for $t$ large and odd $A^t\chi$ is close to the vector $\frac{2}{n} x((1-p),p)$, where $x((1-p),p)_v$ is $1-p$ if $v$ is in $S$ and $p$ if $v$ is in $T$.  
For the lazy walk and $d \in O(1)$, the distribution convereges to uniform here just as it does in the nonbipartite case. 
Anyway, for the (not lazy) random walk how quickly the distribution goes to uniform depends on $\lambda$ and not on $d$. However, $\lambda$ is always asymptotically at least $\frac{2\sqrt{d-1}}{d}$ and there are constructions that achieve this bound.
