- Let X be a finite dimensional normed linear space, real or complex. Show that X
is separable. Hint: Let
_{i=1}^{n}be a basis and define a map from F^{n}to X,θ, as follows. θ≡∑_{k=1}^{n}x_{k}v_{k}. Show θ is continuous and has a continuous inverse. Now let D be a countable dense set in F^{n}and consider θ. - Let α ∈ (0,1]. We define, for X a compact subset of ℝ
^{p},where

and

Show that

is a complete normed linear space. This is called a Holder space. What would this space consist of if α > 1? - Let {f
_{n}}_{n=1}^{∞}⊆ C^{α}where X is a compact subset of ℝ^{p}and supposefor all n. Show there exists a subsequence, n

_{k}, such that f_{nk}converges in C. We say the given sequence is precompact when this happens. (This also shows the embedding of C^{α}into Cis a compact embedding.) Hint: You might want to use the Ascoli Arzela theorem. - Let f :ℝ × ℝ
^{n}→ ℝ^{n}be continuous and bounded and let x_{0}∈ ℝ^{n}. Ifand h > 0, let

For t ∈

, letShow using the Ascoli Arzela theorem that there exists a sequence h → 0 such that

in C

. Next argueand conclude the following theorem. If f :ℝ × ℝ

^{n}→ ℝ^{n}is continuous and bounded, and if x_{0}∈ ℝ^{n}is given, there exists a solution to the following initial value problem. - Let H and K be disjoint closed sets in a metric space, , and let
where

Show g

∈for all x ∈ X, g is continuous, and g equalson H while g equalson K. - Suppose M is a closed set in X where X is the metric space of problem 5 and suppose
f : M →is continuous. Show there exists g : X →such that g is continuous and g = f on M. Hint: Show there exists
and

≤for all x ∈ H. To do this, consider the disjoint closed setsand use Urysohn’s lemma or something to obtain a continuous function g

_{1}defined on X such that g_{1}= −1∕3,g_{1}= 1 ∕3 and g_{1}has values in. When this has been done, letplay the role of f and let g

_{2}be like g_{1}. Obtainand consider

- ↑ Let M be a closed set in a metric space and suppose f ∈ C. Show there exists g ∈ Csuch that g= ffor all x ∈ M and if f⊆, then g⊆. This is a version of the Tietze extension theorem.
- Suppose f : ℝ → ℝ and f ≥ 0 on with f= f= 0 and f< 0 for all x. Can you use a modification of the proof of the Weierstrass approximation theorem for functions on an interval presented earlier to show that for all ε > 0 there exists a polynomial p, such that< ε for x ∈and p≤ 0 for all x? Hint:Let f
_{ε}= f−. Thus there exists δ such that 1 > δ > 0 and f_{ε}< 0 onand. Now consider ϕ_{k}= a_{k}^{k}and try something similar to the proof given for the Weierstrass approximation theorem above. - Suppose f ∈ C
_{0}and also≤ Ce^{−rt}. Let A denote the algebra of linear combinations of functions of the form e^{−st}for s sufficiently large. Thus A is dense in C_{0}. Show that iffor each s sufficiently large, then f

= 0 . Next consider only≤ Ce^{rt}for some r. That is f has exponential growth. Show the same conclusion holds for f iffor all s sufficiently large. This justifies the Laplace transform procedure of differential equations where if the Laplace transforms of two functions are equal, then the two functions are considered to be equal. More can be said about this. Hint: For the last part, consider g

≡ e^{−2rt}fand apply the first part to g. If g= 0 then so is f. - A set S along with an order ≤ is said to be a well ordered set if every nonempty subset of S has a smallest element. Here ≤ is an order in the usual way. If x,y ∈ S, then either x ≤ y or y ≤ x and it satisfies the transitive law: If x ≤ y and y ≤ z, then x ≤ z. Using the Hausdorff maximal theorem, show that every nonempty set can be well ordered. That is, there is an order for which the given set is well ordered. In particular ℚ the rational numbers can be well ordered. However, show that there is no way that the well order can coincide with the usual order on any open interval.
- Verify that Pthe set of all subsets of the natural numbers is uncountable. Thus there exist uncountable sets. Pick an uncountable set Ω. Then you can consider this set to be well ordered, with the order denoted as ≤. Consider
If

= ∅, let Ω_{0}=. If≠∅, let ω_{0}be the first element ofand in this case letThat is ω≠ω

_{0}and ω ≤ ω_{0}. Explain why Ω_{0}is uncountable. Explain why every element of Ω_{0}has a “next” element. Now define a topology in the usual way. In particular, show that sets of the form [α,b),where α is the first element of Ω_{0}is a basis for a topology for Ω_{0}. Verify that a sub-basis is sets of the form [α,b),. Explain why this is a Hausdorff space. Explain why every element of Ω_{0}is preceeded by countably many elements of Ω_{0}and show every increasing sequence converges. Show that this cannot be a separable space. Suppose you have a cover of (a,b] consisting of “sub-basic” open sets. Without loss of generality, all of these have nonempty intersection with. Let p be the first such thatis in the open cover, assuming there are such sets. Thus you could simply useinstead of all the others. If p ≤ a, you are done. If not, then p ∈ (a,b] and so some set of the other kind, [α,q) must contain p and so at most two sets from the open cover contain. Consider the other cases to verify that ( a,b] is compact. Now explain why (a,b) is either equal to (a,b] or. In the second case, verify thatwould be of the form ( a,] which was just shown to be compact thanks to Alexander sub-basis theorem. Is Ω_{0}locally compact? - In the above example, show that Ω
_{0}is not compact. However, show that every sequence has a subsequence which converges. Recall that in any metric space, compactness and sequential compactness are equivalent. Hence one can conclude that there is no metric which will deliver the same topology for Ω_{0}described above. That is to say, this horrible topological space is not metrizable. However, the Riesz representation theorem presented above would hold for this terrible thing.

Download PDFView PDF