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A compactness proof of the interior Schauder estimate can magnify a pair of points where the Hölder seminorm of the Hessian matrix is nearly attained. If the proposed estimate fails, normalize that seminorm to one while lower norms and forcing tend to zero. The separation $r_j$ of the chosen pair tends to zero. Subtract the quadratic Taylor polynomial and divide by $r_j^{2+\alpha }$ after rescaling space by $r_j$.

The resulting functions have uniformly bounded Hessian Hölder seminorms on expanding domains and vanishing Taylor data at zero. Compact embedding of Hölder spaces yields an entire harmonic function whose second derivatives have finite global $\alpha$-Hölder seminorms, with $0<\alpha <1$. The polynomial-growth Liouville theorem for harmonic functions makes those derivatives constant, contradicting the normalized nonzero oscillation. The Simon absorption lemma removes the residual small multiple of the larger-ball seminorm.

For $m\ge 2$, the symmetric Hadamard finite-part integral defining ${\Lambda }_m$ subtracts the Taylor polynomial through degree $m-2$ from a test function before division by $x^m$. It has finite order of a distribution and satisfiesThe lower-order subtraction terms are integrable at infinity, while the remaining odd $1/x$ singularity cancels under symmetric truncation.

Write $\alpha =r+\beta$ with an integer $r\ge 0$ and $0<\beta \le 1$, using $r=\alpha -1$ at integer $\alpha$. For $f\in C^{r,\beta }$, subtract its degree-$r$ Taylor polynomial at $x_0$. For $r\ge 1$, the integral Taylor remainder with $f^{(r)}(t)-f^{(r)}(x_0)$ is bounded using the Hölder seminorm; for $r=0$ this is exactly Hölder continuity. The resulting bound is the displayed estimate. At integer order, the convention is $C^{r,1}$; a $C^{r+1}$ function on a compact closed interval also obeys this estimate.

For equally spaced fixed-design nonparametric regression, independent mean-zero uniformly finite-variance errors, a bounded compactly supported kernel, and a uniformly invertible local Gram matrix, degree $\ell \ge s-1$ gives the displayed bound for $s$ bounded derivatives. Kernel weights have absolute sum $O(1)$ and squared sum $O((nh{)}^{-1})$. Polynomial reproduction removes a degree-$s-1$ Taylor polynomial. Without the degree requirement the general bias of an estimator is only $O(h^{\min (s,\ell +1)})$.

For a nonprincipal Dirichlet character modulo $N$, setThe numerator vanishes at $t=0$, so $H_{\chi }$ is analytic there and exponentially decreasing as $t\to \infty$. Initially for $\mathrm{Re}s>1$, $\Gamma (s)L(\chi ,s)={\int }_0^{\infty }{H}_{\chi }(t)t^{s-1}dt$. This integral already extends holomorphically to $\mathrm{Re}s>0$. Subtracting finite Taylor polynomials near zero, as in meromorphic continuation of a Mellin transform from an asymptotic expansion, continues it with possible simple poles only at nonpositive integers. Zeros of the reciprocal Gamma function cancel these, proving that $L(\chi ,s)$ is an entire function. This does not require a primitive Dirichlet character.

If $f\in L^1(\mathbb{R})$ and $x^{k}f\in L^1(\mathbb{R})$ for $0\le k\le r$, then its Fourier transform is $r$ times continuously differentiable and the displayed formula holds. The dominated convergence theorem permits each derivative under the integral. Thus $r+1$ vanishing moments give a zero Taylor polynomial through degree $r$ at the origin. Unweighted $L^1$ membership only guarantees a continuous function as the Fourier transform.

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.

We use a blow-up compactness proof of an interior Schauder estimate, with Taylor normalization in elliptic blow-up arguments. Fix $n,\mu ,\delta$ and suppose no suitable constant exists. There would be solutions $u_j$ with forcing $f_j$ for whichChoose distinct $x_j,y_j\in B_{1/2}$ such that, for $r_j=|y_j-x_j|$,Then $S_j/A_j<2/\delta$, $N_j/A_j<2/j$, andIn particular $r_j\to 0$. Subtract the quadratic Taylor polynomial of $u_j$ at $x_j$ and rescale:Its domain $\{z:x_j+r_{j}z\in B_1\}$ exhausts ${\mathbb{R}}^n$. The normalization givesTaylor theorem, or integration along line segments from zero, consequently bounds $D^2{v}_j,\nabla v_j,v_j$ on every fixed ball by constants times $|z{|}^{\mu },|z{|}^{1+\mu },|z{|}^{2+\mu }$, respectively. The Hessian matrices are equicontinuous. The Arzela-Ascoli theorem and a diagonal subsequence therefore give $v_j\to v$ in $C^2$ on compact subsets of ${\mathbb{R}}^n$.

