10. Sequences and series of functions: uniform convergence
Pointwise and uniform convergence
We have said a good deal about sequences of numbers. It is natural also to consider a
sequence of functions (f1 , f2 , . . .). A simple example is: fn (x) = xn for each n.
What might one mean by the limit of a sequence of functions? There are different
possible answers to this question. The simplest one is as follows. Let E be some real interval
(possibly the whole real line, or the positive half line).
Definition. We say that (fn ) converges to f pointwise on E if, for each x ∈ E, we have
fn (x) → f (x) as n → ∞.
For each fixed x, [fn (x)] is, of course, a sequence of numbers, so this is just saying
that each such sequence of numbers tends to a limit; it is not really a new idea at all. Note
that, for the statement to have any meaning, it is essential to specify the interval E.
Example 1. Let fn (x) = xn , and let E = [0, 1]. By 1.13, fn → f pointwise on E, where
f (x) =
In this example, each fn is continuous, but f is discontinuous at 1. So a pointwise
limit of continuous functions can fail to be continuous. We now introduce a second type of
convergence that will turn out to overcome this defect.
What would one mean by one function being a
good approximation to another (on a stated interval
E)? A natural answer is as follows. We say that G
ε-approximates f on E if |g(x) − f (x)| ≤ ε for all x ∈ E.
This means that g lies between the dotted lines in the
diagram.
Definition. We say that (fn ) converges to f (as n → ∞) uniformly on E if, given any
ε > 0, there exists n0 such that for all n ≥ n0 , fn ε-approximates f on E. That is,
|fn (x) − f (x)| ≤ ε for all x ∈ E.
10.1. If (fn ) converges to f uniformly on E, then it converges to f pointwise on E.
Proof. This follows at once from the definition with x fixed.
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Another trivial observation is: if fn → f uniformly on E, and E1 ⊂ E, then fn → f
uniformly on E1 .
How do we determine whether convergence is uniform in particular cases? In most
cases, this is done along the lines of one of the two following remarks.
(1) Suppose we can find numbers Mn , tending to 0 as n → ∞, such that |fn (x)−f (x)| ≤
Mn for all x ∈ E. Then fn → f uniformly on E; this statement just rewrites the definition
very slightly. It does not matter whether Mn is the exact greatest value of |fn (x) − f (x)| on
E.
(2) Suppose that there exists c > 0 such that, for each n, we can find a point xn ∈ E
with |fn (xn ) − f (xn )| ≥ c. Then the definition fails with ε = c/2, so (fn ) does not tend to
f uniformly on E.
Example 2. Again, let fn (x) = xn , with f as in Example 1.
This shows that the converse of 10.1 is false: pointwise convergence does not imply
uniform convergence.
Example 3. Let fn (x) =
x
.
n
However, (fn ) does tend to 0 uniformly on any bounded interval [−M, M ], since
fn (x)| ≤ M/n on this interval, and M/n → 0 as n → ∞.
nx
. Show that fn → 0 pointwise on [0, ∞). Does it
1 + n2 x2
tend to 0 uniformly (a) on [0, 1], (b) on [1, ∞)?
Example 4. Let fn (x) =
2
Example 5. Let fn (x) =
x+n
. Show that fn → 0 uniformly on R.
x2 + n2
Consequences of uniform convergence
10.2 PROPOSITION. Let E be a real interval. Suppose that (fn ) is a sequence of
functions, each continuous on E, and that fn → f uniformly on E. Then f is continuous on
E.
Proof. Choose x0 ∈ E (for the moment, not an end point) and ε > 0.
Of course, we already saw in Example 1 that pointwise convergence is not sufficient for
this conclusion.
The result for integration is very easy:
10.3 PROPOSITION. Suppose that (fn ) is a sequence of functions, each continuous
on the interval [a, b], and that fn → f uniformly on [a, b]. Then
Z
b
Z
f = lim
a
n→∞
3
b
fn .
a
Proof.
Note. Under the weaker assumption that each fn is Riemann-integrable on [a, b], one
can show that f is Riemann-integrable and the stated limit holds. (Exercise!)
The next example shows that pointwise convergence is again not sufficient.
Example 6. Let fn be as shown.
Actually, pointwise convergence is sufficient in 10.3 under the additional condition that
there exists M such that |fn (x)| ≤ M for all n and all x ∈ E. This is a major theorem, and
we will not prove it here.
Using both versions of the fundamental theorem of calculus (7.13 and 7.15), we can
derive a result for differentiation. The statement is not quite as straightforward as the
previous two results.
10.4 PROPOSITION. Suppose that (fn ) is a sequence of functions, each having a
continuous derivative on [a, b], and that:
fn → f pointwise on [a, b],
fn0 → g uniformly on [a, b].
Then f is differentiable on (a, b), and f 0 (x) = g(x) for a < x < b. (Similarly for one-sided
derivatives at the end points.)
Proof.
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A simple example shows that the hypothesis fn0 → g cannot be removed in 10.4, even
if fn → f uniformly.
Example 7. Let fn (x) =
on R. However,
fn0 (x)
1
n
sin nx. Then |fn (x)| ≤ 1/n for all x, so fn → 0 uniformly
= cos nx, which (for x 6= 0) does not tend to any limit, so (fn0 ) is not
even pointwise convergent.
Series of functions
Let f1 , f2 , . . . be functions, and let sn = f1 + · · · + fn . If (sn ) tends to a function f
P
uniformly on E, we say that the series ∞
n=1 fn converges to f uniformly on E. A pleasantly
simple sufficient condition is given by the next result.
