Show the following is a metric

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In summary, the conversation discusses proving that ##d(f,g) = \int_{0}^{1}\left | f(x) - g(x) \right | dx## is a distance function, where ##f : [0,1] \rightarrow R## and ##f## is continuous. The conversation also touches on the second property of this function, where it needs to be shown that ##d(f,g) = 0## implies ##f = g##. The key fact needed for this proof is that if ##F(x)## is a continuous function on [0,1] and ##F(a) > 0## for some ##a## in [0,1], then ##\int_{0}^{
  • #1
DotKite
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Homework Statement



Show that ##d(f,g) = \int_{0}^{1}\left | f(x) - g(x) \right | dx## is a distance function. Where ##f : [0,1] \rightarrow R## and ##f## is continuous.

Homework Equations





The Attempt at a Solution


I am stuck on the second property where you have to show d(f,g) = 0 iff f = g. THe left direction is trivial. However d(f,g) = 0 implying f=g is giving me trouble. I have tried contrapositive, but it doesn't seem to be getting me anywhere.
 
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  • #2
DotKite said:

Homework Statement



Show that ##d(f,g) = \int_{0}^{1}\left | f(x) - g(x) \right | dx## is a distance function. Where ##f : [0,1] \rightarrow R## and ##f## is continuous.

Homework Equations


The Attempt at a Solution


I am stuck on the second property where you have to show d(f,g) = 0 iff f = g. THe left direction is trivial. However d(f,g) = 0 implying f=g is giving me trouble. I have tried contrapositive, but it doesn't seem to be getting me anywhere.

The fact you need is that if F(x)>=0 on [0,1], F(a)>0 for some ##a## in [0,1] and F(x) is continuous (very important) then ##\int_{0}^{1}\left | F(x) \right | dx \gt 0##. Can you figure out how to prove that?
 
Last edited:
  • #3
Dick said:
The fact you need is that if F(x)>=0 on [0,1], F(a)>0 for some ##a## in [0,1] and F(x) is continuous (very important) then ##\int_{0}^{1}\left | F(x) \right | dx \ge 0##. Can you figure out how to prove that?

wouldn't I need to show that ##\int_{0}^{1}\left | F(x) \right | dx > 0##
to get the contrapositive?
 
  • #4
Oh I see! ok. NVM. Gonna try to prove it
 
  • #5
DotKite said:
wouldn't I need to show that ##\int_{0}^{1}\left | F(x) \right | dx > 0##
to get the contrapositive?

Yes, of course. Typo. Sorry. I corrected it.
 
Last edited:

1. What is a metric?

A metric is a measurement or standard used to evaluate or assess something. In science, metrics are often used to quantify and compare different variables or phenomena.

2. How do you show that something is a metric?

To show that something is a metric, you must first ensure that it meets the three properties of a metric: it is non-negative, it is symmetric, and it satisfies the triangle inequality. Then, you can use these properties to demonstrate that the given variable or function is indeed a metric.

3. What are the three properties of a metric?

The three properties of a metric are:

  • Non-negativity: The metric must always be equal to or greater than 0.
  • Symmetry: The metric should yield the same result regardless of the order of the inputs.
  • Triangle inequality: The metric between two points should always be less than or equal to the sum of the individual metrics between those points and a third point.

4. Why is it important to establish that something is a metric?

Establishing that something is a metric is important because it allows us to make meaningful comparisons and draw conclusions about different variables or phenomena. By using a common measurement or standard, we can accurately evaluate and communicate our findings.

5. Can a metric be subjective?

No, a metric should not be subjective. It should be a quantifiable and objective measure that can be applied consistently. If a metric is too subjective, it may lead to biased or unreliable results.

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