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The pushforward measure of $\mu$ under $g$ is $g_{\ast }\mu$, defined for $B\in \mathcal{B}$ byThe Lebesgue decomposition theorem says that if $\nu$ and $\mu$ are sigma-finite measures on the same measurable space, then uniquely

Let $f:[0,\infty )\to \mathbb{R}\cup +\infty$ be convex with $f(1)=0$. If $\lambda$ dominates the probability distributions $P,Q$, with densities $p,q$, their f-divergence isusing the lower-semicontinuous perspective convention where $q=0$. This definition is independent of the dominating measure.

The data processing inequality for f-divergences statesTo prove it, take $\lambda =P+Q$ and let $\mathcal{G}=g^{-1}(\mathcal{B})$. If $p=dP/d\lambda$ and $q=dQ/d\lambda$, then the pullbacks of the densities of $g_{\ast }P$ and $g_{\ast }Q$ with respect to $g_{\ast }\lambda$ are respectively ${\mathbb{E}}_{\lambda }[p\mid \mathcal{G}]$ and ${\mathbb{E}}_{\lambda }[q\mid \mathcal{G}]$. The perspective $F(a,b)=bf(a/b)$ of a convex function is jointly convex. Conditional Jensen inequality therefore givesIntegration proves the claim.

The Squared Hellinger distance isIf $P,Q$ have densities $p,q$ with respect to a sigma-finite measure $\mu$, this becomes

Fix any probability distribution $Q$ and set $p_j=P_j(A_j)$, $q_j=Q(A_j)$, and $t_j=H^2(P_j,Q)$. Applying the data processing inequality to the indicator function of $A_j$ gives the Bernoulli Hellinger boundThe hinted inequality impliesThe function $t\mapsto \sqrt{t(1-t/4)}$ is concave on $[0,2]$. Since the $A_j$ form a set partition, ${\sum }_j{q}_j=1$, and Jensen inequality givesTaking the infimum over $Q$ proves the stated inequality.

A linear estimator in nonparametric regression at $x_0$ has the form ${\hat{m}}_n(x_0)={\sum }_i{W}_i(x_0)Y_i$, where the weights may depend on the design, $x_0$, $p$, $h$, and $K$, but not on the responses. Putand define the local polynomial Gram matrixWhen ${\lambda }_0>0$, the weighted least squares normal equations have the unique solutionTherefore ${\hat{m}}_n(x_0)={\sum }_i{W}_i{Y}_i$, where the effective kernel weight is

The identityshows that these weights exactly reproduce at $x_0$ every multivariate polynomial of total degree at most $p$. This is the required polynomial reproduction property of local polynomial regression.

Only grid points with $\Vert x_i-x_0{\Vert }_{\infty }\le h$ have nonzero weight. There are at most $(2n_{1}h+1{)}^d\le (4n_{1}h{)}^d=4^{d}n{h}^d$ such points because $n_{1}h\ge 1/2$. On this support,so the operator norm bound $\Vert B_n^{-1}{\Vert }_{\mathrm{op}}={\lambda }_0^{-1}$ givesBecause the errors are independent random variables with the stated variance bounds,Thus ${\gamma }_1=d$.

Let $T_{x_0}$ be the Multivariate Taylor polynomial of $m$ at $x_0$ through total degree ${\beta }_0$. The assumed Hölder continuity of the derivatives andgive the Taylor remainder boundPolynomial reproduction cancels $T_{x_0}$ in the bias. The same support count and weight bound giveConsequentlyso ${\gamma }_2=\beta$.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.