Thread: Metric Spaces

Problem:

Let X = {the set of all continuous functions f: [a,b] to R}. Let d(f,g) = the integral from a to b of |f(x) - g(x)|dx. Show that d is a metric on X, and therefore, X,d is a metric space.

I also have a hint: Recall that if h(x) is greater than or equal to 0 on [a,b] then the integral from a to b of h(x)dx is greater than or equal to 0; also if the integral from a to b of h(x)dx = 0 for h(x) greater than or equal to 0 then h(x) = 0 for all x in [a,b]

Here are the three conditions for a metric space:

1. d(x,y) = 0 if and only if x=y
Meaning that the distance from function to itself is 0

2. d(x,y) = d(y,x)
Reflexive property

3. d(x,y) is less than or equal to d(x,z) + d(z,y)
Triangle Inequality

I have no clue where to even start on this problem so any help would be greatly appreciated!

Originally Posted by hannahs

Problem:

Let X = {the set of all continuous functions f: [a,b] to R}. Let d(f,g) = the integral from a to b of |f(x) - g(x)|dx. Show that d is a metric on X, and therefore, X,d is a metric space.

I also have a hint: Recall that if h(x) is greater than or equal to 0 on [a,b] then the integral from a to b of h(x)dx is greater than or equal to 0; also if the integral from a to b of h(x)dx = 0 for h(x) greater than or equal to 0 then h(x) = 0 for all x in [a,b]

Here are the three conditions for a metric space:

1. d(x,y) = 0 if and only if x=y
Meaning that the distance from function to itself is 0

we need to show that

and that

these are trivial, since |a| = 0 iff a = 0. here we can take a = f(x) - g(x)

2. d(x,y) = d(y,x)
Reflexive property

Again, by the definition of the absolute value function, it is obvious that hence

3. d(x,y) is less than or equal to d(x,z) + d(z,y)
Triangle Inequality

Let be continuous, then and we have that

by the Triangle Inequality

thus we have shown d is a metric

Thank you so much!