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Write $R(v)={\int }_{\mathbb{R}}v(u{)}^{2}du$. The kernel density estimator with bandwidth $h>0$ isSufficient conditions are that $K$ be a nonnegative symmetric probability density function with $R(K)<\infty$ and finite nonzero second moment ${\mu }_2(K)=\int u^{2}K(u)du$, and that $h\to 0$, $nh\to \infty$. Symmetry and the second-moment assumption give $\int uK(u)du=0$. The expectation is a convolution:Taylor's theorem with integral remainder givesThe zeroth-order term integrates to $f(x)$, and the first-order term vanishes. The second derivative is bounded and continuous, so dominated convergence, with dominating function proportional to $u^{2}K(u)$, yieldsCombining this with the supplied leading variance gives the asymptotic mean squared errorWhen $f(x)>0$ and $f^{\prime \prime }(x)\ne 0$, its derivative is $-R(K)f(x)/(n{h}^2)+{\mu }_2(K{)}^2{f}^{\prime \prime }(x{)}^2{h}^3$. It changes sign exactly once, giving the pointwise optimal kernel bandwidthPositivity of $f(x)$ is needed for this positive interior optimum. The printed condition $f^{\prime \prime }(x)\ne 0$ alone does not ensure it. For instance, $f(x)=x^2\varphi (x)$ is a normalized bounded probability density function with bounded continuous square-integrable second derivative, but $f(0)=0$ and $f^{\prime \prime }(0)=2\varphi (0)>0$. At this point the displayed leading variance vanishes and the two-term AMSE has no positive interior minimizer; finer variance terms are needed to determine an appropriate bandwidth. The formal formula there gives zero, which does not satisfy the bandwidth assumptions.

For integrated error, the same expansion holds in $L^2$: translation continuity of $f^{\prime \prime }\in L^2$ applied to the integral remainder givesIndeed Minkowski's integral inequality bounds the norm of the difference, divided by $h^2$, by the integral of $u^{2}K(u){\int }_0^1(1-s)\Vert f^{\prime \prime }(\cdot -shu)-f^{\prime \prime }{\Vert }_{2}dsdu$; each translated difference tends to zero and is bounded by $2\Vert f^{\prime \prime }{\Vert }_2$. This proves the claimed integrated bias expansion without assuming a uniform pointwise remainder over the whole real line.

If $K_h(x)=h^{-1}K(x/h)$, direct integration of the variance givesThe last term is $O(n^{-1})$ because $f$ is bounded and integrable, hence square integrable, and convolution with a probability density function does not increase its $L^2$ norm. It is negligible compared with $1/(nh)$. Thus the asymptotic mean integrated squared error and its minimizer areHere $R(f^{\prime \prime })>0$: otherwise the continuous second derivative vanishes everywhere, making $f$ affine, impossible for a probability density function on the whole real line.

Now take the standard normal density $f=\varphi$. Differentiation gives ${\varphi }^{\prime \prime }(x)=(x^2-1)\varphi (x)$. Gaussian integration givesThe kernel constants cancel from the bandwidth ratio. At $x\ne \pm 1$,Put $t=x^2$. The logarithmic derivative of $e^{t/2}/(t-1{)}^2$ is $1/2-2/(t-1)$. On $0\le t<1$ the function increases, giving its minimum at $t=0$, with value one. On $t>1$ it decreases up to $t=5$ and then increases, giving its minimum $e^{5/2}/16<1$. Thus the global minimizers are $x=\pm \sqrt{5}$, and the normal-density local-to-global bandwidth ratio isThe ratio diverges at the two inflection points $\pm 1$, where the second-order pointwise bias approximation vanishes and needs a higher-order replacement.

