November 21, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework10
Let L : S → T be a linear map from vector spaces S to T. Let V

Put V = 2V

so that V is a solution to LX = 2Y _{1} - 7Y _{2}.

Let V be as above and set V = V + 6W - 2V

Hence V is another solution to LX = Y _{2}.

Let f(x) ∈ C([a,b]) and put

Because e^{u} ≥ 1 + u for all u, then in particular we have e^{f-f} ≥ 1 + f -f. Thus we have the following sequence

Since uniform convergence implies pointwise convergence, the proofs below omit any mention of pointwise convergence if the given sequence is proven to be uniformly convergent.

Let ε > 0 be given. Choose integer N so that Nε > 1. Since 0 ≤|sinx|≤ 1 for all x, then

for all n ≥ N. Hence f_{n} → 0 uniformly.

This sequence converges pointwise to

since for x = 0 f_{n}(x) = 1 for all n. And furthermore, for any ε > 0, at a fixed x_{0} ∈ (0,1] we can choose integer N such that
N > so that

for all n ≥ N

This sequence of functions, however, does not converge uniformly. Because each f_{n} is continuous on [0,1],
converging uniformly would imply that f above is continuous, but it’s not, due to the simple discontinuity at
x = 0.

For each n,

which is zero only at x = ±. Since f_{n}(x) is positive on (0,∞) and negative on (-∞,0), we have that f_{n}
has a minimum at - and a maximum at . Because the minimum and maximum for each n are
f_{n}(-) = -1∕2 and f_{n}() = 1∕2, respectively, we have

for all n ∈ ℕ. Hence, given any ε > 0, choosing integer N so that Nε > 1 gives us |f_{n}(x)| < ε for all n ≥ N and all x ∈ ℝ.
Thus f_{n} → 0 uniformly.

Let {f

Yes. For a fixed x

Let ε > 0 be given. Because all {f

and

for all x ∈ [0,1] and n ≥ N. Hence we have

This problem is identical to a problem on the previous homework.

Denote this map by f. This map is not a contraction map because there is no α < 1 such that

| (7.1) |

for all x,y ∈ [0,∞). This is because the difference between d(x,y) and d(f(x),f(y)) is

which can be made arbitrarily close to zero by making x and y large enough. This is problematic because the contraction condition in 7.1 implies that

must be true; which is not the case when the difference on the righthand side above can be made arbitrarily close to zero.

Denote this mapping by f. The failure here is that the metric space (0,1] is not complete. In particular, the method of successive approximations fails for this mapping since the sequence {x

Let

Define {f

Since f(x) is a summation that converges for all x, then f(x) < ∞, and thus f(x) is continuous since it is the composition of continuous functions. Furthermore, because

then f′(x) = by the Fundamental Theorem of Calculus.

For any complex number z = x + iy and integer n we have n

Thus for any sequence bounded by M and all z ∈ ℂ in the set {z = x + iy | x ≥ c} where c > 1, we have

so that ∑
_{n=1}^{∞} converges.

For x

Therefore since (1 - 1∕n)^{n} → 1∕e as n →∞ then f_{n} can be made arbitrarily large and so cannot converge
uniformly.

Problem ten has already proven this.