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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 a differentiable function, a local linear approximation replaces nearby values by its value plus its derivative times the input change. The error divided by that change tends to zero. It uses the first-order Taylor polynomial; the statement concerns a limit near the specified point, not a guaranteed approximation far from it.

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.

Expand the fifth-degree Taylor polynomial $p_5$ in Chebyshev polynomials, using the identities for powers obtained from part (a):Delete the last two Chebyshev terms and convert back to powers. The economized cubic isSince $|T_j|\le 1$,Equality holds at $x=-1$, where both deleted terms have negative sign. Thus the economization changes the polynomial by at most $7/160$, exactly in supremum norm on $[-1,1]$.

For the target itself, $|\log (1+x)-p_3(x)|\le |\log (1+x)-p_5(x)|+7/160$ wherever $x>-1$. The logarithm is singular at $-1$, so its total uniform error on all of $(-1,1]$ is infinite; the finite estimate concerns the additional polynomial change, or the increase on any smaller interval where the original error is bounded.

pleiotropy means that a single gene or genetic variant influences more than one distinguishable phenotypic trait. Its evolutionary consequence depends on the directions and magnitudes of those effects and on how they affect genotype fitness. It must be distinguished from genetic linkage, in which different causal genetic loci travel together, and from epistasis, in which the effect of one allele depends on another. pleiotropy can constrain adaptation, facilitate coordinated change, and maintain trade-offs; its sign is not intrinsically negative.

For a small change in a phenotypic trait vector $\mathbf{z}$, write its effect as $\mathbf{a}$ and define the phenotypic selection gradient $\mathbf{\beta }={\nabla }_{\mathbf{z}}\log w$. A first-order Taylor polynomial givesA mutation that improves one component can therefore be harmful overall if its other effects oppose sufficiently strong natural selection. For an illustrative two-trait case, $\mathbf{\beta }=(1,1)$ and $\mathbf{a}=(0.2,-0.3)$ give $\Delta \log w\approx -0.1$. With $\mathbf{a}=(0.2,0.1)$, the same approximation gives $0.3$. What matters is the full effect vector, not whether the mutation has been called “large” or “pleiotropic”. Near an optimum the first-order term can vanish, making curvature important.

This geometric constraint is explicit in Fisher's geometric model. Put the optimum at the origin and let $\mathbf{z}$ denote the current displacement. A step $\mathbf{a}$ improves an isotropic quadratic measure of genotype fitness exactly whenA step toward improvement in one component may damage many other components. In models where a step has random direction across many independent selected dimensions, the set of favorable directions can become restricted: this motivates a cost of complexity. But the prediction depends on how total step size, per-trait effect size and the number of affected phenotypic traits scale. Organismal complexity is not automatically equal to the number of dimensions affected by every mutation.

At the population level, the multivariate breeder's equation iswhere $\mathbf{G}$ is the additive genetic covariance matrix. This is a local approximation under the relevant inheritance and weak-selection assumptions.^[ref] The off-diagonal entries cause correlated responses. For example,There is abundant genetic variance for each phenotypic trait separately, yet their negative genetic covariance slows simultaneous increase. For $\mathbf{\beta }=(1,-1{)}^{\mathsf{T}}$, the response instead becomes $(1.8,-1.8{)}^{\mathsf{T}}$: the same covariance facilitates this direction. A singular $\mathbf{G}$ can make an instantaneous response unavailable, but present covariance does not establish an eternal constraint. Moreover, genetic linkage can also produce genetic covariance; estimating $\mathbf{G}$ alone does not demonstrate pleiotropy.

antagonistic pleiotropy arises when effects favor different outcomes in different components of genotype fitness, environments, sexes or life stages. An allele that increases early reproduction but reduces late survival can be favored if its early benefit receives greater reproductive weighting. This provides a possible route to senescence, without implying that deterioration is itself selected for. Sex-specific benefits and costs can produce sexual antagonism. Such conflicts may maintain variation under some combinations of dominance, demography and environmental variation, but opposite effects alone do not guarantee a protected polymorphism. The invasion success of each alternative when rare must be checked.

pleiotropy can also promote adaptation by coordinating changes that work well together. A developmental regulator can change a group of interacting structures; a single physiological change can improve several measures of performance. natural selection may favor developmental modularity, tissue-specific regulation or gene duplication when they reduce costly coupling. The pelvic enhancer example in question 2 illustrates how cis-regulatory evolution can localize an effect; it does not establish a universal privilege of regulatory over coding mutations.

