[proofplan]
We use Hoeffding's decomposition to split $U_n-\theta$ into its first-order projection and canonical degenerate U-statistics of orders $2,\dots,m$. The first-order projection is exactly an average of i.i.d. centered random variables, so the classical [central limit theorem](/theorems/521) gives the Gaussian limit with variance $m^2\zeta_1$. The higher-order canonical terms have variances of order $n^{-s}$ in order $s$, and hence become negligible after multiplication by $\sqrt n$. Slutsky's theorem then transfers the limiting distribution from the first projection to the full U-statistic.
[/proofplan]
[step:Decompose the U-statistic into Hoeffding projection terms]
For each $s\in\{1,\dots,m\}$, let $h_s:E^s\to\mathbb R$ denote the $s$-th canonical Hoeffding kernel associated to $h$, with $h_1$ equal to the first projection in the theorem statement. Thus each $h_s$ is symmetric, square-integrable, and canonical in the sense that for $s\ge2$,
\begin{align*}
\mathbb E[h_s(x_1,\dots,x_{s-1},X_s)] = 0
\end{align*}
for every fixed $(x_1,\dots,x_{s-1})\in E^{s-1}$ for which the conditional expression is defined.
For $n\ge s$, define the order-$s$ U-statistic generated by $h_s$ as
\begin{align*}
U_{n,s}:=\binom{n}{s}^{-1}\sum_{1\le i_1<\cdots<i_s\le n} h_s(X_{i_1},\dots,X_{i_s}).
\end{align*}
By the Hoeffding decomposition for square-integrable U-statistics (citing a result not yet in the wiki: [Hoeffding decomposition for U-statistics](/theorems/6334)),
\begin{align*}
U_n-\theta
=
\sum_{s=1}^m \binom{m}{s}U_{n,s}.
\end{align*}
Since $U_{n,1}=n^{-1}\sum_{i=1}^n h_1(X_i)$, this becomes
\begin{align*}
U_n-\theta
=
\frac{m}{n}\sum_{i=1}^n h_1(X_i)
+
\sum_{s=2}^m \binom{m}{s}U_{n,s}.
\end{align*}
[/step]
[step:Apply the classical central limit theorem to the first projection]
The random variables $h_1(X_1),h_1(X_2),\dots$ are i.i.d. real-valued random variables. Moreover,
\begin{align*}
\mathbb E[h_1(X_1)] = 0,
\qquad
\operatorname{Var}(h_1(X_1))=\zeta_1\in(0,\infty),
\end{align*}
where finiteness follows from the square-integrability of $h$ and [Jensen's inequality](/theorems/9) applied to the [conditional expectation](/page/Conditional%20Expectation) defining $h_1$.
By the classical [central limit theorem](/theorems/1848) (citing a result not yet in the wiki: Lindeberg-Levy Central Limit Theorem),
\begin{align*}
\frac{1}{\sqrt n}\sum_{i=1}^n h_1(X_i)
\xrightarrow{d}
\mathcal N(0,\zeta_1).
\end{align*}
Multiplying by the constant $m$ gives
\begin{align*}
\sqrt n\,\frac{m}{n}\sum_{i=1}^n h_1(X_i)
=
m\,\frac{1}{\sqrt n}\sum_{i=1}^n h_1(X_i)
\xrightarrow{d}
\mathcal N(0,m^2\zeta_1).
\end{align*}
[/step]
[step:Show that every degenerate Hoeffding term is negligible at scale $\sqrt n$]
Fix $s\in\{2,\dots,m\}$. Let $\mathcal I_{n,s}$ denote the set of all subsets $I\subset\{1,\dots,n\}$ with $|I|=s$. For $I=\{i_1<\cdots<i_s\}\in\mathcal I_{n,s}$, define the real-valued [random variable](/page/Random%20Variable)
\begin{align*}
Y_I := h_s(X_{i_1},\dots,X_{i_s}).
\end{align*}
Then
\begin{align*}
U_{n,s}=\binom{n}{s}^{-1}\sum_{I\in\mathcal I_{n,s}}Y_I.
\end{align*}
If $I,J\in\mathcal I_{n,s}$ and $I\ne J$, then $I\setminus J$ is nonempty. Conditioning on all random variables indexed by $J$ together with those indexed by $I\cap J$, and then integrating over one index in $I\setminus J$, the canonical property of $h_s$ gives
\begin{align*}
\mathbb E[Y_IY_J]=0.
\end{align*}
Therefore only the diagonal terms contribute to the second moment. First,
\begin{align*}
\mathbb E[U_{n,s}^2]=\binom{n}{s}^{-2}\sum_{I,J\in\mathcal I_{n,s}}\mathbb E[Y_IY_J].
\end{align*}
Using $\mathbb E[Y_IY_J]=0$ for $I\ne J$, this reduces to
\begin{align*}
\mathbb E[U_{n,s}^2]=\binom{n}{s}^{-2}\sum_{I\in\mathcal I_{n,s}}\mathbb E[Y_I^2].
\end{align*}
Since each $Y_I$ has the same distribution as $h_s(X_1,\dots,X_s)$ and $|\mathcal I_{n,s}|=\binom{n}{s}$, we get
\begin{align*}
\mathbb E[U_{n,s}^2]=\binom{n}{s}^{-1}\mathbb E[h_s(X_1,\dots,X_s)^2].