Assume the consistency regularity implicit in the requested rate, for example $u\in C^4$ on the closed square. Bounded fourth derivatives and Taylor's theorem give a uniform residual ${\tau }_{ij}={\Delta }_{h}u(ih,jh)-f(ih,jh)$ with $|{\tau }_{ij}|\le C_0{h}^2$. Subtracting the exact grid values from the numerical equations gives ${\Delta }_{h}e=-\tau$, with zero boundary error.

Let $A=-h^2{\Delta }_h$ be the unscaled positive Dirichlet matrix. Its eigenvectors are products of discrete sine vectors, which form an orthogonal basis; direct substitution gives eigenvaluesSince $\sin (\pi h/2)\ge h$ for $0<h\le 1$, ${\lambda }_{11}\ge 8h^2$. Therefore $\Vert A^{-1}{\Vert }_2\le 1/(8h^2)$. The error equation is $Ae=h^2\tau$, and its residual has norm $\Vert \tau {\Vert }_2\le M{C}_0{h}^2\le C_{0}h$. Combining the bounds yieldsThis is the unweighted grid error for the five-point Poisson formula. The unweighted norm loses one power of $h$ because the grid has order $h^{-2}$ points; the area-weighted norm is second-order accurate. A regularity assumption is needed for the uniform truncation estimate and hence for this rate; the question does not specify it explicitly.

Start with Taylor's theorem with integral remainder, written on the whole interval using truncated powers:For the Peano kernel theorem, require that $L$ be a linear error functional, annihilate all polynomials of degree at most $k$, and permit interchange with this integral. These conditions hold, in particular, for finite linear combinations of point evaluations of $f$ and its derivatives through order $k$, with the truncated-power derivatives interpreted almost everywhere in $\theta$; dominated integration justifies the interchange. Another sufficient setting is a bounded functional on a differentiability space where the truncated-power integral is convergent in its norm. Polynomial exactness alone, without linearity and an integration-interchange justification, is not sufficient.

Applying $L$ annihilates the Taylor polynomial and givesThis is the unnormalized Peano kernel convention used by the question; including $1/k!$ within the kernel instead removes the factor outside the integral.

For the stated derivative error, constants and linear and quadratic polynomials are annihilated, so $k=2$. Acting on $(x-\theta {)}_{+}^2$ givesThus the sharp central-secant derivative error satisfiesHence $c=1/6$ works. It is the minimum: for $f(x)=x^3$, $L(f)=3-4=-1$ and $\Vert f^{\prime \prime \prime }{\Vert }_{\infty }=6$, so no smaller constant can satisfy the inequality. ThereforeThe sign-definite kernel explains why a constant third derivative attains the bound exactly.

DefineIt contains zero, and $X\in \mathfrak{h}$ implies $aX\in \mathfrak{h}$ for every real scalar $a$, by replacing $t$ with $at$.

For $X,Y\in \mathfrak{h}$, we prove closure under addition using the Lie product formula in a Lie group. In the local exponential chart, let $Z(s)=\log (\exp (sX)\exp (sY))$. Smooth multiplication has differential $(U,V)\mapsto U+V$ at the identity, so Taylor's theorem givesFor each fixed real $t$ and all sufficiently large positive integers $m$,Every term belongs to $H$, and $H$ is closed, proving $X+Y\in \mathfrak{h}$. Thus $\mathfrak{h}$ is a real vector subspace. It is also closed: if $X_j\to X$ with $X_j\in \mathfrak{h}$, continuity of the Exponential map of a Lie group gives $\exp (tX)\in H$ for each $t$.

For the Lie bracket, use the Adjoint representation of a Lie group. Conjugating a one-parameter subgroup givesConsequently ${\mathrm{Ad}}_{\exp (sX)}Y\in \mathfrak{h}$. For $s\ne 0$, linearity puts $({\mathrm{Ad}}_{\exp (sX)}Y-Y)/s$ in $\mathfrak{h}$. Its limit as $s\to 0$ is ${\mathrm{ad}}_{X}Y=[X,Y]$. Closedness of $\mathfrak{h}$ therefore gives $[X,Y]\in \mathfrak{h}$, proving that the Lie algebra of a closed subgroup is a Lie subalgebra of $\mathfrak{g}$.