Empirical measurement requires care. Screens of yeast, nematode and mouse mutants found strongly modular relationships between genes and measured phenotypic traits, challenging assumptions of universal equal-effect pleiotropy.^[ref] A deletion can nevertheless have more severe and widespread effects than a naturally segregating regulatory allele; unmeasured environments and phenotypic traits hide effects, whereas multiple correlated measurements can inflate apparent dimensionality. To test causal pleiotropy, introduce the same precise genetic variant into matched genetic backgrounds, measure multiple prespecified phenotypic traits across environments and life stages, and use recombinant or edited lines to exclude a neighboring causal genetic locus. Assay genotype fitness as well as convenient laboratory phenotypic traits. The evolutionary constraint is the alignment between available heritable effects and current selection, not the mere count of affected phenotypic traits.

The first-order multivariable Taylor expansion isTaking $(x_0,y_0)=(2,0)$ gives the linear approximationTo assess its significant figures, the quadratic Taylor polynomial supplies the correction $h^2+4hk+2k^2=-0.0001$ for $h=0.03$, $k=-0.01$. Direct evaluation gives $2.03^2{e}^{-0.01}=4.0798963599\dots$. Thus the estimate is correct to four significant figures: both round to $4.080$, whereas to five significant figures the exact value is $4.0799$ and the approximation is $4.0800$. The unexpectedly small quadratic correction results partly from cancellation of its mixed and unmixed terms.

Use the standard knot-domain convention in which the spline space contains every polynomial of degree at most $k-1$ on $[a,b]$, with completed or clamped endpoint knots. The De Boor–Fix functional extracts a local spline coefficient, and its Hahn-Banach theorem extension is linear on continuous local data with the stated bound. Only that bound and spline reproduction are needed here.

For $x$ in a knot cell $[t_j,t_{j+1}]$, support and Gram bandwidth of B-splines implies that only indicescan contribute, restricted also to $1\le i\le n$. Every corresponding local support $[t_i,t_{i+k}]$ lies insideNonnegativity and subpartition of unity for B-splines therefore giveTaking the supremum over the cell proves the local stability estimate for spline quasi-interpolation:Endpoint limits follow by continuity for $k\ge 2$; the order-one version uses the usual one-sided interval convention.

Let $h={\max }_i|t_{i+1}-t_i|$ and $f\in C^k[a,b]$. Choose any point $y_j$ in the stencil and form its degree-$k-1$ Taylor polynomial $p_j$. The Taylor theorem remainder givesThe polynomial is itself a spline, so $Q{p}_j=p_j$. By linearity and the local stability estimate,Taking the maximum over cells yieldsThe constant depends only on $k$ and the supplied functional bound, not on mesh ratios or the number of knots.

The endpoint convention is essential to the global assertion. The domain can be the basic knot interval, with any needed exterior data extended smoothly, or the full approximation interval can use clamped endpoint knots. If one instead takes a finite, unclamped collection on the entire outer support interval $[t_1,t_{n+k}]$, every order-$k\ge 2$ basis function vanishes at $t_1$, so $Q1(t_1)=0$ and uniform approximation to the constant one there is impossible. This does not affect the local estimate; it explains the usual boundary completion implicit in the global spline approximation statement.

The Taylor formula with integral remainder isFor $n=0$, this is the fundamental theorem of calculus. Integration by parts givesRearranging and applying mathematical induction proves the displayed expansion. The oriented integral also covers $x<a$.

