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We first prove the compact-domain Schauder fixed-point theorem. Let $K$ be a nonempty compact convex set in a Banach space $X$, and let $T:K\to K$ be a continuous map. The Brouwer fixed-point theorem states that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point. By the homeomorphism proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact convex set, including a singleton.

For $\epsilon >0$, use the supplied Schauder approximation lemma to obtain a finite-dimensional linear subspace $L_{\epsilon }$ and a continuous map $T_{\epsilon }:K\to K\cap L_{\epsilon }$ with $\Vert Tx-T_{\epsilon }x\Vert \le \epsilon$. The set $K_{\epsilon }=K\cap L_{\epsilon }$ is nonempty because it contains $T_{\epsilon }(K)$. It is a compact convex set, since a finite-dimensional linear subspace is closed. Restricting $T_{\epsilon }$ to $K_{\epsilon }$ and applying the Brouwer fixed-point theorem gives $x_{\epsilon }=T_{\epsilon }{x}_{\epsilon }$, and thusTake ${\epsilon }_j=1/j$. The compactness of $K$ yields a convergent subsequence $x_{{\epsilon }_{j_m}}\to x_{\ast }\in K$. The continuity of $T$ and the preceding estimate giveFor completeness, this also proves the version on a nonempty closed bounded convex set $C$ when $T(C)$ is relatively compact. Let $K=\overline{\mathrm{conv}}T(C)\subset C$. The closed convex hull of a compact set in a Banach space is compact. To verify the fact needed here, cover $T(C)$ by finitely many balls of radius $\delta$. Replacing every term of a convex combination by a centre changes it by at most $\delta$ in norm. The convex hull of those finitely many centres is compact in a finite-dimensional linear subspace, so it has a finite net of radius $\delta$. Thus $\mathrm{conv}T(C)$ is totally bounded. Its closure is complete and totally bounded, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a fixed point in $C$.

Write $Ph=\mathbb{E}h(X)$ and $P_{n}h=n^{-1}{\sum }_{i}h(X_i)$ for the expectation and empirical measure. A sufficient bracketing of a function class condition is that, for every $\epsilon >0$, finitely many brackets $[l_j,u_j]$ cover $\mathcal{H}$, with measurable integrable endpoints and $P(u_j-l_j)<\epsilon$. The uniform strong law from finite L1 bracketing then givesHere the observations are independent and identically distributed. For an uncountable class, one either assumes a measurable supremum, for example through a pointwise separable function class, or formulates the conclusion as a pathwise bound on a common probability-one event.

For the parameterized class, the Heine-Borel theorem makes $\Theta$ compact. Put $M(x)={\sup }_{\theta \in \Theta }|q(\theta ,x)|$. A countable dense subset of $\Theta$ gives the same supremum because of continuity, so $M$ is measurable. Define the modulusThe supremum is measurable by taking a countable dense subset of the compact set of admissible pairs. For each $x$, uniform continuity on $\Theta$ gives $w_{\delta }(x)\to 0$. Also $w_{\delta }\le 2M$, and $PM<\infty$. Thus the dominated convergence theorem gives $P{w}_{\delta }\to 0$.

Choose a finite net of radius $\delta$ ${\theta }_1,\dots ,{\theta }_N$ in $\Theta$. The lower and upper envelopes of $q(\theta ,\cdot )$ over each closed ball $\Theta \cap \overline{B}({\theta }_j,\delta )$ are measurable integrable function brackets; the same countable-dense-set argument applies within each such compact ball. Their widths are at most $w_{2\delta }$. They cover the whole class, so its bracketing of a function class condition follows. Moreover, dominated convergence shows that $\theta \mapsto Pq(\theta ,\cdot )$ is continuous. Both empirical and population functions therefore have a supremum over a common countable dense parameter set. Applying the uniform strong law from finite L1 bracketing proves uniform almost-sure convergence over the entire compact parameter set, rather than merely convergence at each fixed parameter.

For the exponential family,where $A={\sup }_{\Theta }|\theta |$ and $B={\sup }_{\Theta }|K(\theta )|$. Thus a weak sufficient condition is bounded $K$ on $\Theta$, $f_0(X)>0$ almost surely, and integrability of $|X|+|\log f_0(X)|$ under the sampling law. No positive lower bound on $f_0$ over the whole real line is required. Zeros off the sampling support are harmless: choose arbitrary finite versions of the log-densities on that common null set when applying the function-class theorem.

If the sampling law is $f({\theta }_0,\cdot )$, one convenient assumption is that $\Theta$ is a compact subset of the interior of the finite domain of the cumulant function of an exponential family, andOn that interior, $K$ is continuous, hence bounded on $\Theta$. Finiteness of $K$ at ${\theta }_0\pm a$ for some $a>0$ implies ${\mathbb{E}}_{{\theta }_0}{e}^{a|X|}<\infty$, hence ${\mathbb{E}}_{{\theta }_0}|X|<\infty$. The displayed integral gives the remaining integrability. If the observations come from an arbitrary unrelated law, assumptions on $K$ and $f_0$ alone cannot control that law's tails; the sampling-law integrability must then be stated explicitly.

Let $m=\mathbb{E}\Vert X_1{\Vert }_2$ and $q=\sqrt{n}K\ge 2$. The Lipschitz continuity of cosine and the Cauchy-Schwarz inequality implyThe centered expectation term contributes $nm$; omitting it would leave the Lipschitz bound incomplete. In particular $S_n$ has continuous sample paths on the compact cube, so its maximum exists and is measurable.

PutChoose a Cartesian finite net $G\subset [-K,K{]}^d$ whose coordinate spacings are at most $\eta$, including the endpoints. Every point of the cube is within Euclidean norm distance $\sqrt{d}\eta$ of $G$, andOn the event $L_n\le n{q}^a$, replacing any $u$ by a nearby grid point changes $S_n$ by at most $\sqrt{dn}$. Since $R>8\sqrt{d}$ and $\log (n{K}^2)\ge \log 4$, this is less than $\frac{R}{4}\sqrt{n\log (n{K}^2)}$. ThereforeThe Markov inequality bounds the first event by $2m{q}^{-a}$. At a fixed $v$, the summands are centered, independent and identically distributed, and have absolute value at most $2$. The supplied Hoeffding inequality, with $M=2$, bounds each grid event byA union bound now givesThe choice of $a$ balances the cost of a large random Lipschitz constant against the covering number of the cube: $d(1+a)-R^2/64=-a$. Consequently $C=2m+2\cdot 4^d$ works, withThis uses only the first moment $m$ and has a constant independent of $n,K,R$. The finite net is deterministic even though the Lipschitz constant is random.

Suppose $|Z(u)-Z(v)|\le L\Vert u-v\Vert$ with a random finite $L$. On $L\le \ell$, a finite net with radius $\eta$ bounds the full supremum by the grid maximum plus $\ell \eta$. The Markov inequality controls the exceptional event using $\mathbb{E}L$, and a union bound controls the grid maximum. Choosing the net spacing balances these two costs.