At an interior statistical parameter $\theta$, local asymptotic normality means that there are random vectors ${\Delta }_{n,\theta }$ and a symmetric nonnegative Fisher information matrix $I_{\theta }$ such that, for each fixed $h\in {\mathbb{R}}^p$,A common stronger definition requires this expansion also for $h_n\to h$. The fixed-$h$ version suffices here. A regular identifiable model usually has positive definite Fisher information; nonsingularity is unnecessary for the contiguity conclusion.

Here are sufficient differentiability conditions for a proof by Taylor's theorem. Suppose the local probability density functions have common support, ${\ell }_{\vartheta }(y)=\log f(\vartheta ,y)$ is twice continuously differentiable near $\theta$, differentiation can pass through the normalization integral twice, the score function $s_{\theta }=\nabla {\ell }_{\theta }$ has finite second moment, and the local Hessian matrix is bounded in operator norm by an integrable envelope of a function class. These hypotheses implyThe first identity differentiates $\int f_{\theta }=1$ once; the second uses ${\nabla }^2{f}_{\theta }=f_{\theta }({\nabla }^2{\ell }_{\theta }+s_{\theta }{s}_{\theta }^T)$ and differentiates twice. The central limit theorem gives ${\Delta }_{n,\theta }=n^{-1/2}{\sum }_i{s}_{\theta }(Y_i)\stackrel{d}{\to }{N}_p(0,I_{\theta })$. Taylor's integral remainder givesThe weak law of large numbers and the continuity and integrable-envelope hypotheses make the bracket converge in probability to $-I_{\theta }/2$. Indeed, the expectation of the supremum of the Hessian matrix difference over a shrinking neighborhood tends to zero by dominated convergence, and Markov inequality controls its empirical average. This proves local asymptotic normality and explains the role of both derivatives.

Write $Q_n◃P_n$ for contiguity of probability measures: for every measurable sequence $A_n$, $P_n(A_n)\to 0$ implies $Q_n(A_n)\to 0$. Mutual contiguity requires this in both directions. A useful form of Le Cam's first lemma is the following: if the Radon-Nikodym derivative $L_n$ of the absolutely continuous part of $Q_n$ relative to $P_n$ converges in distribution under $P_n$ to $L$, with $L>0$ almost surely and $\mathbb{E}L=1$, then the two sequences are mutually contiguous. This includes the usual case $L_n=d{Q}_n/d{P}_n$. The mean-one condition ensures uniform integrability of $L_n$ and vanishing mass of any singular part; positivity ensures the reverse implication.

Take $P_n=P_{\theta }^n$ and $Q_n=P_{\theta +h/\sqrt{n}}^n$. Their likelihood ratio on the support of $P_n$ satisfies, by local asymptotic normality and the continuous mapping theorem,This limit is positive and $\mathbb{E}L=e^{-v/2}{e}^{v/2}=1$, including $v=0$. The conditions of Le Cam's first lemma therefore hold, giving mutual contiguity of the product sampling laws. The products, rather than the one-observation laws, are the relevant sequences.

For any $\epsilon >0$, let $A_n=\{\Vert {\hat{\theta }}_n-\theta \Vert >\epsilon \}$. Consistency under $P_{\theta }^n$ makes $P_{\theta }^n(A_n)\to 0$, and contiguity gives $Q_n(A_n)\to 0$. Thus the estimator remains consistent for $\theta$ under each fixed local alternative. Since $\Vert \theta +h/\sqrt{n}-\theta \Vert \to 0$, it is also consistent for the moving parameter in the sense that $\Vert {\hat{\theta }}_n-(\theta +h/\sqrt{n})\Vert \to 0$ in probability under $Q_n$.