For $x\ne 0$, the chain rule and product rule giveThusBecause Gaussian decay dominates every power, $p(1/x)e^{-1/x^2}/x\to 0$ as $x\to 0$. Hence $f^{\prime }(0)=0$, and the derivative formula extends continuously to zero. Repeating the argument replaces $p$ by another polynomial at each step: $p_{j+1}(t)=2t^3{p}_j(t)-t^2{p}_j^{\prime }(t)$. All derivative difference quotients at zero tend to zero, so the polynomial multiples of a Gaussian flat function are smooth functions withThe Taylor polynomial about zero is therefore identically zero for every degree, andA nonzero polynomial has only finitely many roots of a polynomial, so $p(1/x)\ne 0$ for every sufficiently small nonzero $x$. At each such fixed $x$, this remainder never tends to zero. The flat function is therefore smooth but not real analytic at zero.

Set $r=k-1$, with $k\ge 2$ for the final linear-reproduction assertion, and use the normalized knot polynomial ${\psi }_i={\omega }_i/r!$. Differentiate the Marsden identity $r-j$ times with respect to the auxiliary variable $x$, for $0\le j\le r$. This givesThe Taylor polynomial expansion of an arbitrary polynomial $p$ of degree at most $r$ about $x$ is exact:Substitute the differentiated identity into each term and collect the B-spline coefficients. The result isThis is the normalized Marsden dual functional. There is no extra factorial outside the sum, because it is already included in ${\psi }_i$.

To prove independence of $x$ directly, differentiate the finite sum. The contribution from differentiating ${\psi }_i$ in term $j$ cancels the contribution from differentiating $p$ in term $j-1$. Explicitly,The two uncancelled endpoint terms vanish because $p^{(r+1)}={\psi }_i^{(r+1)}=0$. Thus the coefficient functional is constant in the auxiliary variable, as required.

The leading two terms of the knot polynomial giveFor a linear polynomial $p(t)=A+Bt$, only $j=0,1$ survive, whenceThe coefficient points $t_i^{\ast }$ are the Greville abscissae. Inserting these coefficients proves exact linear reproduction:All identities hold on the basic knot interval $[t_k,t_{n+1}]$, including the endpoints with the standard continuous endpoint convention. Repeated knots cause no difficulty in these polynomial differentiations.

One precise version of the Taylor theorem with Lagrange remainder is as follows. Let $n\ge 0$, let $f$ and its derivatives through order $n$ be continuous on the closed interval between distinct points $a$ and $x$, and suppose $f^{(n+1)}$ exists throughout its interior. Then some $\xi$ strictly between $a$ and $x$ satisfiesHere the endpoint derivatives are understood as restrictions of derivatives defined near the interval, or as their continuous one-sided extensions. No continuity of $f^{(n+1)}$ is required. For $x=a$ the equality is immediate with zero remainder.

To prove the formula, write the Taylor polynomial as $P_n(t)={\sum }_{k=0}^n{f}^{(k)}(a)(t-a{)}^k/k!$ and defineThen $h(a)=h(x)=0$, and $h^{(j)}(a)=0$ for $j=1,\dots ,n$. The Rolle theorem gives a zero ${\xi }_1$ of $h^{\prime }$ strictly between $a$ and $x$. If $n\ge 1$, apply the Rolle theorem to $h^{\prime }$ between $a$ and ${\xi }_1$, using $h^{\prime }(a)=0$, to get a zero ${\xi }_2$ of $h^{\prime \prime }$. Repeating produces a point ${\xi }_{n+1}$ strictly between $a$ and $x$ with $h^{(n+1)}({\xi }_{n+1})=0$. Each application is justified by continuity on the relevant closed interval and differentiability in its interior; the same construction works when $x<a$.

Since $P_n$ has degree at most $n$,Substituting this value of $K$ into its definition proves the displayed Taylor theorem with Lagrange remainder.

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 Arzelà-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. Arzelà-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.