[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*}[/step]

[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: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]