The rescaled equation ison each compact set. Thus $v$ is harmonic and is smooth. Every entry of its Hessian matrix is harmonic and has a global Hölder seminorm at most $2/\delta$. The allowed Liouville lemma for globally Hölder harmonic functions makes each entry constant. Since $D^{2}v(0)=0$, all these constants are zero.

On the other hand, $e_j=(y_j-x_j)/r_j$ is a unit vector and $|D^2{v}_j(e_j)|=1$ by construction. A further subsequence has $e_j\to e$; the local $C^2$ convergence implies $|D^{2}v(e)|=1$, a contradiction. Hence the desired constant exists, andThe global Hölder seminorm on the limiting Hessian matrix, rather than a bound on its magnitude throughout space, is what makes this Liouville argument work.

Use the paper's classical norm convention: $|u{|}_{2;U}$ is the derivative supremum norm through order two, not an $L^2$ norm. Write $H(u;U)=[D^{2}u{]}_{\alpha ;U}$ and $Q(u,f)=|u{|}_{2;B_1}+|f{|}_{0,\alpha ;B_1}$, where the latter is the full Hölder norm of the forcing.

Fix $0<\delta <1$. If the asserted estimate failed, rescaling the functions by their global Hessian matrix Hölder seminorm would produce $u_j,f_j$ withIn particular $Q(u_j,f_j)\to 0$. Choose $x_j,y_j\in B_{1/2}$ for whichThe supremum norm of $D^2{u}_j$ tends to zero, so $r_j=|y_j-x_j|\to 0$. Let $P_j$ be the quadratic Taylor polynomial of $u_j$ at $x_j$, and defineThe domains contain $B_{1/(2r_j)}$, and $U_j,D{U}_j,D^2{U}_j$ vanish at zero. MoreoverThe right side is bounded by $[f_j{]}_{\alpha }|z{|}^{\alpha }$ and tends to zero locally uniformly. The normalized Hessian matrix bound controls $|D^2{U}_j(z)|\le |z{|}^{\alpha }$; integration along line segments then bounds $D{U}_j,U_j$ on every compact Euclidean ball. Arzela-Ascoli theorem and a diagonal subsequence argument give convergence in $C^2$ on compact subsets to an entire harmonic function $U$, with $[D^{2}U{]}_{\alpha ;{\mathbb{R}}^n}\le 1$ and $D^{2}U(0)=0$.

After a further subsequence $z_j=(y_j-x_j)/r_j\to z_{\ast }$ with $|z_{\ast }|=1$. The normalization ensures $|D^{2}U(z_{\ast })|\ge \delta /2$. But each second derivative of $U$ is harmonic and has finite global Hölder seminorm with exponent below one. The supplied Liouville theorem makes each derivative constant; it is the globally Hölder case of the polynomial-growth Liouville theorem for harmonic functions. Its value at zero makes it zero, a contradiction. This blow-up compactness proof of an interior Schauder estimate establishes

To absorb the term on the larger Euclidean ball, rescale this estimate to Euclidean balls contained in $B_s$. For $1/2\le r<s<1$, pairs separated by less than $(s-r)/2$ are controlled in a Euclidean ball centred at their first point; farther pairs are controlled directly by $\Vert D^{2}u{\Vert }_{\infty }$. ThusThis is the covering step in the Simon absorption lemma. Take $r_j=1-2^{-j-1}$ and fix $\delta <2^{-2-\alpha }$. Iterating makes the remainder ${\delta }^{j}H(u;B_{r_j})$ tend to zero, since the global seminorm is finite; the error terms form a convergent geometric series. Consequently

For the requested failure of the weaker estimate, take $u_m=m^{-2}\sin (m{x}_1)$ and $f_m=-\sin (m{x}_1)$. The derivative supremum norm $|u_m{|}_2$ and $\Vert f_m{\Vert }_{\infty }$ remain bounded, but comparison at $0$ and $\pi e_1/(2m)$, both in $B_{1/2}$ for large $m$, givesThis proves the necessity of Hölder forcing for Schauder estimates. Every member of the sequence is smooth; failure comes from the increasing frequency, not from any lack of individual regularity.

Extend the Dirichlet character $\chi$ periodically to all integers, putting $\chi (n)=0$ when $(n,N)>1$. Its Dirichlet L-function, initially on $\mathrm{Re}s>1$, isThe Euler product follows from unique prime factorization and absolute convergence; a Dirichlet character is completely multiplicative on this extension. Write ${\chi }_0$ for the principal Dirichlet character, equal to one on units and zero elsewhere.

