Math 3100: Sequences and Series
Fall 2014, Dr. Klipper
In-Depth Examples for Week 12’s Start
Pointwise Convergence Vs. Uniform Convergence
Near the beginning of Section 3.1 in the textbook, we get two very important definitions
for sequences of functions. Suppose A ⊆ R, and we have a function f and a sequence of
functions {fn } all defined on A. We say that fn → f pointwise on A when
∀x ∈ A, ∀ > 0, ∃N ∈ N, ∀n > N, |fn (x) − f (x)| < whereas fn → f uniformly on A when
∀ > 0, ∃N ∈ N, ∀n > N, ∀x ∈ A, |fn (x) − f (x)| < Uniform convergence is the stronger definition: it implies pointwise convergence. The
only difference in the definitions is that with pointwise convergence, the value N may depend
on the point x used in the limit, whereas with uniform convergence, N may not depend on
x. A rough way to say this is that with pointwise convergence, you only need fn (x) → f (x)
for every x independently, and different x values may produce limits that converge at very
different rates. However, with uniform convergence, the error measurement |fn (x) − f (x)|
should be able to be bounded by something independent of x.
In this handout, we’ll go through a couple parts of Exercise 3.1.4 from the textbook in
detail, to provide a model for the kinds of arguments you might use in deciding if fn → f
uniformly or non-uniformly. There are a few things you should keep in mind:
• When fn does not converge uniformly to f , it’s generally because no matter which
n you use, there is some point xn which makes |fn (xn ) − f (xn )| particularly large
compared to the distances |fn (x) − f (x)| for most other values of x. I like to think
of xn as a “straggler” that is taking a longer time than usual to converge to its limit.
By finding suitable choices of xn , you may be able to make |fn (xn ) − f (xn )| have a
nonzero limit as a function of n.
• When trying to see whether fn → f uniformly, try and find a maximum (or at least,
an upper bound) for the distance |fn (x) − f (x)| over all x values in A. (If fn or f is
piecewise, this might be a little subtle.) If you can find a bound |fn (x) − f (x)| ≤ Mn ,
where Mn doesn’t depend on x and Mn → 0, then fn → f uniformly.
– Why? For any > 0, pick N large enough so that n > N ⇒ |Mn − 0| < , and
then |fn (x) − f (x)| ≤ Mn < for all x ∈ A.
• The domain A is very important! The N in uniform convergence can depend on A, so if
fn doesn’t converge to f uniformly on a set A, sometimes you can find a slightly smaller
subset B ⊂ A and get uniform convergence on B. For instance, maybe |fn (x) − f (x)|
has a convenient maximum value on B which it doesn’t have on A. We’ll see this occur
sometimes in class and in this document.
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First Example: nx/(1 + n2 x2 ) on [0, ∞)
This is from part (b) of Exercise 3.1.4: say A = [0, ∞), and for each x ≥ 0,
nx
fn (x) =
1 + n 2 x2
The exercise asks what the pointwise limit is (i.e. what f (x) is) and then asks whether the
convergence is uniform on A. For each part of this exercise, we’ll present a little brainstorming and then give a proper solution (written with a similar level of detail you’d want in your
homework).
The Pointwise Limit:
Brainstorming: When we study pointwise limits, think of leaving x fixed (i.e. temporarily
pretending it is constant) and focusing just on what happens when n → ∞. Here, fn (x) has
n1 in the numerator and n2 in the denominator. Since n2 dominates n1 , this suggests that
the limit is 0.
However, we should be careful here: if x is 0, then n2 x2 = 0, and hence there are no
n2 ’s in the denominator of fn (0)! We treat x = 0 as a special case, but then we see that
fn (0) = 0/(1 + 0) = 0 anyway.
Solution: For every x ∈ A, we claim lim fn (x) = 0 as n → ∞, i.e. f (x) = 0 for all x ∈ A.
First, note that when x = 0, fn (0) = 0, so lim 0 = 0. For x 6= 0, the informal argument for
this is that the leading terms of the fraction fn (x) form the ratio nx/(n2 x2 ) = 1/(nx), so
fn (x) ∼ 1/(nx) → 0 as n → ∞.
To be more formal, factor n out of the numerator and n2 out of the denominator:
fn (x) =
n
(x/n)
1
(x/n)
·
= ·
2
2
2
n (1/n ) + x
n (1/n2 ) + x2
When x 6= 0, the limit is 0 · 0/(0 + x2 ) = 0. (Note: When x = 0, the expression 0/(0 + x2 ) becomes a 0/0 indeterminate form. That’s
why x = 0 needs to be a special case.)
Whether fn → f uniformly:
Brainstorming: The standard procedure is to figure out how big the distance |fn (x)−f (x)|
can get. We want to maximize this difference and hence bound it with a formula which does
not use x. When the convergence is uniform, that bound will go to 0 as a function of n
alone. Since f (x) = 0 here, and each fn (x) is positive, |fn (x) − f (x)| is just fn (x).
