# Linear Mapping in Subsets

Question 1.

1) Suppose (V, | * |) is a normed space. If x, y E V and r is a positive real number, show that the open r-balls Br(x) and Br(x + y) in V are homeomorphic.

2) Suppose V and W are two normed spaces. If A : V ---> W is a linear map, then show that it is continuous at every point v E V if and only if it is continuous at 0 E V.

3) Suppose A: (V, | * |) ---> (W, | * |w) is a linear map between normed spaces, and there is a number R E R such that |A(v)|w <= R|v|_v for all v E V. Explain why A is continuous.

Question 2

Let (0,1) denote the open unit interval in R, and C(0,1) the set of all continuous functions (0,1) ---> R. Is C(0,1) a subset of B((0,1),R) the set of all bounded functions on (0,1)? Is C(0,1) a normed space with the sup-norm | * | given by |f| = sup_(t E (0,1){|f(t)|}?

Question 3

1) For a compact topological space, and Y a compact subset of X. The inclusion i: Y --->X gives a function i* : B(X,R) ---> B(Y,R), from the bounded functions on X to the bounded functions on Y by i*(f) = f in i for each f E B(X,R). Explain why i* is subjective.

2) Using the sup norm | * | on both these sets of bounded functions, for a function f E B(X,R), what, is any, is the relation between |f| and |i* (f)|. Is i* continuous?

3) For a compact topological space X, denoted by C(X) the banach space of continuous functions on X with the usual sup-norm.Following the idea of part (1), explain why a continuous function alpha: X --->Z between two compact topological spaces gives a function alpha*:C(Z) ---> C(X)

4) Explain why alpha* of part (3) is a linear map

5) Explain why alpha* of part (3) is continuous.

6) If I: X ---> X is the identity function, show that I*, as in part (3), is the identity function C(X) ---> C(X).

7) If alpha: X_1 ---> X_2 and beta: X_2 ---> X_3 are continuous functions of compact topological spaces, explain why (beta is in alpha)* = alpha* is in beta*

8) Hence prove that if gama: X --> Z is a homeomorphism of compact topological spaces, gama*: C(Z) ---> C(X) is a homeomorphism.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see attached Document for detailed solution.

Problem #1

1. Proof:

We consider the map, such that for any, we have. Since, then we have

Thus . The map is well-defined.

Then I show that is continuous. We consider any and , we can find some , such that . Then for any , we have , then we have

Therefore, is continuous.

Similarly, we can define with , then we can also prove that is continuous.

Therefore, and are homeomorphism.

2. Proof:

From 1, we know that and are homeomorphism, then we have a continuous function and is also continuous. Since is a linear map, if is continuous at , then is also continuous; if is continuous at 0, then is also continuous.

3. Proof:

For any and , we can find some , such that for all and , we have

Therefore, is continuous.

Problem #2

First, I claim that is NOT a subset of . Here is a counter example. is continuous on and thus . But is not bounded and thus .

Second, I claim that is a normed spaced under sup norm. It is easy to see that for any . To show is a norm, we also need to verify the following three features.

(1) For any ,

(2) For any , we have . Thus

(3) If , then for all and thus

Therefore, is a normed space.

Problem #3

1. Proof:

We have the inclusive map . We need to show that is surjective. We consider any bounded function , then for any , is bounded. We note that , then we have is bounded. Now we define if , then is extended to a bounded function on or , and thus . Therefore, is surjective.

2. Proof:

I show that . For any , and thus

Then I show that is continuous. We consider any . For any , we can find , such that for all with , we have

Therefore, is continuous.

3. We consider a continuous function , where and are two compact topological spaces. We can define , such that for any , we have .

4. Proof:

I show that is a linear map. For any and , we have

Therefore, is a linear map.

5. Similar to the proof of 2, is continuous.

6. Proof:

Compared with 3, we change to the identity function , then . Then . Thus is also an identity function on .

7. Proof:

8. Proof:

If is a homeomorphism of compact topological spaces, according to the results of 4 and 5, is a linear and continuous. Along with the result of 6, is also a homeomorphism.

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