HW 1
Uniform convergence. This section reviews some ideas about uniform conver- gence.
(1) For n∈Nletfn: (0,1)→Rbe the function given by the formula fn(x) = n
1 +nx (x∈(0,1)).
Show that {fn} is pointwise convergent but not uniformly convergent on (0,1).
Solution: We have fn(x) = 1
1
n +x (n∈N, x∈(0,1))
Since 1n ↓ 0 as n → ∞, it is clear that for each x ∈ (0,1), fn(x) ↑ x1 as n→ ∞. This shows that{fn} is pointwise convergent to the function f: (0,1)→Rgiven byf(x) =x−1.
Now, forn∈N, |f(x)−fn(x)|= x(1+nx)1 . Supposefn →f uniformly.
Fix >0. By uniform convergence of{fn}tof on (0,1) we can findN∈N such that |f(x)−fn(x)| < for allx ∈ (0, 1) and all n ≥ N. However, clearly limx→0+ 1
x(1+N x) = ∞, i.e., |f(x)−fN(x)| → ∞ as x ↓ 0. This contradicts the earlier conclusion that|f(x)−fN(x)|< for allx∈(0, 1).
Thus{fn}does not converge uniformly.
(2) Show that {fn}, where fn: [0,1]→Ris the map fn(x) =xn, is not uni- formly convergent.
Solution: Note that{fn}converges pointwise to the the functionf: [0,1]→ Rgiven by
f(x) =
(0 if 0≤x <1 1 ifx= 1.
Suppose{fn} converges uniformly. Then it must converge tof. By prob- lem (3), the uniform limit of continuous functions is continuous, and hence f is continuous. This is a contradiction, sincef clearly has a discontinuity
atx= 1.
(3) Come up with a definition of uniform convergence for a sequence of func- tions {fn} on a set A taking values in a normed linear space W over R.
Show that ifA= [0,1], and if{fn} is a sequence of continuous W-valued functions on [0,1] which converges uniformly to f: [0, 1]→W, then f is continuous.
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Solution: Definition:{fn} is said to converge uniformly tof:A →W if for every >0, there existsN ∈N, depending only onand not onx∈A, such thatkfn(x)−f(x)k< for alln≥N, and allx∈A.
Suppose A = [0,1] and {fn}, f are as in the statement of the prob- lem. Given a positive real number , there exists N ∈ N such that kfn(x)−f(x)k < 3 for all n ≥ N and all x ∈ [0,1]. In particular kfN(x)−f(x)k < 3 for all x ∈ [0,1]. Let a ∈ [0,1] be a point. We wish to show f is continuous at a. Since fN is continuous, there exists δ >0 such thatkfN(x)−fN(a)k< 3 whenever|x−a|< δ. Thus
kf(x)−f(a)k ≤ kf(x)−fN(x)k+kfN(x)−fN(a)k+kfN(a)−f(a)k
≤ 3+
3 + 3 =
whenever|x−a|< δ. This proves that f is continuous at a. Since awas
an arbitrary point of [0,1], we are done.
Normed spaces. LetV andW be normed linear spaces overR.
(4) On Rn show thatk k∞≤ k k2.
Solution: For (x, y)∈R2, we have|x| ≤p
x2+y2 and|y| ≤p
x2+y2. This means max{|x|,|y|} ≤ k(x, y)k2, i.e k(x, y)k∞≤ k(x, y)k2. (5) Let {vn}be a sequence inR2, sayvn= (xn, yn). GiveR2 thek k∞ norm.
Show that limn→∞vn=vif and only if limn→∞xn=xand limn→∞yn=y wherev= (x, y).
Solution: Suppose limn→∞vn = v in (R2,k k∞). This means given > 0 there exists N ∈ N such that kvn−vk∞ < for n ≥ N. Thus max{|xn−x|,|yn−y|}< forn≥N. Thus|xn−x|< and|yn−y|<
forn≥N, i.e., limn→∞xn =xand limn→∞yn=y.
Conversely, suppose limn→∞xn = xand limn→∞yn = y. Then given > 0 there exists N1, N2 ∈ N such that |xn−x| < for n ≥ N1 and
|yn−y| < for n ≥ N2. Let N = maxN1, N2. Then |xn−x| < and
|yn−y| < for n ≥ N. It follows that max{|xn−x|,|yn−y|} < for n≥N, i.e.kvn−vk∞< forn≥N. This means limn→∞vn=v.
(6) Use what you know from analysis on Rto come up with a definition of a Cauchy sequence inV. When would you say V is complete? Is Rn with k k∞ complete?
Solution: One defines a sequence{vn}inV to be Cauchy if for every >0 there existsN ∈N such that kvn−vmk< whenevern, m≥N. We say V is complete if every Cauchy sequence inV is convergent inV.
Suppose{vk}k∈N is a Cauchy sequence in (Rn,k k∞). Say vk= (xk1, . . . , xkn)
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for each k ∈ N. Then for each > 0 there exists N ∈ N such that kvk−vlk∞= max1≤j≤n|xkj−xlj|< for allk, l≥N. Thus|xkj−xlj|<
for everyj ∈ {1, . . . , n} and every k, l≥N. It follows that{xkj}k∈N is Cauchy for everyj∈ {1, . . . , n}. Since these are Cauchy sequences of real numbers, and Ris complete (proved last semester), these are convergent sequences. Let xj = limk→∞xkj and v = (x1, . . . , xn). For each > 0 we have Nj ∈N such that|xkj−xj|< for k≥Nj, j ∈ {1, . . . , n}. Let N = max1≤j≤nNj. Then |xkj−xj|< for k≥ N, j ∈ {1, . . . , n}. This means max1≤j≤n|xkj−xj|< for k≥N, i.e.kvk−vk∞< for k≥N. In other words limk→∞vk =v in (Rn,k k∞).
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