To maximize fn (x), one approach is to use calculus! When you compute fn0 (x), you get
(n − n3 x2 )/(1 + n2 x2 )2 . The denominator is never zero, so the only critical values occur when
n3 x2 = n, i.e. x2 = 1/n2 and x = ±1/n (but only x = 1/n is in the domain [0, ∞)). You can
check that the absolute maximum occurs at xn = 1/n, and fn (1/n) = n(1/n)/(1 + n2 /n2 ) =
1/2. This shows the biggest distance |fn (x) − f (x)| is 1/2, which does not go to 0 as n → ∞!
The convergence is not uniform. (In the description on the first page, xn = 1/n is the
“straggler” which is the furthest from the limit. Graph a few of these fn ’s with a calculator
to see where the absolute maxes occur!)
Solution: We prove fn does not converge uniformly to f on [0, ∞). For any n ∈ N, we
set xn = 1/n and note that fn (1/n) = n(1/n)/(1 + n2 (1/n)2 ) = 1/(1 + 1) = 1/2. Thus,
|fn (xn ) − f (xn )| = 1/2 − 0 = 1/2.
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If the convergence were uniform, then for any > 0, |fn (x) − f (x)| < for all x simultaneously when n is large enough. However, if we use = 1/2, then no matter how big n is,
the choice of xn = 1/n produces |fn (xn ) − f (x)| = 1/2 − 0, which is not less than = 1/2. (Note that I didn’t actually need to use calculus in the final proof. I didn’t need to prove
that xn = 1/n really was the absolute max; the fact that fn (xn ) was constantly 1/2 was all
I needed to finish the argument.)
Going further with this:
There are a couple ways this example could be altered to make the convergence uniform.
One way is to make the denominator rise at a faster rate, which changes the bound we get
on |fn (x) − f (x)| to something which does go to 0 as n → ∞!
Variant: For instance, let’s consider
gn (x) =
nx
for all x ∈ [0, ∞)
1 + n 3 x2
The only difference between fn (x) and gn (x) is the power of n in the denominator. A very
similar proof to the one given above shows that gn (x) → 0 for all x ∈ [0, ∞) as well.
However, if you try to maximize |gn (x)| = gn (x), you get gn0 (x) = (n − n4 x2 )/(1 + n3 x2 )2 .
The critical value occurs when x2 = 1/n3 , so x = 1/n3/2 . If we call xn = 1/n3/2 , you can check
that gn0 is positive to the left of xn and negative to the right of xn , so the absolute maximum
3/2
3/2
3
−3/2 2
occurs
) ) =√n−1/2 /(1 + 1) =
√ at x = xn . Its value is gn (1/n ) = n(1/n )/(1 + n (n
1/(2 n). Unlike the value 1/2 we got from the exercise, the bound 1/(2 n) does go to 0
when n is large!
Solution: We prove gn → 0 uniformly on [0, ∞). Let > 0 be given. We will find N so
that for all n > N and all x ≥ 0, |gn (x) − g(x)| < .
First, we note that since gn (x) ≥ 0 for each x ≥ 0, |gn (x) − g(x)| = gn (x) − 0. We
maximize gn (x); we find that
gn0 (x) =
n · (1 + n3 x2 ) − nx · (2n3 x2 )
n + n4 x2 − 2n4 x2
n − n4 x2
=
=
(1 + n3 x2 )2
(1 + n3 x2 )2
(1 + n3 x2 )2
The denominator is never zero since 1 + n3 x2 ≥ 1 for all x. The numerator is zero when
n4 x2 = n, so x2 = 1/n3 and x = ±n−3/2 (though only the positive root is in domain). We
let xn = n−3/2 ; check that gn0 (x) > 0 for x < xn and gn0 (x) < 0 for x > xn , so the absolute
maximum of gn (x) occurs at x = xn .
As a result, for all x ≥ 0,
gn (x) ≤ gn (xn ) =
n−1/2
n−1/2
n(n−3/2 )
=
=
1 + n3 (n−3/2 )2
1+1
2
Since n−1/2 /2 → 0 as n → ∞, we may choose N ∈ N so that for all n > N , |n−1/2 /2| < .
Therefore, we have |gn (x) − g(x)| = gn (x) ≤ gn (xn ) = n−1/2 /2 < , as desired. 3
Second Example: xn /(1 + xn ) on [0, ∞)
This is from part (c) of Exercise 3.1.4: say A = [0, ∞), and for each x ≥ 0,
fn (x) =
xn
1 + xn
The exercise asks what the pointwise limit is (i.e. what f (x) is) and then asks whether the
convergence is uniform on A.
The Pointwise Limit:
Brainstorming: This really comes down to knowing what lim(xn ) is. This is a geometric
sequence limit, so the value of the limit depends on whether |x| < 1. Since A only has
nonnegative numbers, this means our cases depend on whether x < 1.
When x < 1, that limit is 0. When x = 1, 1n is just 1. Lastly, when x > 1, lim(xn ) = ∞,
so we get an ∞/∞ indeterminate form in xn /(1 + xn ). However, since the numerator and
denominator both feature the same dominant term xn , the limit should be 1.
Solution: The limit of xn depends on whether |x| < 1, |x| = 1, or |x| > 1. Our limit f (x)
will be found in each case.
