When you have a function which is a limit of a sequence of functions, when can you say the
derivative of the limit function is the limit of the derivatives of the functions in the sequence?
The following theorem seems to be one of the best results available. It is based on the
mean value theorem. First of all, recall Definition 6.9.6 on Page 298 listed here for
convenience.
Definition 7.11.1Let
{fn}
be a sequence of functions defined on D. Then
{fn}
issaid to converge uniformly to f if it convergespointwise to f and for every ε > 0 there existsN such that for all n ≥ N
|f (x)− fn(x)| < ε
for all x ∈ D.
To save on notation, denote by
||k|| ≡ sup{|k(ξ)| : ξ ∈ D }.
Then
||k + l|| ≤ ||k||+||l|| (7.18)
(7.18)
because for each ξ ∈ D,
|k(ξ)+ l(ξ)| ≤ ||k||+ ||l||
and taking sup yields 7.18. From the definition of uniform convergence, you see that f_{n}
converges uniformly to f is the same as saying
lim ||fn − f|| = 0.
n→∞
Now here is the theorem. Note how the mean value theorem is one of the principal parts of
the argument.
Theorem 7.11.2Let
(a,b)
be a finiteopen interval and let f_{k} :
(a,b)
→ ℝ bedifferentiable and suppose there exists x_{0}∈
(a,b)
such that
{fk (x0)} converges,
{fk′} converges uniformly to a function g on (a,b).
Then there exists a function f defined on
(a,b)
such that
fk → f uniformly,
and
f′ = g.
Proof: Let c ∈
(a,b)
and define
{ fn(x)−fn(c)if x ⁄= c
gn (x,c) ≡ f′(xc−)c if x = c .
n
Also let
||h|| ≡ sup {|h (x)| : x ∈ (a,b)}.
Thus h_{k}→ h uniformly means
||hk − h||
→ 0.
Claim 1: For each c, x → g_{n}
(x,c)
converges uniformly to a continuous function h_{c}, on
(a,b)
and h_{c}
(c)
= g
(c)
.
Proof: First note that each x → g_{n}
(x,c)
is continuous. Next consider the claim about
uniform convergence. Let x≠c. Then by the mean value theorem applied to the function
x → f_{n}