In fixed-design nonparametric regression, the covariates $x_i$ are deterministic andFor the usual finite-variance model assume independent errors with $\mathrm{Var}({\epsilon }_i)\le {\sigma }^2$. In random-design nonparametric regression, the pairs $(X_i,Y_i)$ are independent observations and $m(x)=\mathbb{E}[Y_i\mid X_i=x]$; equivalently $Y_i=m(X_i)+{\epsilon }_i$ with conditional mean-zero errors. Conditional variances are generally assumed bounded for risk bounds.

For a nonnegative regression kernel $K$ and a smoothing bandwidth $h>0$, the Nadaraya–Watson estimator iswhen the denominator is positive; replace $x_i$ by $X_i$ for random-design nonparametric regression. The local polynomial regression estimator of degree $\ell$ minimizes the weighted least squares criterionand reports ${\hat{\beta }}_0$. If the weighted Gram matrix is invertible, this minimizer is unique. The Nadaraya-Watson estimator is exactly local polynomial regression of degree zero. Higher degrees reproduce nonconstant polynomials and reduce boundary bias of an estimator.

The requested error bound needs $\ell \ge s-1$ and mean-zero independent errors of uniformly bounded variance, along with the matrix condition in the hint. The degree requirement is missing from the printed question. To see the problem, take $\ell =0$, $K=1_{[-1,1]}$, $m(t)=t$, zero errors, $x=0$, and $h=n^{-1/4}$. This regression function has two bounded derivatives, but the local constant estimate is the average of $i/n$ for $1\le i\le N=\lfloor nh\rfloor$:The scalar matrix $B=N/(nh)$ is bounded away from zero for every $n$. Nevertheless $(nh{)}^{-1/2}+h^2=n^{-3/8}+n^{-1/2}=o(h)$, so the printed conclusion with $s=2$ fails. A bounded-variance error hypothesis is also essential; otherwise an expectation bound need not even be finite.

Under the qualified hypotheses, use $U(t)=(1,t,\dots ,t^{\ell }/\ell !{)}^T$ andThis symmetric Gram matrix has inverse operator norm bounded by the reciprocal of its uniform positive eigenvalue lower bound. If $K$ is supported in $[-R,R]$, only $|x_i-x|\le Rh$ contribute. The equally spaced design and $nh\ge 1$ imply that there are at most $2Rnh+1\le Cnh$ such points. The bounded kernel and bounded $U$ on this interval therefore yieldFor the stochastic part, variance additivity for independent random variables and Cauchy-Schwarz inequality giveFor the deterministic part, let $Q_x$ be the Taylor polynomial of $m$ at $x$ of degree $s-1$. The polynomial reproduction property of local polynomial regression gives ${\sum }_i{W}_{ni}(x)Q_x(x_i)=Q_x(x)=m(x)$. By Taylor's theorem, $|m(x_i)-Q_x(x_i)|\le C|x_i-x{|}^s\le C{h}^s$ on the kernel window. ThusCombining the two parts provesThe argument also applies at the endpoints when the assumed matrix bound holds. With arbitrary degree, the generally valid smoothness term is $h^{\min (s,\ell +1)}$; extra symmetry can improve some interior biases but does not fix the general boundary counterexample.

Put $r=k-1$. The linear-reproduction request requires $k\ge 2$, which we use for that final step. Since ${\psi }_i={\omega }_i/r!$, the useful normalized Marsden dual functional isThere is no further factor $1/r!$ outside this sum: it has already been included in ${\psi }_i$.

To derive the formula directly from Marsden's identity, Taylor's theorem for the polynomial $p$ around the arbitrary point $x$ givesOn the other hand,Differentiate Marsden's identity $r-j$ times in $x$, multiply by $(-1{)}^j{p}^{(j)}(x)/r!$, and sum over $j$. The left side becomes $p(t)$, while the right side becomesOnly finite sums and derivatives of polynomials are involved.