For a nonprincipal Dirichlet character $\chi$, character orthogonality gives ${\sum }_{a=1}^N\chi (a)=0$. Explicitly, multiplication by a unit $u$ with $\chi (u)\ne 1$ permutes the unit residues and multiplies this sum by $\chi (u)$, forcing it to vanish. For $t>0$ setThe numerator is $O(t)$ at zero and the denominator is $Nt+O(t^2)$, so $H_{\chi }$ is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for $\mathrm{Re}s>1$, absolute convergence justifiesThis Mellin transform integral is holomorphic for $\mathrm{Re}s>0$, locally uniformly in $s$, and division by the Gamma function proves the requested analytic continuation to the left of the line one.

In fact, the same argument proves Mellin continuation of a nonprincipal Dirichlet L-function to the entire plane. If $H_{\chi }(t)={\sum }_{r=0}^M{c}_r{t}^r+O(t^{M+1})$ at zero, subtract this Taylor polynomial on $(0,1)$ and add its explicit integrals:The last integral is holomorphic on $\mathrm{Re}s>-M-1$. Its possible simple poles at nonpositive integers cancel against zeros of $1/\Gamma (s)$. Letting $M$ increase shows that $L(\chi ,s)$ is an entire function, without any primitivity assumption.

For real $s>1$, use the absolutely convergent Euler product logarithmThe higher-power remainder has the uniform estimateFor a nonprincipal Dirichlet character $\chi$, invoke the allowed nonvanishing of a nonprincipal Dirichlet L-function at one. Its holomorphy and nonvanishing give a holomorphic logarithm on a small disk about one. On the connected real interval $1<s<1+\eta$, $B_{\chi }(s)$ differs from this logarithm by a fixed element of $2\pi i\mathbb{Z}$: the difference is continuous with exponential one. Thus the prime-character sum near one is bounded. This branch argument is needed for complex-valued Dirichlet characters.

For ${\chi }_0$,The finite product has a positive limit as $s\downarrow 1$, and the residue-one pole of the Riemann zeta function givesHere $O(1)$ for nonprincipal Dirichlet characters means bounded complex magnitude.

Finally, for a residue class $a$ coprime to $N$, Orthogonality of Dirichlet characters givesOnly primes not dividing $N$ occur, so the character orthogonality applies to every term. This sum diverges as $s\downarrow 1$. A finite collection of primes would give a bounded sum, a contradiction. Every reduced residue class contains infinitely many primes. This is the Dirichlet theorem on primes in arithmetic progressions; the coprimality hypothesis is essential.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

For $0<\alpha \le 1$, uniform Lipschitz-$\alpha$ means that one constant $M$ satisfies $|f(x)-f(y)|\le M|x-y{|}^{\alpha }$ for every $x,y\in [0,1]$. The resulting Hölder space has normFor the subsequent estimates at general $\alpha >0$, use the higher-order Hölder class convention: write $\alpha =r+\beta$ with $r\ge 0$ an integer and $0<\beta \le 1$, and require $f\in C^r([0,1])$ with $f^{(r)}$ uniformly Lipschitz-$\beta$. ThusAt integer $\alpha =k$, this uses $C^{k-1,1}$; the alternative classical $C^k$ convention is stronger on this compact closed interval and also suffices for the estimates.

For exponents above one, a first-difference inequality alone forces the function to be constant. Partition $[x,y]$ into $n$ equal pieces and use the triangle inequality: $|f(y)-f(x)|\le M|y-x{|}^{\alpha }{n}^{1-\alpha }\to 0$. Consequently a nontrivial higher-order definition is needed for the full range of exponents in the later wavelet estimates. Equivalently, higher-order Hölder class gives a local Taylor polynomial with remainder bounded by $C\Vert f{\Vert }_{C^{\alpha }}|x-x_0{|}^{\alpha }$.

Use localization and cancellation of the assumed interval-adapted wavelet basis: at scale $j$, every detail wavelet has support diameter at most $B{2}^{-j}$, L1 norm at most $B^{\prime }{2}^{-j/2}$, and annihilates polynomials of degree less than $q$. These bounds also apply to the modified boundary wavelets; a restriction of an arbitrary whole-line wavelet without the cancellation-preserving boundary construction would not suffice. The finitely many coarse scaling functions can be treated separately.

Write $\alpha =r+\beta$ with $0<\beta \le 1$ as above. Because $\alpha <q$, we have $r<q$. Choose $x_0$ in the support of a function of ${\psi }_{j,n}$, and let $T_{x_0}$ be the degree-$r$ Taylor polynomial of $f$. The Hölder-Taylor remainder bound givesFor $r=0$ this is the Hölder seminorm bound. For $r\ge 1$, the integral Taylor remainder is bounded using $f^{(r)}(t)-f^{(r)}(x_0)$, whose magnitude is at most $[f^{(r)}{]}_{C^{0,\beta }}|t-x_0{|}^{\beta }$.

