Real Analysis Grinshpan The derivative of the limit vs the limit of derivatives ( )0 0 Question: Under what conditions lim fn = lim f ?(∗) n!1 n!1 n The conditions must, in particular, assume or imply that fn are differentiable, converge to a 0 differentiable function, and fn converge as well. Here are a few caveats. The uniform limit of differentiable functions on an interval need not be differentiable. p Example: x2 + 1=n ⇒ jxj on the entire real line. Even if the uniform limit function is differentiable, the sequence of derivatives need not converge. p p Example: sin(nx)= n ⇒ 0 on the entire real line, but n cos(nx) is nowhere convergent. Finally, a sequence of functions whose derivatives converge uniformly may diverge everywhere. − n 0 ⇒ Example: fn(x) = ( 1) diverges, and yet fn 0: In fact, the uniform convergence of derivatives is almost strong enough to guarantee (∗). But the preceding example shows that additional assumptions are needed. In all statements below the underlying set, I; is a closed interval. 0 Lemma 1: If the derivatives fn converge uniformly on I and if fn converge at some point x0 2 I; then fn converge uniformly on all of I: Proof of Lemma 1: Let fn(x0) be convergent for some x0 2 I: For each x 2 I; we have by the triangle inequality and by the mean value theorem applied to fn − fm : jfn(x) − fm(x)j ≤ j(fn − fm)(x) − (fn − fm)(x0)j + jfn(x0) − fm(x0)j ≤ j 0 − 0 jj j j − j 2 fn(ξ) fm(ξ) I + fn(x0) fm(x0) ; for some ξ = ξmn(x) I: j 0 − 0 j " j − j " Given " > 0; choose an index k so that fn fm < 2jIj on I and fn(x0) fm(x0) < 2 ; for 0 all n; m > k: This is possible because fn is uniformly Cauchy and fn(x0) is Cauchy. Then jfn(x) − fm(x)j < " for all n; m > k: This shows that fn is uniformly Cauchy. ! 0 2 Lemma 2: If fn f and if fn converge uniformly on I; then, for any x0 I; fn(x) − fn(x0) f(x) − f(x0) ⇒ ; x =6 x0: x − x0 x − x0 Proof of Lemma 2: Fix any x0 2 I: By the mean value theorem, − − − − − fm(x) fm(x0) − fn(x) fn(x0) (fm fn)(x) (fm fn)(x0) − 0 0 − 0 = = (fm fn) (ξ) = fm(ξ) fn(ξ); x − x0 x − x0 x − x0 − 2 0 fn(x) fn(x0) for some ξ = ξmn(x) I: Thus, since fn is uniformly Cauchy, so is : x − x0 fn(x) − fn(x0) f(x) − f(x0) And since fn ! f; we must have ⇒ on I n fx0g: x − x0 x − x0 We are now ready to give sufficient conditions for (∗): Theorem: Let fn be defined and differentiable on a closed interval I and let the derivatives 0 fn be uniformly convergent on I: Then, if fn converge at some point of I; they converge uniformly on all of I to a differentiable limit, and ( )0 0 lim fn = lim f : n!1 n!1 n ⇒ 0 ⇒ 2 Proof of Theorem: Let fn f (Lemma 1) and fn g: Fix any x0 I: 0 We will show that f (x0) exists and agrees with g(x0): Let " > 0 be given. Write − − − − f(x) f(x0) − ≤ f(x) f(x0) − fn(x) fn(x0) fn(x) fn(x0) − 0 0 − g(x0) + fn(x0) + fn(x0) g(x0) : x − x0 x − x0 x − x0 x − x0 − − j 0 − j " f(x) f(x0) − fn(x) fn(x0) " 6 Select n : fn(x0) g(x0) < and < ; x = x0 (Lemma 2). 3 x − x0 x − x0 3 − fn(x) fn(x0) − 0 " j − j Choose δ > 0 so that fn(x0) < for 0 < x x0 < δ: x − x0 3 Then, for any x with 0 < jx − x j < δ; 0 f(x) − f(x0) " " " − g(x0) < + + = ": x − x0 3 3 3 It follows that f(x) − f(x ) lim 0 = g(x ); x 2 I: ! 0 0 x x0 x − x0 Example: Consider the sequence fn(x) = sin(x=n) + cos(x=n): The sequence converges pointwise to f(x) ≡ 1: Is the convergence uniform? While we may address this question directly, 0 j 0 j ≤ it is clear that the sequence of derivatives fn(x) satisfies fn(x) 2=n and so converges to 0 uniformly on all of R: So fn ⇒ 1 on all of R; by the preceding theorem..