Independence of the auxiliary point does not require an assumption about uniqueness of an expansion. Differentiate the formula for ${\lambda }_i$ itself:The first summand at $j=r$ and the second at $j=0$ vanish because both polynomials have degree at most $r$. All remaining terms cancel after shifting the index by one. Therefore ${\lambda }_i(p)$ is a constant in $x$, and it is visibly a linear functional of $p$.

For $p(t)=a+bt$, only $j=0,1$ remain. The leading two coefficients of the monic knot polynomial giveIt follows thatSubstituting into the expansion proveson the basic knot interval. Thus the Greville abscissae are exactly the sampling coefficients that reproduce linear polynomials; the constant case also gives the subpartition of unity for B-splines with equality on this interval.

From $f^{\prime \prime }=f$, differentiating successively gives $f^{(2r)}=f$ and $f^{(2r+1)}=f^{\prime }$ whenever those derivatives have been constructed. Starting with twice differentiability, this relation supplies the next two derivatives at each stage. Thus $f$ is infinitely differentiable, and its origin series isFor a fixed $x$, both $f$ and $f^{\prime }$ are bounded on the compact interval between zero and $x$, say by $M_x$. Every higher derivative is one of these two functions. Taylor's theorem, with the remainder just proved, therefore bounds the order-$n$ remainder by $M_x|x{|}^n/n!$, which tends to zero: its consecutive positive-term ratio is $|x|/(n+1)$. Hence the series converges to $f(x)$ for every real $x$. It also gives $f(x)=A\cosh x+B\sinh x$ by the usual hyperbolic-function series.

At any other centre $a$, apply the proved origin result to $g(t)=f(a+t)$, which also satisfies $g^{\prime \prime }=g$. Its even derivatives at zero are $f(a)$, and its odd ones are $f^{\prime }(a)$. ConsequentlyThis proves convergence at every centre without assuming that smoothness alone implies equality to a Taylor series.

The unheaded definitions use the complex structure $J$ on the real tangent bundle. Compatibility means $g(Ju,Jv)=g(u,v)$. Its fundamental Hermitian form is $\omega (u,v)=g(Ju,v)$; the compatibility identity makes this real, alternating and of type $(1,1)$. The metric is a Kähler metric precisely when $d\omega =0$. In complex dimension one a real three-form is zero, so every compatible metric is a Kähler metric.

Now assume closedness and prove the Kähler normal holomorphic coordinates condition. Start with holomorphic coordinates $w$ centered at the point and make a complex-linear change so that the positive Hermitian coefficient matrix satisfies $h(0)=I$. Closedness of the $(1,1)$ form givesDefine $a_{i\bar{j},k}={\partial }_{w_k}{h}_{i\bar{j}}(0)$; it is symmetric in $i,k$. Choose new coordinates implicitly byThe derivative of this holomorphic map at zero is the identity, so the holomorphic inverse function theorem makes $z$ valid local coordinates. In the new coordinates,At zero, $h^{\prime }=I$, and differentiating givesHermitian symmetry makes all antiholomorphic first derivatives zero as well. Taylor's theorem for a smooth function now yieldsThus condition (a) implies condition (b), with the exact factor $i/2$ in the fundamental-form convention retained.

For an $N+1$ times continuously differentiable function, Taylor's theorem giveswhere $\xi$ lies between $x_0$ and $x$. An infinite Taylor series represents the function when its remainder tends to zero; this holds locally for an analytic function, rather than for every smooth function automatically.

The Taylor expansion about $a$ has coefficients given by the successive derivatives at $a$:When $f$ is $N+1$ times continuously differentiable on the interval between $a$ and $x$, Taylor's theorem gives $R_N(x)=f^{(N+1)}(\xi )(x-a{)}^{N+1}/(N+1)!$ for an intermediate point $\xi$. If the remainders tend to zero, this yields the infinite Taylor series $f(x)={\sum }_{k=0}^{\infty }{f}^{(k)}(a)(x-a{)}^k/k!$. The three functions below are real analytic near zero after taking the continuous extension in part (i).

