# 8.5 Combining the techniques

To get the maximum power out of the techniques of this chapter, it is sometimes necessary to combine the various techniques. Before proceeding to a general result in this direction, we record a simple fact. Recall (8.36).

###### Lemma 8.41

If $q$ and $q^{*}$ are conjugate exponents with $2\leq q\leq\infty$, then

 $f$

Proof. Apply Hölder’s inequality

 $\|gh\|_{1}\leq\|g\|_{p}\,\|h\|_{p^{*}}$

with

 $g=|f|^{(q-2)/(q-1)},\ \ \ h=|f|^{2/(q-1)},\ \ \ p=\frac{q-1}{q-2}.~{}\ \ \rule% {4.3pt}{4.3pt}$
###### Theorem 8.42

Suppose that a continuous-time reversible chain satisfies

 $0 (8.75)

for some constants $C,T,D$ satisfying $CT^{-D}\geq e$. If $c\geq 0$, then

 $\sqrt{{\hat{d}}(2t)}=\max_{i}\|P_{i}(X_{t}\in\cdot)-\pi(\cdot)\|_{2}\leq e^{2-c}$

for

 $t\geq T+{\textstyle\frac{1}{2}}\tau_{l}\log\left[\log(CT^{-D})-1\right]+c\tau_% {2},$

where $\tau_{2}$ is the relaxation time and $\tau_{l}$ is the log-Sobolev time.

Proof. ¿From Lemma 8.11 and a slight extension of (8.34), for any $s,t,u\geq 0$ and any initial distribution we have

 $\|P(X_{s+t+u}\in\cdot)-\pi(\cdot)\|_{2}\leq\|P(X_{s}\in\cdot)\|_{q^{*}}\,\|{% \bf P}_{t}\|_{q^{*}\to 2}\,e^{-u/\tau_{2}}$

for any $1\leq q^{*}\leq\infty$. Choose $q=q(t)=1+e^{2t/\tau_{l}}$ and $q^{*}$ to be its conjugate. Then, as in the proof of Theorem 8.26(a),

 $\|{\bf P}_{t}\|_{q^{*}\to 2}=\|{\bf P}_{t}\|_{2\to q}\leq 1.$

According to Lemma 8.41, (8.39), and (8.75), if $0 then

 $\|P(X_{s}\in\cdot)\|_{q^{*}}\leq\|P(X_{s}\in\cdot)\|^{2/q}_{2}\leq N(s)^{2/q}% \leq(Cs^{-D})^{2/q}.$

Now choose $s=T$. Combining everything so far,

 $t,u\geq 0$

The final idea is to choose $t$ so that the first factor is bounded by $e^{2}$. ¿From the formula for $q(t)$, the smallest such $t$ is

 ${\textstyle\frac{1}{2}}\tau_{l}\log\left[\log(CT^{-D})-1\right].$

With this choice, the theorem follows readily.

###### Example 8.43

Random walk on a $d$-dimensional grid.

Return one last time to the walk of interest in Example 8.5. Example 8.21 showed that (8.75) holds with

 $D=d/4,\ \ \ C=e(2^{14}d^{2}m^{2})^{d/4}=e(2^{7}dm)^{d/2},\ \ \ T=dm^{2}/32.$

Also recall $\tau_{2}\leq\frac{1}{2}dm^{2}$ from (8.49) and $\tau_{l}\leq 2dm^{2}$ from Example 8.40. Plugging these into Theorem 8.42 with $c=2$ yields

 $\leq 5m^{2}d\log d$

xxx Finally of right order of magnitude.