10.5 PROPOSITION (the “M-test”). Suppose that, for each n, we have |fn (x)| ≤ Mn
P∞
P∞
for all x ∈ E, and that
n=1 Mn is convergent. Then the series
n=1 fn is uniformly
convergent on E.
Proof. By the comparison test, for each x ∈ E, the series
P∞
n=1
fn (x) is convergent:
let its sum be f (x).
P∞
Note that if
n=1
fn is uniformly convergent on E (with sum f ), and each fn is
continuous on E, then, by 10.2, so is f .
Example 8. If
P∞
n=1
|an | is convergent, then
P∞
n=1
an cos nx is uniformly convergent on
R, since
Example 9.
P∞
n=0
xn /n2 is uniformly convergent on [−1, 1], since
Example 10. Recall that the Riemann zeta function is defined for x > 1 by: ζ(x) =
P∞
n=1 1/n
1+δ
1/n
x
. This series is uniformly convergent on [1 + δ, ∞) for any δ > 0, since 1/nx ≤
on this interval. It follows that the zeta function is continuous for all x > 1.
Example 11. Recall that the power series x −
5
x3 x5
+
− · · · converges to tan−1 x for x
3
5
in [−1, 1]. In fact, we showed in 8.2 that if sn (x) is the sum of n terms, then |sn (x)−tan−1 x| ≤
1/(2n + 1) on this interval. So the series is uniformly convergent to tan−1 x on [−1, 1], but
not by the M-test!
10.6 PROPOSITION. Let
P∞
n=0
an xn be a power series. Suppose that
P∞
n=0
|an |rn is
convergent for some r > 0. Then the power series is uniformly convergent on [−r, r]. In
particular, this holds for any r < R, where R is the radius of convergence.
Proof.
Note that this result says that, for any ε > 0, the sum of the series can be εapproximated on [−r, r] by a polynomial, the sum of a suitable number of terms. The
simplest possible example shows that we can not say this for the open interval (−R, R):
Example 12. The geometric series
P∞
n=0
xn . For this series, R = 1, and the sum of the
series is 1/(1 − x), which is unbounded on (−1, 1), so certainly cannot be ε-approximated
by a polynomial there.
By 10.6 and 10.2, we have re-proved continuity of power series functions (3.11), seemingly with no effort! Without very much effort, we can also re-prove the harder theorem on
differentiation of power series (4.13):
P∞
an xn converges for |x| < R, with
P
n−1
.
sum f (x). Then f is differentiable on (−R, R), with f 0 (x) = ∞
n=1 nan x
10.7 THEOREM. Suppose that the power series
Proof. By Lemma 4.11, the series
P∞
n=1
n=0
nan xn−1 also converges for |x| < R: let its sum
be g(x). For each n, let
sn (x) = a0 + a1 x + · · · + an xn ,
so that
s0n (x) = a1 + 2a2 x + · · · + nan xn−1 .
Choose r < R. By 10.6, sn (x) → f (x) uniformly on [−r, r]. Also, since the series for
g(x) is also a power series, s0n (x) → g(x) uniformly on [−r, r]. By 10.4, it follows that f is
differentiable on (−r, r), with f 0 (x) = g(x) for |x| < r. But this holds for every r < R, so in
fact it holds for all x with |x| < R.
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A continuous function that is nowhere differentiable
We have presented a number of consequences of uniform convergence that might seem
fairly routine. We now describe one that most people find quite amazing! It is a “fractal”
construction, first thought of by Weierstrass in the 1880’s.
For each n ≥ 0, define
fn (x) = inf{|x − k/4n | : k ∈ Z}.
To clarify what this means, write dn = 1/4n . Then fn (x) = x on [0, 21 dn ] and fn (x) = dn − x
on [ 12 dn , dn ]. These values are then repeated on each interval [kdn , (k + 1)dn ].
The function fn is continuous,
since it comprises straight-line pieces,
and 0 ≤ fn (x) ≤
1
d
2 n
for all x. By
P∞
the M-test, the series
n=0 fn (x) is
uniformly convergent on R, defining a
continuous function f (x).
Let sn =
f0 + f1 + · · · + fn . The diagram shows
s0 , s 1 , s 2 .
We show that f is not differentiable at any point. Choose any x. For each r, let
ar = 4r [f (x + dr ) − f (x)],
br = 4r [f (x) − f (x − dr )].
Note that for all n ≥ r, we have fn (x ± dr ) = fn (x). Let Ar be the union of all intervals
[4kdr , (4k + 1)dr ] and [(4k + 2)dr , (4k + 3)dr ], and let Br = R \ Ar . A bit of thought shows
that if x ∈ Ar , then x and x + dr lie under a single line segment of the graph of fr−1 , and of
fn for 0 ≤ n ≤ r − 1 (the diagram helps!). Each such line segment has gradient ±1, so for
0 ≤ n ≤ r − 1, we have 4r [fn (x + dr ) − fn (x)] = ±1. Now ar is the sum of r such terms, so
it is an integer, even if r is even and odd if r is odd.
In the same way, if x ∈ Br , then br is even if r is even, odd if r is odd. Now define cr
to be ar if x ∈ Ar and br if x ∈ Br . Then cr is even if r is even and odd if r is odd, so does
not tend to a limit as r → ∞. But if f were differentiable at x, then (ar ) and (br ), hence
also (cr ), would converge to f 0 (x).
There is no hope of adequately drawing the graph of f ! It has a strict local minimum at
every point of the form k/4r . Any function you can “draw” is really piecewise differentiable.
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