Real Analysis

Grinshpan

The derivative of the limit vs the limit of derivatives

( )′ ′ Question: Under what conditions lim fn = lim f ?(∗) n→∞ n→∞ n

The conditions must, in particular, assume or imply that fn are differentiable, converge to a ′ 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. √ Example: x2 + 1/n ⇒ |x| on the entire real line.

Even if the uniform limit function is differentiable, the sequence of derivatives need not converge. √ √ 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 ′ ⇒ 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.

′ Lemma 1: If the derivatives fn converge uniformly on I and if fn converge at some point x0 ∈ I, then fn converge uniformly on all of I.

Proof of Lemma 1: Let fn(x0) be convergent for some x0 ∈ I. For each x ∈ I, we have by the triangle inequality and by the mean value theorem applied to fn − fm :

|fn(x) − fm(x)| ≤ |(fn − fm)(x) − (fn − fm)(x0)| + |fn(x0) − fm(x0)| ≤ | ′ − ′ || | | − | ∈ fn(ξ) fm(ξ) I + fn(x0) fm(x0) , for some ξ = ξmn(x) I. | ′ − ′ | ε | − | ε Given ε > 0, choose an index k so that fn fm < 2|I| on I and fn(x0) fm(x0) < 2 , for ′ all n, m > k. This is possible because fn is uniformly Cauchy and fn(x0) is Cauchy. Then |fn(x) − fm(x)| < ε for all n, m > k. This shows that fn is uniformly Cauchy.  → ′ ∈ 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 ≠ x0. x − x0 x − x0 Proof of Lemma 2: Fix any x0 ∈ I. By the mean value theorem,

− − − − − fm(x) fm(x0) − fn(x) fn(x0) (fm fn)(x) (fm fn)(x0) − ′ ′ − ′ = = (fm fn) (ξ) = fm(ξ) fn(ξ), x − x0 x − x0 x − x0 − ∈ ′ 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 \{x0}.  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 ′ 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 ( )′ ′ lim fn = lim f . n→∞ n→∞ n

⇒ ′ ⇒ ∈ Proof of Theorem: Let fn f (Lemma 1) and fn g. Fix any x0 I. ′ We will show that f (x0) exists and agrees with g(x0). Let ε > 0 be given. Write

− − − − f(x) f(x0) − ≤ f(x) f(x0) − fn(x) fn(x0) fn(x) fn(x0) − ′ ′ − g(x0) + fn(x0) + fn(x0) g(x0) . x − x0 x − x0 x − x0 x − x0

− − | ′ − | ε f(x) f(x0) − fn(x) fn(x0) ε ̸ Select n : fn(x0) g(x0) < and < , x = x0 (Lemma 2). 3 x − x0 x − x0 3

− fn(x) fn(x0) − ′ ε | − | Choose δ > 0 so that fn(x0) < for 0 < x x0 < δ. x − x0 3 Then, for any x with 0 < |x − x | < δ, 0

f(x) − f(x0) ε ε ε − g(x0) < + + = ε. x − x0 3 3 3 It follows that f(x) − f(x ) lim 0 = g(x ), x ∈ 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, ′ | ′ | ≤ 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.