# Metric Spaces

• Nov 26th 2007, 08:52 PM
hannahs
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!
• Nov 26th 2007, 09:04 PM
Jhevon
Quote:

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 $\int_a^b |f(x) - g(x)|~dx = 0 \implies f(x) = g(x)$

and that $f(x) = g(x) \implies \int_a^b |f(x) - g(x)|~dx = 0$

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

Quote:

2. d(x,y) = d(y,x)
Reflexive property
Again, by the definition of the absolute value function, it is obvious that $|f(x) - g(x)| = |g(x) - f(x)|$ hence $\int_a^b |f(x) - g(x)|~dx = \int_a^b |g(x) - f(x)|~dx$

Quote:

3. d(x,y) is less than or equal to d(x,z) + d(z,y)
Triangle Inequality
Let $h(x): [a,b] \to \mathbb{R}$ be continuous, then $h \in X$ and we have that

$\int_a^b |f(x) - g(x)|~dx = \int_a^b |f(x) - h(x) + h(x) - g(x)|~dx$ $\le \int_a^b|f(x) - h(x)|~dx + \int_a^b |h(x) - g(x)|~dx$ by the Triangle Inequality

thus we have shown d is a metric
• Nov 26th 2007, 09:11 PM
hannahs
Thank you so much! :)