Homework 1 Solutions - Math 321,Spring 2015
1a Pointwise convergence: for a fix x ∈ [0, ∞) If x = 0, fn (x) = 0 for all n so it converges
to 0. If x 6= 0, we have
nx
= lim
n→∞ 1 + n2 x2
n→∞
lim
x
n
1
n2
+
x2
=
0
=0
0 + x2
The convergence is not uniform on [0, ∞): If fn converges uniformly to f then f must
be identically zero (by above and uniqueness of limit). i.e.
nx → 0, n → ∞
sup 2 2
x∈[0,∞) 1 + n x
nx
That is for any > 0 there is N such that for all n ≥ N, | 1+n
2 x2 | ≤ for every
x > 0 (a single works for all x). However, we have fn (1/n) = 1/2 for all n so the
sequence cannot converges uniformly on this domain. (when speaking about uniform
convergence, don’t forget to state the domain where the sequence converges uniformly
on).
Observe that any interval containing 0 or has 0 as a limit point will contain n1 for all
n sufficiently large and fn (1/n) = 1/2 so the sequence cannot converge to 0 uniformly
on this domain. Now consider intervals of the form [a, ∞), a > 0 then for all x in this
interval
nx nx 1 1
1 + n2 x2 ≤ n2 x2 = nx ≤ an
where a/n → 0, n → ∞ independent of x. In the same way, the sequence also
converges uniformly on (−∞, a], a < 0 and hence also on any intervals that are
subsets of them. In summary, the sequence converges uniformly on any interval not
containing 0 as a limit point.
1b Compact intervals are of the form [a, b] (a, b are finite). First we have
ˆ x/n
ˆ x/n
n
nt
|x − n log(1 + x/n)| = n−
dt = dt
t+1
t+1
0
0
nt ≤ |x/n| sup t∈[0,x/n] t + 1
x2
≤ |x/n| sup |nt| =
n
t∈[0,x/n]
Now for all x ∈ [a, b].
|ex − (1 + x/n)n | = |ex − en log(1+x/n) | = ex |1 − en log(1+x/n)−x |
≤ eb |e0 − en log(1+x/n)−x |
1
2
Now if N is sufficiently large then for all n ≥ N , |n log(1 + x/n) − x| ≤ x2 /n <
(a2 + b2 )/n ≤ δ for any δ > 0. Since f (x) = ex is continuous at x = 0 , for any
> 0 there is a δ > 0 such that |c| < δ implies |e0 − ec | < . Hence for x ∈ [a, b], N
sufficiently large,
|ex − (1 + x/n)n | ≤ eb as required (eb is just a constant, you may choose /eb instead of if you like.)
There is another way to do using Arithmetic-Geometric mean inequality: For a
fixed x, consider large enough n such that 1+ nx is positive (when consider convergence
of sequence it is enough to consider only sufficiently large n). Applying A.M.-G.M.
to 1 + x/n, ..., 1 + x/n (n copies) and 1 :
1+
1 + (1 + x/n) + ... + (1 + x/n)
x
=
≥ ((1 + x/n)n )1/(n+1)
n+1
n+1
This implies (1 + x/(n + 1))n+1 ≥ (1 + x/n)n is increasing in n for a fixed x. Then
we can apply Dini’s theorem (Theorem 7.13 in the textbook).
Note: You may want argue that |(1 + x/n)n − ex | attains a maximum at some
x = c ∈ [a, b] and let n → ∞ however you change n, c might change.
2 We show the contraspositive: If fn → f uniformly on X and xn → x, then fn (xn ) →
f (x). Since fn → f uniformly, we have that f is continous. Now
ρ(fn (xn ), f (x)) ≤ ρ(fn (xn ), f (xn )) + ρ(f (xn ), f (x))
Let > 0. Since fn → f uniformly on X, there is N1 such that if n ≥ N1 then
ρ(fn (xn ), f (xn )) < (when we change n we change the point but the convergence is
uniform so this is fine). Since f is continuous, there is N2 such that if n > N2 then
ρ(f (xn ), f (x)) < . Take N > N1 , N2 then for all n > N, ρ(fn (xn ), f (x)) ≤ 2.
Note: You may want to argue ρ(fn (xn ), f (x)) ≤ ρ(fn (xn ), fn (x)) + ρ(fn (x), f (x))
and argue that that since fn is continuous then ρ(fn (xn ), fn (x)) is small for large
n. But this does not seem to work since when you change n your function fn also
changes.
Also an 6→ a, n → ∞ does not imply that you can find > 0 such that aN > for all
N or for all sufficiently large N but it means that you can find an infinite subsequence
ni such that ani > for all ni
3 We have that fn0 continuous and fn0 → g uniformly on [a, b] and there is a point
x0 , limn→∞ fn (x0 ) exists (actually enough to assume that there is only one such point
x0 which we will do here.) We want to show fn → f uniformly for some differentiable
function f (we don’t have f yet we have to show that it does exist or find it).
Call limn→∞ fn (x0 ):=a. Let
ˆ x
f (x) :=
g(t)dt + a
x0
3
(Question: How can one comes up with this formula from the assumptions given
and the Fundamental Theorem of Calculus- FTC) Show fn → f uniformly: For any
x ∈ [a, b],
ˆ x
ˆ x
0
g(t)dt − a
fn (t)dt + fn (x0 ) −
|fn (x) − f (x)| = x0
ˆ xx0
≤
|fn0 (t) − g(t)|dt + |a − fn (x0 )|
x0
≤ sup |fn0 (x) − g(x)|(b − a) + |a − fn (x0 )|
x∈[a.b]
fn0
And we are done as
→ g uniformly and fn (x0 ) → a. Also since g is continuous, f
is differenceble on [a, b] with f 0 = g (FTC).
Note There is a stronger version of this question in the textbook (Theorem 7.17) which
fn0 are no longer assumed to be continuous and hence cannot apply Fundamental
Theorem of Calculus).
4a ^
¨
4b Suppose fn → f uniformly on X, gn → g uniformly on X and |fn (x)| ≤ M1 , |gn (x)| ≤
M2 all x and for all n . Choose N1 such that n ≥ N1 implies |fn (x) − f (x)| < for
all x ∈ X and N2 such that n ≥ N2 implies |gn (x) − g(x)| < for all x ∈ X. If
N > max{N1 , N2 } then
|fn (x) − gn (x) − f (x)g(x)| = |fn (x) − gn (x) − fn (x)g(x) + fn (x)g(x) − f (x)g(x)|
≤ |fn (x) − gn (x) − fn (x)g(x)| + |fn (x)g(x) − f (x)g(x)|
≤ M1 + M2 Since M1 , M2 are fixed constant M1 + M2 is arbritarily small and this holds for all
X hence fn gn → f g uniformly.
4c Choose e.g. fn (x) = x, gn (x) = 1/n on R