[proofplan]
We prove the estimate by constructing an admissible sequence of finite nets at entropy levels matching the cardinality constraints in the definition of $\gamma_2$. The net radii are chosen as entropy numbers $e_k$, where the covering number first drops below $2^{2^k}$. The resulting chaining cost is bounded by a dyadic sum $\sum_k 2^{k/2}e_k$, and a monotone slicing argument compares this sum to Dudley's entropy integral. The endpoint cases of zero diameter and infinite entropy integral are disposed of separately.
[/proofplan]
[step:Remove the endpoint cases where the estimate is immediate]
Let $D:=\operatorname{diam}(T,d)\in[0,\infty)$. Since $(T,d)$ is [totally bounded](/page/Totally%20Bounded), $D<\infty$.
If $D=0$, then $d(s,t)=0$ for all $s,t\in T$. Choose a point $t_0\in T$ if $T\ne\varnothing$ and set $A_k:=\{t_0\}$ for every $k\in\mathbb N\cup\{0\}$. If $T=\varnothing$, take $A_k:=\varnothing$. In either case the admissible chaining cost is $0$, so $\gamma_2(T,d)=0$, and the desired inequality follows.
Assume from now on that $D>0$. Define
\begin{align*}
I:=\int_0^D\sqrt{\log N(T,d,\varepsilon)}\,d\mathcal L^1(\varepsilon).
\end{align*}
If $I=\infty$, the asserted bound is immediate for any positive constant $C$. Thus we assume $I<\infty$.
[/step]
[step:Define entropy radii at the admissible cardinality levels]
For each integer $k\ge 1$, define the entropy radius $e_k\in[0,D]$ by
\begin{align*}
e_k:=\inf\{\varepsilon>0:N(T,d,\varepsilon)\le 2^{2^k}\}.
\end{align*}
Also set $e_0:=D$.
The sequence $(e_k)_{k\ge 0}$ is nonincreasing, because the thresholds $2^{2^k}$ increase with $k$. Since $(T,d)$ is totally bounded, $N(T,d,\varepsilon)<\infty$ for every $\varepsilon>0$, and therefore $e_k\to 0$ as $k\to\infty$.
For each $k\ge 1$ and each parameter $\delta>0$, the definition of $e_k$ gives
\begin{align*}
N(T,d,e_k+\delta)\le 2^{2^k}.
\end{align*}
Hence there exists a finite set $A_k^\delta\subset T$ such that
\begin{align*}
|A_k^\delta|\le 2^{2^k}
\end{align*}
and
\begin{align*}
\sup_{t\in T}d(t,A_k^\delta)\le e_k+\delta.
\end{align*}
Choose an arbitrary point $t_0\in T$ and set $A_0^\delta:=\{t_0\}$. Then $|A_0^\delta|=1\le 2^{2^0}$.
[/step]
[step:Convert the admissible nets into a preliminary chaining bound]
Call a sequence $(B_k)_{k\ge 0}$ of subsets of $T$ admissible if $|B_0|\le 1$ and $|B_k|\le 2^{2^k}$ for every $k\ge 1$. The defining formula for the $\gamma_2$ functional is
\begin{align*}
\gamma_2(T,d):=\inf_{(B_k)}\sup_{t\in T}\sum_{k=0}^{\infty}2^{k/2}d(t,B_k),
\end{align*}
where the infimum is over all admissible sequences. Since $(A_k^\delta)_{k\ge 0}$ is admissible, for every $\delta>0$,
\begin{align*}
\gamma_2(T,d)\le \sup_{t\in T}\sum_{k=0}^{\infty}2^{k/2}d(t,A_k^\delta).
\end{align*}
For $k=0$, since $A_0^\delta=\{t_0\}$ and $D=\operatorname{diam}(T,d)$,
\begin{align*}
d(t,A_0^\delta)\le D=e_0.
\end{align*}
For $k\ge 1$, the construction gives
\begin{align*}
d(t,A_k^\delta)\le e_k+\delta.
\end{align*}
Therefore, for each integer $m\ge 1$,
\begin{align*}
\gamma_2(T,d)\le e_0+\sum_{k=1}^{m}2^{k/2}(e_k+\delta)+\sup_{t\in T}\sum_{k=m+1}^{\infty}2^{k/2}d(t,A_k^\delta).
\end{align*}
Instead of estimating the infinite tail with a fixed $\delta$, choose $\delta_k>0$ for each $k\ge 1$ such that
\begin{align*}
\sum_{k=1}^{\infty}2^{k/2}\delta_k\le D.
\end{align*}
Applying the construction with $\delta=\delta_k$ at level $k$ gives an admissible sequence $(A_k)_{k\ge 0}$ satisfying
\begin{align*}
\sup_{t\in T}\sum_{k=0}^{\infty}2^{k/2}d(t,A_k)\le e_0+\sum_{k=1}^{\infty}2^{k/2}e_k+D.
\end{align*}
Since $e_0=D$, we obtain
\begin{align*}
\gamma_2(T,d)\le 2D+\sum_{k=1}^{\infty}2^{k/2}e_k.
\end{align*}
[guided]
The $\gamma_2$ functional is an infimum over admissible sequences. An admissible sequence is a sequence of subsets whose cardinalities obey the dyadic growth rule
\begin{align*}
|A_0|\le 1
\end{align*}
and
\begin{align*}
|A_k|\le 2^{2^k}
\end{align*}
for $k\ge 1$. Our entropy radii $e_k$ were chosen exactly so that radius $e_k$ is the limiting scale at which a net with this allowed cardinality exists.
