# The crux of the matter is that error function e n x f

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The crux of the matter is that error functionen(x) =f(x)-pn(x),x2[a, b],(2.39)has toequioscillateat leastn+2 points, between +kenk1and-kenk1. Thatis, there arekpoints,x1, x2, . . . , xk, withkn+ 2, such thaten(x1) =±kenk1en(x2) =-en(x1),en(x3) =-en(x2),...en(xk) =-en(xk-1),(2.40)for if not, it would be possible to find a polynomial of degree at mostn,with the same sign at the extrema ofen(at mostnsign changes), and usethis polynomial to decrease the value ofkenk1. This would contradict thefact thatpnis a best approximation. This is easy to see forn= 0 as it isimpossible to find a polynomial of degree 0 (a constant) with one change ofsign. This is the content of the next result.Theorem 4.The erroren=f-pnhas at least two extremax1andx2in[a, b]such that|en(x1)|=|en(x2)|=kenk1anden(x1) =-en(x2)for alln0.Proof.The continuous function|en(x)|attains its maximumEn=kenk1inat least one pointx1in [a, b]. SupposeEn=en(x1) and thaten(x)>-Enfor allx2[a, b]. Then,m= minx2[a,b]en(x)>-Enand we have some roomto decreaseEnby shifting downena suitable amountc. In particular, if takecas one half the gap between the minimummofenand-En,c=12(m+En)>0,(2.41)
28CHAPTER 2.FUNCTION APPROXIMATIONand subtract it toenwe have-En+cen(x)-cEn-c.(2.42)Therefore,ken-ck1=kf-(pn+c)k1=En-c < Enbutpn+c2Pnsothis is impossible sincepnis a best approximation. A similar argument canused whenen(x1) =-En.Before proceeding to the general case, let us look at then= 1 situation.Suppose there only two alternating extremax1andx2fore1as describedin (2.40). We are going to construct a linear polynomial that has the samesign ase1atx1andx2and which can be used to decreaseE1.Supposee1(x1) =E1ande1(x2) =-E1.Sincee1is continuous, we can find smallclosed intervalsI1andI2, containingx1andx2, respectively, and such thate1(x)>E12for allx2I1,(2.43)e1(x)<-E12for allx2I2.(2.44)ClearlyI1andI2are disjoint sets so we can choose a pointx0between thetwo intervals. Then, it is possible to find a linear polynomialqthat passesthroughx0and that is positive inI1and negative inI2. We are now goingto find a suitable constant>0 such thatkf-p1-qk1< E1.Sincep1+q2P1this would be a contradiction to the fact thatp1is a bestapproximation.LetR= [a, b]\(I1[I2) andE01= maxx2R|e1(x)|.ClearlyE01< E1.Choosesuch that0<<12kqk1(E1-E01).(2.45)OnI1, we have0<q(x)<12kqk1(E1-E01)q(x)12(E1-E01)< e1(x).(2.46)Therefore|e1(x)-q(x)|=e1(x)-q(x)< E1,for allx2I1.(2.47)
2.3.BEST APPROXIMATION29Similarly, onI2, we can show that|e1(x)-q(x)|< E1. Finally, onRwehave|e1(x)-q(x)||e1(x)|+|q(x)|E01+12(E1-E01)< E1.(2.48)Therefore,ke1-qk1=kf-(p1+q)k1< E1, which contradicts the bestapproximation assumption onp1.