Taylor's theorem expands about $a$ in powers of $h=x-a$. With the usual differentiability assumptions, its finite form with Lagrange remainder iswhere $\xi$ lies between $a$ and $x$. If the remainder tends to zero as $n\to \infty$, this yields the Taylor series $f(x)={\sum }_{k\ge 0}{f}^{(k)}(a)(x-a{)}^k/k!$. Smoothness alone does not guarantee that infinite-series equality.

The first four terms form the cubic Taylor polynomial about $x_0$:With a continuous fourth derivative nearby, Taylor's theorem gives a remainder $O((x-x_0{)}^4)$. Continuous derivatives through third order suffice for the weaker remainder $o((x-x_0{)}^3)$. These are local approximations; equality with an infinite Taylor series requires additional convergence information.

The cubic Taylor polynomial about $a$ isIf $f$ has a continuous fourth derivative near $a$, Taylor's theorem gives $f(x)$ equal to this polynomial plus $O((x-a{)}^4)$. With continuous derivatives only through third order, the corresponding remainder is $o((x-a{)}^3)$. An equality with an infinite Taylor series would require additional convergence information, rather than just the existence of these four terms.

A cubic version of Taylor's theorem requires the corresponding higher differentiability, rather than only one derivative. For example, if $f$ has four continuous derivatives on the interval between $a$ and $x$, thenwhere the Lagrange form of the Taylor remainder isfor some $\xi$ strictly between $a$ and $x$ when they differ. At $x=a$, the remainder is zero. These hypotheses apply to both functions used below.

With sufficiently many derivatives, Taylor's theorem gives a finite approximation and a controlled remainder. For example, if $f$ is $N+1$ times continuously differentiable between $a$ and $a+\delta$, thenThus derivatives at the nearby base point determine the polynomial approximation; a bound on the next derivative bounds its error. The smallness of $\delta$ makes this useful, provided the derivatives remain controlled.

The statement's bare assumption of differentiability alone guarantees only $f(a+\delta )=f(a)+f^{\prime }(a)\delta +o(\delta )$. Higher-order formulas need higher regularity. Even smoothness does not by itself guarantee equality with an infinite Taylor series; that stronger conclusion requires the remainder to tend to zero.

For $n\ge 3$, Taylor's theorem after $n$ terms, counting the constant term, isUnder the usual hypothesis that $f$ is $n-1$ times continuously differentiable on the interval joining $a$ and $x$, and $n$ times differentiable in its interior, the Taylor theorem with Lagrange remainder givesThis is a finite expansion with a controlled remainder; it does not assume an infinite Taylor series converges. For $x=a$ the remainder is zero.

With the first $n$ terms counted from the constant term, Taylor's theorem giveswhere $\xi$ lies between $a$ and $x$. Continuous derivatives through order $n$ on the joining interval are a sufficient hypothesis for this Taylor theorem with Lagrange remainder. The Taylor remainder measures the error of the finite Taylor polynomial; convergence of the corresponding infinite Taylor series is a separate question.

The first $n$ terms of Taylor's theorem, with the Taylor remainder in mean-value form, arewhere $\xi$ lies between $a$ and $x$ for $x\ne a$. Sufficient hypotheses are continuous derivatives through order $n-1$ on the interval between the endpoints and an $n$th derivative in its interior. At $x=a$ the remainder is zero. The finite Taylor polynomial ends at degree $n-1$; this identity does not require convergence of an infinite Taylor series.

The finite form of Taylor's theorem iswhere $\xi$ lies between $a$ and $x$ when $x\ne a$. Sufficient hypotheses are continuous derivatives up to order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The first $n$ terms end at degree $n-1$; the Taylor remainder uses the $n$th derivative. At $x=a$ the remainder is zero. This finite Taylor polynomial identity does not require convergence of an infinite Taylor series.