\end{align*}
Since $h_s$ is square-integrable, define
\begin{align*}
A_s:=\mathbb E[h_s(X_1,\dots,X_s)^2]<\infty.
\end{align*}
Then
\begin{align*}
\mathbb E[(\sqrt n\,U_{n,s})^2]
=
n\binom{n}{s}^{-1}A_s.
\end{align*}
Because $s\ge2$, we have $n\binom{n}{s}^{-1}\to0$, and hence
\begin{align*}
\sqrt n\,U_{n,s}\to0
\end{align*}
in $L^2$, therefore also in probability.
[guided]
Fix one order $s\in\{2,\dots,m\}$. The goal is to prove that the corresponding Hoeffding term is too small to affect the $\sqrt n$ limit. Define $\mathcal I_{n,s}$ to be the set of all subsets $I\subset\{1,\dots,n\}$ with $|I|=s$. For $I=\{i_1<\cdots<i_s\}\in\mathcal I_{n,s}$, define
\begin{align*}
Y_I := h_s(X_{i_1},\dots,X_{i_s}).
\end{align*}
With this notation,
\begin{align*}
U_{n,s}=\binom{n}{s}^{-1}\sum_{I\in\mathcal I_{n,s}}Y_I.
\end{align*}
We compute the second moment of $U_{n,s}$. Expanding the square gives
\begin{align*}
\mathbb E[U_{n,s}^2]
=
\binom{n}{s}^{-2}
\sum_{I,J\in\mathcal I_{n,s}}\mathbb E[Y_IY_J].
\end{align*}
The important point is that canonical degeneracy kills every off-diagonal covariance. Indeed, if $I\ne J$, then at least one index belongs to $I$ but not to $J$. Choose such an index $i\in I\setminus J$. Condition on all random variables except $X_i$. With the remaining $s-1$ arguments of $h_s$ fixed, the canonical property gives conditional mean zero in the $X_i$ variable:
\begin{align*}
\mathbb E[Y_I\mid (X_k)_{k\ne i}]=0.
\end{align*}
Since $Y_J$ is measurable with respect to the random variables $(X_k)_{k\ne i}$, the tower property gives
\begin{align*}
\mathbb E[Y_IY_J]
=
\mathbb E\left[Y_J\,\mathbb E[Y_I\mid (X_k)_{k\ne i}]\right]
=
0.
\end{align*}
Thus the only surviving terms in the double sum are the terms with $I=J$. Hence
\begin{align*}
\mathbb E[U_{n,s}^2]=\binom{n}{s}^{-2}\sum_{I\in\mathcal I_{n,s}}\mathbb E[Y_I^2].
\end{align*}
Since there are $\binom{n}{s}$ such sets $I$, and each $Y_I$ has the same distribution as $h_s(X_1,\dots,X_s)$, we obtain
\begin{align*}
\mathbb E[U_{n,s}^2]=\binom{n}{s}^{-1}\mathbb E[h_s(X_1,\dots,X_s)^2].
\end{align*}
Define the finite constant
\begin{align*}
A_s:=\mathbb E[h_s(X_1,\dots,X_s)^2].
\end{align*}
Then
\begin{align*}
\mathbb E[(\sqrt n\,U_{n,s})^2]
=
n\binom{n}{s}^{-1}A_s.
\end{align*}
For fixed $s\ge2$, the binomial coefficient grows like a constant multiple of $n^s$, so $n\binom{n}{s}^{-1}\to0$. Therefore
\begin{align*}
\mathbb E[(\sqrt n\,U_{n,s})^2]\to0.
\end{align*}
This proves $\sqrt n\,U_{n,s}\to0$ in $L^2$, and convergence in $L^2$ implies convergence in probability.
[/guided]
[/step]
[step:Combine the negligible terms and apply Slutsky's theorem]
Define the remainder random variable $R_n$ by
\begin{align*}
R_n:=\sum_{s=2}^m \binom{m}{s}U_{n,s}.
\end{align*}
Using the inequality
\begin{align*}
\left(\sum_{s=2}^m a_s\right)^2 \le (m-1)\sum_{s=2}^m a_s^2
\end{align*}
for [real numbers](/page/Real%20Numbers) $a_2,\dots,a_m$, we get
\begin{align*}
\mathbb E[(\sqrt n\,R_n)^2]\le (m-1)\sum_{s=2}^m \binom{m}{s}^2\mathbb E[(\sqrt n\,U_{n,s})^2].
\end{align*}
Each summand on the right tends to $0$ by the preceding step, and there are only finitely many values of $s$. Hence
\begin{align*}
\sqrt n\,R_n\to0
\end{align*}
in $L^2$, therefore in probability.
From the decomposition,
\begin{align*}
\sqrt n\,(U_n-\theta)
=
m\,\frac{1}{\sqrt n}\sum_{i=1}^n h_1(X_i)
+
\sqrt n\,R_n.
\end{align*}
The first term converges in distribution to $\mathcal N(0,m^2\zeta_1)$, and the second term converges in probability to $0$. By Slutsky's theorem (citing a result not yet in the wiki: Slutsky's Theorem),
\begin{align*}
\sqrt n\,(U_n-\theta)
\xrightarrow{d}
\mathcal N(0,m^2\zeta_1).
\end{align*}
This is the claimed central limit theorem.
[/step]
