uniform

Analysis II (Michaelmas 2009) A.G. [email protected] Term by term integration and differentiation Sometimes the calculus one needs to do involves functions which cannot be defined in a traditional way by a formula, but only in terms of convergent series of ‘elementary’ functions. This then poses a question: When is the formal term by term integration or differentiation of a series of functions valid (i.e. will give the same result as the integration or differentiation applied directly to the sum of the series)? Earlier we proved the following. Theorem 1. Suppose that fn: [a, b]→R, for each k= 1,2, . . ., is integrable on [a, b]and that fn(x)→f(x)uniformly on [a, b]asn→ ∞ . Then the limit function f(x)is integrable on [a, b]and/integraltextb af(x)dx= lim n→∞/integraltextb afn(x)dx. This can be easily converted into a version for the series. Theorem 1/prime(Term by term integration) .Suppose that uk: [a, b]→R, for each k= 1,2, . . ., is integrable on [a, b]and/summationtext∞ k=1uk(x)converges uniformly on [a, b]. Then the sum f(x) =/summationtext∞ k=1uk(x)is integrable on [a, b]and/integraltextb af(x)dx=/summationtext∞ k=1/integraltextb auk(x). Proof. Putsn(x) =/summationtextn k=1uk(x) and apply Theorem 1 to the sequence ( sn). Thus uniformly convergent series can be integrated term by term. What about term by term differentiation? Here the situation is somewhat less elegant than with integration and there is a good reason for that. Very informally , the integration is a ‘bounded’ operation whereas the differentiation is not. Ifu(x) is ‘small’, say |u(x)|< εfor each x, then the integral |/integraltext1 0u(x)dx|< εis also ‘small’, but the derivative |u/prime(x)|may be arbitrary ‘large’, consider e.g. u(x) =εsin(x/ε2) and let ε→0. The next theorem is essentially a result for integration in disguise (as you will see from the proof). We can assume very little on the initial series, but the term by term differentiated series must satisfy a strong condition of being uniformly convergent. Theorem 2 (Term by term differentiation) .Suppose that uk: [a, b]→R, for each k= 1,2, . . ., has continuous derivative on [a, b](at the end-points aandbthis means one-sided derivative). Suppose further that (i) the series/summationtext∞ k=1uk(x0)converges at some point x0∈[a, b]and (ii) the series of derivatives/summationtext∞ k=1u/prime k(x)converges uniformly on [a, b], tof(x) =/summationtext∞ k=1u/prime k(x)say. Then (1) the series/summationtext∞ k=1uk(x)converges at every x∈[a, b]and the sum F(x) =/summationtext∞ k=1uk(x)is differentiable with F/prime(x) =f(x)for each x∈[a, b]; (2) moreover, the convergence of/summationtext∞ k=1uk(x)toF(x)is uniform on [a, b]. Proof. (1) As each u/prime kis continuous and/summationtext∞ k=1u/prime kis uniformly convergent, we have fcontinuous on [a, b].1Therefore, each u/prime kandfare integrable on [ a, b]. Let x∈[a, b]. Applying Theorem 1/prime tou/prime kandfon [x0, x] (or on [ x, x0] ifx < x 0) we obtain ∞/summationdisplay k=1/integraldisplayx x0u/prime k(x)dx=/integraldisplayx x0f(x)dx. (1) 1because we proved earlier that the limit of a uniformly convergent sequence of continuous functions is continuous 1 The left-hand side of (1) may be written as ∞/summationdisplay k=1/integraldisplayx x0u/prime k(x)dx= lim n→∞n/summationdisplay k=1/integraldisplayx x0u/prime k(x)dx = lim n→∞n/summationdisplay k=1/parenleftbig uk(x)−uk(x0)/parenrightbig = lim n→∞/parenleftbiggn/summationdisplay k=1uk(x)−n/summationdisplay k=1uk(x0)/parenrightbigg (2) The lim n→∞/summationtextn k=1uk(x0) exists by hypothesis (i), therefore the series F(x) = lim n→∞/summationtextn k=1uk(x) converges for each x∈[a, b]. Thus a function F: [a, b]→Ris well-defined and from (1) and (2) we obtain F(x)−F(x0) =/integraldisplayx x0f(x)dx. Differentiation in xgives F/prime(x) =f(x) for x∈[a, b] asfis continuous. (2) Let ε >0. Then there is N1such that whenever n≥m≥N1we have /vextendsingle/vextendsingle/vextendsingle/vextendsinglen/summationdisplay k=muk(x0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle<ε 2, applying the Cauchy criterion to the convergent series/summationtext∞ k=1uk(x0). Also there is N2such that whenever n≥m≥N2we have /vextendsingle/vextendsingle/vextendsingle/vextendsinglen/summationdisplay k=mu/prime k(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle<ε 2(b−a),for each x∈[a, b], applying the Cauchy criterion to the uniformly convergent series of functions/summationtext∞ k=1u/prime k(x). PutN= max {N1, N2}andg(x) =/summationtextn k=muk(x). Then for x∈[a, b] we can write g(x)− g(x0) =g/prime(ξ)(x−x0) with some ξ∈[x0, x]⊆[a, b] (or ξ∈[x, x0]⊆[a, b]), using the Mean Value Theorem from Analysis I. We then obtain |g(x)| ≤ |g(x0)|+|g(x)−g(x0)|=|g(x0)|+|g/prime(ξ)| · |x−x0|<ε 2+ε 2(b−a)(b−a) =ε. Thus for n≥m≥Nand for each a≤x≤bwe have |/summationtextn k=muk(x)| ≤εwhich is the Cauchy criterion for uniform convergence of/summationtext∞ k=1uk(x) as we had to prove. Once again, it is now easy to obtain a version of the result concerning differentiation for the sequences of functions. Corollary 1. Suppose that fn: [a, b]→R, for each n= 1,2, . . ., has continuous derivative on [a, b]andfn(x)→f(x)point-wise for each x∈[a, b]. Iff/prime n(x)→ϕ(x)uniformly on [a, b]then the function fis differentiable with f/prime(x) =ϕ(x)for each x∈[a, b]. Moreover, f/prime n→funiformly on[a, b]. Proof. After putting u1=f1,un=fn−fn−1forn >1, the result follows from Theorem 2. Remark 1.The condition that the derivatives u/prime kin Theorem 2 (resp. f/prime nin Corollary 1) are continuous can be dropped. The result then is still true but the proof is longer. You can find details in K¨ orner’s book (although he still assumes the continuity) or in Rudin’s book. Remark 2.One more remark for interest is that the condition of uniform continuity in Theorem 1, while certainly sufficient, is in fact too strong. It would suffice to assume e.g. that |fn(x)| ≤1 for all nandx∈[a, b]. The proof is not obvious and requires some more advanced theory of Lebesgues integral and measure; you can learn it in Part II. 2