I am really desperate for someone to answer all 4 parts of this question (d, e, f, and g) (I have already posted this question twice and no one has answered all 4 parts sufficiently). This is due in one day and I just really have no idea what to do for these 4 parts. This is for a classical analysis course. Please be as detailed as possible in your answer. Thank you so so much in advance.

This is the question (and its 4 parts) that I am needing help with is below:

Below is some information from the textbook chapter that could help you solve the 4 parts of this problem:

4. When X is noncompact, we have defined our metric p on the space C(X,Y) of bounded continuous function and not on the space C(X,Y) of all continuous func- tions. As mentioned in the text, the reason is that for unbounded, continuous functions, p(f,g) = sup{dy (f(x), g(x))|x X} may be oo, and a metric can not take infinite values. Restricting ourselves to C6(X,Y) is one way of overcoming this problem. Another method is to change the metric on Y such that it never occurs. We shall now take a look at this alternative method. If (Y, d) is a metric space, we define the truncated metric d by: d(x,y) if d(x, y) 1 d) Show that the truncated metric d is complete if and only if the original metric is complete. e) Show that a set K CY is compact in (Y,d) if and only if it is compact in (Y,d). f) Show that for a metric space (X, dx), a function f: X + Y is continuous with respect to d if and only if it is continuous with respect to d. Show the same for functions g: Y + X g) For functions f, ge C(X,Y), define (f,g) = sup{d(f(x), g(x)| E X} Show that is a metric on C(X,Y). Show that is complete if d is. 4.6. Spaces of bounded, continuous functions The spaces of bounded functions that we worked with in the previous section are too large for many purposes. It may sound strange that a space can be too large, but the problem is that if a space is large, it contains very little information - just knowing that a function is bounded, gives us very little to work with. Knowing that a function is continuous contains a lot more information, and for that reason we shall now turn to spaces of continuous functions. It's a bit like geography; knowing that a person is in France contains much more information than knowing she's in Europe. As before, we assume that (x,dx) and (Y, dy) are metric spaces. We define C6(X,Y)= {f: X+Y|f is continuous and bounded} to be the collection of all bounded and continuous functions from X to Y. As C6(X,Y) is a subset of B(X,Y), the metric p(f,g) = sup{dy (f(x), g(x))|x E X} that we introduced on B(X,Y) is also a metric on C(X,Y). We make a crucial observation: Proposition 4.6.1. C(X,Y) is a closed subset of B(X,Y). Proof. By Proposition 3.3.7 it suffices to show that if {fn} is a sequence in C6(X,Y) that converges to an element f e B(X,Y), then f e C6(X,Y). Since by Proposition 4.5.2 {fr} converges uniformly to f, Proposition 4.2.4 tells us that f is continuous and hence in Co(X,Y). The next result is a more useful version of Theorem 4.5.3 Theorem 4.6.2. Let (X,dy) and (Y,dy) be metric spaces and assume that (Y, dy) is complete. Then (C6(X,Y),p) is also complete. Proof. Recall from Proposition 3.4.4 that a closed subspace of a complete space it itself complete. Since B(X,Y) is complete by Theorem 4.5.3 and C(X,Y) is a closed subset of B(X,Y) by the proposition above, it follows that C6(X,Y) is complete. The reason why we so far have restricted ourselves to the space C6(X,Y) of bounded, continuous functions and not worked with the space of all continuous functions, is that the supremum p(f,g) = sup{dy(f(x), g(x))| X E X} can be infinite when f and g are just assumed to be continuous. As a metric is not allowed to take infinite values, this creates problems for the theory, and the simplest solution is to restrict ourselves to bounded, continuous functions. Sometimes this is a small nuisance, and it is useful to know that the problem doesn't occur when X is compact: Proposition 4.6.3. Let (X, dx) and (Y, dy) be metric spaces, and assume that X is compact. Then all continuous functions from X to Y are bounded. Proof. Assume that f: X + Y is continuous, and pick a point a E X. It suffices to prove that the function h(x) = dy(f(x), f(a)) is bounded, and this will follow from the Extreme Value Theorem 3.5.10 if we can show that it is continuous. By the Inverse Triangle Inequality 3.1.4 |h(x) h(y)= \dy (f(x), a) dy(f(y), a) 1 d) Show that the truncated metric d is complete if and only if the original metric is complete. e) Show that a set K CY is compact in (Y,d) if and only if it is compact in (Y,d). f) Show that for a metric space (X, dx), a function f: X + Y is continuous with respect to d if and only if it is continuous with respect to d. Show the same for functions g: Y + X g) For functions f, ge C(X,Y), define (f,g) = sup{d(f(x), g(x)| E X} Show that is a metric on C(X,Y). Show that is complete if d is. 4.6. Spaces of bounded, continuous functions The spaces of bounded functions that we worked with in the previous section are too large for many purposes. It may sound strange that a space can be too large, but the problem is that if a space is large, it contains very little information - just knowing that a function is bounded, gives us very little to work with. Knowing that a function is continuous contains a lot more information, and for that reason we shall now turn to spaces of continuous functions. It's a bit like geography; knowing that a person is in France contains much more information than knowing she's in Europe. As before, we assume that (x,dx) and (Y, dy) are metric spaces. We define C6(X,Y)= {f: X+Y|f is continuous and bounded} to be the collection of all bounded and continuous functions from X to Y. As C6(X,Y) is a subset of B(X,Y), the metric p(f,g) = sup{dy (f(x), g(x))|x E X} that we introduced on B(X,Y) is also a metric on C(X,Y). We make a crucial observation: Proposition 4.6.1. C(X,Y) is a closed subset of B(X,Y). Proof. By Proposition 3.3.7 it suffices to show that if {fn} is a sequence in C6(X,Y) that converges to an element f e B(X,Y), then f e C6(X,Y). Since by Proposition 4.5.2 {fr} converges uniformly to f, Proposition 4.2.4 tells us that f is continuous and hence in Co(X,Y). The next result is a more useful version of Theorem 4.5.3 Theorem 4.6.2. Let (X,dy) and (Y,dy) be metric spaces and assume that (Y, dy) is complete. Then (C6(X,Y),p) is also complete. Proof. Recall from Proposition 3.4.4 that a closed subspace of a complete space it itself complete. Since B(X,Y) is complete by Theorem 4.5.3 and C(X,Y) is a closed subset of B(X,Y) by the proposition above, it follows that C6(X,Y) is complete. The reason why we so far have restricted ourselves to the space C6(X,Y) of bounded, continuous functions and not worked with the space of all continuous functions, is that the supremum p(f,g) = sup{dy(f(x), g(x))| X E X} can be infinite when f and g are just assumed to be continuous. As a metric is not allowed to take infinite values, this creates problems for the theory, and the simplest solution is to restrict ourselves to bounded, continuous functions. Sometimes this is a small nuisance, and it is useful to know that the problem doesn't occur when X is compact: Proposition 4.6.3. Let (X, dx) and (Y, dy) be metric spaces, and assume that X is compact. Then all continuous functions from X to Y are bounded. Proof. Assume that f: X + Y is continuous, and pick a point a E X. It suffices to prove that the function h(x) = dy(f(x), f(a)) is bounded, and this will follow from the Extreme Value Theorem 3.5.10 if we can show that it is continuous. By the Inverse Triangle Inequality 3.1.4 |h(x) h(y)= \dy (f(x), a) dy(f(y), a)