# Show the following is a metric

## 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.

## 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 doesnt seem to be getting me anywhere.

Dick
Science Advisor
Homework Helper

## 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.

## 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 doesnt 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?

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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?

Oh I see! ok. NVM. Gonna try to prove it

Dick
Science Advisor
Homework Helper
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.

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