The vanishing moments remove $T_{x_0}$ from the inner product. Applying the remainder estimate and the L1 norm bound on the localized support yields the wavelet coefficient decay for Hölder functionsThusHere $C$ depends on the fixed wavelet family, the exponent and boundary construction, but not on $f,j,n$. The proof uses the assumed regularity and moments; it does not identify the minimal Daubechies wavelet order with its differentiability order, which need not coincide.

Repeated integration by parts inproduces the first $k+1$ terms of the Taylor polynomial and leaves $f(x)$, proving the integral remainder in Taylor theoremApply the linear functional ${\lambda }_k$ to this identity. It annihilates the Taylor polynomial, and commuting it with integration gives the Peano kernel theorem

For the midpoint quadrature rule on $[0,1]$, $e[1]=e[x]=0$, so $k=1$. Its Peano kernel isFor $f(x)=x^2$, the formula givesagreeing with ${\int }_0^1{x}^{2}dx-(1/2{)}^2=1/3-1/4$.

Write $K(u,f)=\Vert u{\Vert }_{C^2(B_1)}+\Vert f{\Vert }_{C^{0,\alpha }(B_1)}$; the paper's $|u{|}_{2;B_1}$ denotes a $C^2$ norm, not an $L^2$ norm. We use a blow-up compactness proof of an interior Schauder estimate. If the estimate failed for fixed $\delta >0$, normalization would produceChoose $x_j,y_j\in B_{1/2}$ with $d_j=|y_j-x_j|$ and $|D^2{u}_j(y_j)-D^2{u}_j(x_j)|\ge \frac{1}{2}{d}_j^{\alpha }$. Since $\Vert D^2{u}_j{\Vert }_{\infty }\to 0$, necessarily $d_j\to 0$. Subtract the quadratic Taylor polynomial $P_j$ at $x_j$ and setThe domains expand to ${\mathbb{R}}^n$. At zero, $w_j,D{w}_j,D^2{w}_j$ vanish; the Hölder seminorm of $D^2{w}_j$ is at most ${\delta }^{-1}$. Integration along segments gives uniform $C^{2,\alpha }$ bounds on each fixed ball. Alsolocally uniformly, since $[f_j{]}_{\alpha }\to 0$. By compact embedding of Hölder spaces and a diagonal subsequence, $w_j$ converges locally in $C^{2,\beta }$ to an entire harmonic function $w$, with $[D^{2}w{]}_{\alpha }\le {\delta }^{-1}$ and $D^{2}w(0)=0$. The unit vectors $(y_j-x_j)/d_j$ have a convergent subsequence; its limit $e$ satisfies $|D^{2}w(e)|\ge 1/2$.

Each second derivative of $w$ is an entire harmonic function with finite global Hölder seminorm. The supplied polynomial-growth Liouville theorem for harmonic functions, at growth exponent $\alpha <1$, makes it constant, and its value at zero makes it zero. This contradicts the nonzero Hessian at $e$, proving the estimate with the small $\delta$ term.

To remove that term, rescale the estimate to all contained balls and apply the Simon absorption lemma. Explicitly, with $d(x)=1-|x|$ and $d_{xy}=\min (d(x),d(y))$, defineFor $|x-y|<d_{xy}/8$, apply the rescaled estimate on $B_{d(x)/2}(x)$; its larger-ball seminorm is controlled by $M$ because all points there have boundary distance at least $d(x)/2$. For more separated pairs use $2\Vert D^{2}u{\Vert }_{\infty }$. The result is $M\le C_0\delta M+C_{\delta }K(u,f)$. Choose $C_0\delta <1/2$ and use $d_{xy}\ge 1/2$ on $B_{1/2}$ to obtainThe forcing's Hölder seminorm cannot be discarded. For integers $k\to \infty$, takeTheir $C^2$ and forcing supremum norms stay bounded, but the $11$ entry of the Hessian matrix differs by two at $x_1=\pm \pi /(2k)$, giving $[D^2{u}_k{]}_{\alpha ;B_{1/2}}\ge 2(k/\pi {)}^{\alpha }\to \infty$.

Let $T_q\phi (x)={\sum }_{k=0}^q{\phi }^{(k)}(0)x^k/k!$ denote the Taylor polynomial at zero. On $0<|x|<1$, the integrand defining the Hadamard finite-part reciprocal-power distribution isThe first term is odd, so it cancels under symmetric truncation. The Taylor remainder in the second term has absolute value at most $|x{|}^m\Vert {\phi }^{(m)}{\Vert }_{\infty }/m!$. The integral near zero therefore has a finite limit. For $|x|\ge 1$, the original terms are integrable: $x^{-m}\phi$ has compact support, and each subtracted power has exponent $k-m\le -2$.