• When x < 1, xn → 0 as n → ∞, so xn /(1 + xn ) → 0/(1 + 0) = 0.
• When x = 1, xn = 1, so fn (x) = 1/(1 + 1) = 1/2 and hence has limit 1/2.
• When x > 1, xn → ∞. We factor xn to obtain
xn
1
1
xn
=
=
→
=1
n
n
n
n
1+x
x (1 + 1/x )
1 + 1/x
1+0
since 1/xn → 0 as n → ∞.
Thus, the pointwise limit is


0
f (x) = 1/2


1
if x < 1
if x = 1
if x > 1
Whether fn → f uniformly:
Brainstorming: Every fn curve is continuous, but f is not! The values of f (x) jump from
0 to 1/2 and then to 1 for x values near 1. If fn were to converge uniformly to f , we’d need
similar behavior as well in each fn : the values of fn (x) would be very close to 0 when x < 1,
then they’d jump to 1/2 when x = 1, then they’d be close to 1 for x > 1. However, each fn
is continuous; the values of fn (x) smoothly go from 0 to 1/2 as x goes from 0 to 1.
To illustrate what’s going on, here’s a figure showing the graphs of f1 through f4 . (The
graph of f4 is the one with the steepest slope passing through the point (1, 1).) Two thick
solid lines show pieces of f (x) for x < 1 and for x > 1. The dashed horizontal lines are away from y = f (x), and on each fn , there is a solid dot in (0, 1) whose y-value lies outside
of the dashes. Notice that as n gets bigger, these points move further to the right, getting
closer to x = 1.
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We will use those points to show that the convergence is not uniform. For these points,
although fn (x) should approach a limit of 0 as n → ∞, those points are closer to 1/2; they
are not within of the limit 0. If we call xn the x-value of the point on the nth graph, we
need |fn (xn ) − f (xn )| to be bigger than , so we get fn (xn ) > 1/2. (Here, xn is the “straggler”
mentioned in the first page’s comments.) We can either come up with xn explicitly, or we
can use the fact that fn (xn ) ≈ 1/2 when xn ≈ 1 and rely on continuity to establish xn ’s
existence. We’ll do the latter.
Solution: We will prove fn does not converge uniformly to f on [0, ∞). Let = 1/4; for
each n, we will find some xn ≥ 0 such that |fn (xn ) − f (xn )| ≥ for all n. (This means that
no matter how large n is, it’s impossible to ensure |fn (x) − f (x)| < for every x.)
Because fn is continuous and fn (1) = 1n /(1 + 1n ) = 1/2, we know that |fn (x) − 1/2| < whenever |x − 1| < δ for some choice of δ > 0. Pick any value xn between 1 − δ and 1, so
xn < 1 and fn (xn ) > 1/2 − = 1/4. Therefore, |fn (xn ) − f (xn )| = fn (xn ) − 0 > 1/4 = , as
desired. Going further with this:
Our convergence is not uniform on [0, ∞), and the problem comes from taking xn points
close to 1 whose y-values are out of the range f (x) ± . (In other words, those points fall
out of the “-stripe” drawn in the picture above.) However, if we adjust the domain A to
keep away from x = 1, then we can get uniform convergence! Let’s prove that for any b with
0 < b < 1, fn does converge uniformly to f on [0, b]!
Brainstorming: The problem with the original exercise was that no matter how high n
went, we could always find an xn point which made fn (xn ) close to 1/2. Thus, we could
never force fn (xn ) to go below 1/4. In effect, our sup of the range on [0, 1] was always 1/2,
which does not converge to 0 as n → ∞.
However, now that we only allow x ≥ b, the situation changes! Each fn graph is strictly
increasing (show that by computing fn0 (x), for instance), so the sup of the range occurs at the
rightmost end of x = b, i.e. fn (x) ≤ fn (b). However, since b < 1, this sup fn (b) does converge
to 0 as n → ∞! Put another way: on [0, 1], we only have the bound |fn (x) − 0| ≤ 1/2, but
on [0, b], we have |fn (x) − 0| ≤ fn (b) = bn /(1 + bn ). This latter bound can be made small by
taking n large enough.
Solution: Let b ∈ (0, 1). We prove that fn → f uniformly on [0, b]. Let > 0; we find
some N so that for all n > N and for all x ∈ [0, b], |fn (x) − f (x)| < , which is the same as
saying fn (x) − 0 ≤ since x < 1.
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First, note that for any n and any x ∈ [0, b], we have fn (x) ≤ fn (b) since fn is strictly
increasing (proof left to reader). We’ve seen that as n → ∞, fn (b) → f (b) = 0, so there is
some N such that for all n > N , |fn (b) − 0| < . Therefore, when n > N , fn (x) ≤ fn (b) < ,
as desired. Note: Try and imitate this argument to also see that the convergence is uniform on
[a, ∞) for any a > 1! There, you’ll instead show that |fn (x) − f (x)| < by showing
|fn (x) − f (x)| = 1 − fn (x) ≤ 1 − fn (a) and use the fact that fn (a) → 1 as n → ∞.
We say that the convergence is uniform “away from 1”; as long as our interval maintains
a fixed distance away from x = 1, we get uniform convergence!
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