There is a small technical point: because $e_k$ is defined as an infimum, a net of radius exactly $e_k$ need not exist. This is why we choose small positive errors $\delta_k>0$. For every $k\ge 1$, there is a finite set $A_k\subset T$ such that
\begin{align*}
|A_k|\le 2^{2^k}
\end{align*}
and
\begin{align*}
\sup_{t\in T}d(t,A_k)\le e_k+\delta_k.
\end{align*}
Choose $A_0=\{t_0\}$ for some $t_0\in T$. Since $D$ is the diameter of $T$, every $t\in T$ satisfies
\begin{align*}
d(t,A_0)=d(t,t_0)\le D=e_0.
\end{align*}
Now choose the errors so that their weighted total is harmless:
\begin{align*}
\sum_{k=1}^{\infty}2^{k/2}\delta_k\le D.
\end{align*}
For example, any positive summable sequence with sufficiently small weighted sum works. Then the admissible-chain cost satisfies
\begin{align*}
\sup_{t\in T}\sum_{k=0}^{\infty}2^{k/2}d(t,A_k)\le e_0+\sum_{k=1}^{\infty}2^{k/2}(e_k+\delta_k).
\end{align*}
Using $e_0=D$ and the chosen bound on the errors gives
\begin{align*}
\sup_{t\in T}\sum_{k=0}^{\infty}2^{k/2}d(t,A_k)\le 2D+\sum_{k=1}^{\infty}2^{k/2}e_k.
\end{align*}
Taking the infimum over all admissible sequences can only reduce the left-hand side, so
\begin{align*}
\gamma_2(T,d)\le 2D+\sum_{k=1}^{\infty}2^{k/2}e_k.
\end{align*}
[/guided]
[/step]
[step:Bound the dyadic entropy sum by the entropy integral]
For each $k\ge 1$ and each $\varepsilon<e_{k-1}$, the definition of $e_{k-1}$ implies
\begin{align*}
N(T,d,\varepsilon)>2^{2^{k-1}}.
\end{align*}
Thus, for $\varepsilon\in(e_k,e_{k-1})$,
\begin{align*}
\sqrt{\log N(T,d,\varepsilon)}\ge \sqrt{2^{k-1}\log 2}.
\end{align*}
Integrating over the interval $(e_k,e_{k-1})$ with respect to $\mathcal L^1$ gives
\begin{align*}
\int_{e_k}^{e_{k-1}}\sqrt{\log N(T,d,\varepsilon)}\,d\mathcal L^1(\varepsilon)\ge \sqrt{\log 2}\,2^{(k-1)/2}(e_{k-1}-e_k).
\end{align*}
Summing over $k\ge 1$ and using the disjointness of the intervals $(e_k,e_{k-1})$ yields
\begin{align*}
I\ge \sqrt{\log 2}\sum_{k=1}^{\infty}2^{(k-1)/2}(e_{k-1}-e_k).
\end{align*}
Since $e_k\downarrow 0$, write
\begin{align*}
e_k=\sum_{j=k+1}^{\infty}(e_{j-1}-e_j).
\end{align*}
All summands are nonnegative, so Tonelli's theorem for series permits interchanging the sums:
\begin{align*}
\sum_{k=1}^{\infty}2^{k/2}e_k=\sum_{j=2}^{\infty}(e_{j-1}-e_j)\sum_{k=1}^{j-1}2^{k/2}.
\end{align*}
For every $j\ge 2$,
\begin{align*}
\sum_{k=1}^{j-1}2^{k/2}\le \frac{2^{j/2}}{\sqrt 2-1}.
\end{align*}
Therefore
\begin{align*}
\sum_{k=1}^{\infty}2^{k/2}e_k\le \frac{\sqrt 2}{\sqrt 2-1}\sum_{j=2}^{\infty}2^{(j-1)/2}(e_{j-1}-e_j).
\end{align*}
Combining this with the lower bound for $I$ gives
\begin{align*}
\sum_{k=1}^{\infty}2^{k/2}e_k\le \frac{\sqrt 2}{(\sqrt 2-1)\sqrt{\log 2}}\,I.
\end{align*}
[/step]
[step:Absorb the diameter term into the entropy integral and conclude]
The preceding steps already give the estimate with the harmless additive term $2D$. To absorb it into the same universal form, note that for every $0<\varepsilon<D/2$, one ball of radius $\varepsilon$ cannot cover $T$, so
\begin{align*}
N(T,d,\varepsilon)\ge 2.
\end{align*}
Therefore
\begin{align*}
I\ge \int_0^{D/2}\sqrt{\log 2}\,d\mathcal L^1(\varepsilon)=\frac{D\sqrt{\log 2}}{2}.
\end{align*}
Thus
\begin{align*}
2D\le \frac{4}{\sqrt{\log 2}}I.
\end{align*}
Together with the dyadic estimate, this yields
\begin{align*}
\gamma_2(T,d)\le \left(\frac{4}{\sqrt{\log 2}}+\frac{\sqrt 2}{(\sqrt 2-1)\sqrt{\log 2}}\right)I.
\end{align*}
Setting
\begin{align*}
C:=\frac{4}{\sqrt{\log 2}}+\frac{\sqrt 2}{(\sqrt 2-1)\sqrt{\log 2}}
\end{align*}
gives
\begin{align*}
\gamma_2(T,d)\le C\int_0^{\operatorname{diam}(T,d)}\sqrt{\log N(T,d,\varepsilon)}\,d\mathcal L^1(\varepsilon).
\end{align*}
The constant $C$ is numerical and independent of $(T,d)$, so the theorem follows.
[/step]
