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I'm supposed to show that If X is the set of all functions on the interval $[a,b]$ and $\displaystyle d(f,g)= \int^{b}_{a}|f(x)-g(x)|dx\,$, then $(X, d)$ is a metric space.

But I don't think it is. The problem I'm having is that in order for $(X,d)$ to be a metric space, it must be true that $d(f,g) = 0$ iff $f=g$. But d could also be $0$ if, for instance,

$f(x) - g(x)$ is odd and is symmetrical across $x=(a+b)/2$

Am I right about this or am I missing something?

Sorry, about the format by the way. I don't know how to type the actual symbols on here!

math_man
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Zachary F
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  • this is a metric space! integrand is positive so integral will be zero only if integrand is zero – Mathronaut Sep 26 '14 at 19:20
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    Just a small note: not every function on $[a,b]$ is (Lebesgue) integrable. So $(X,d)$ as defined cannot be a metric space, since the "metric" cannot be applied to every element in $X$. The answer given below certainly clarifies how to prove that the set of integrable functions on $[a,b]$ is a metric space, with the above metric. – Stromael Sep 26 '14 at 19:36
  • @Stromael's note is not small. The problem is unclear. If it is "the set of all functions" then $d$ does not make sense. If restricting to Lebesgue integrable functions, then it is still not a metric but not for the reason indicated. You can have $f$ such that $\int|f| = 0$ but $f$ is not everywhere $0$. If restricting to continuous functions, or instead working with equivalence classes of integrable functions, it can define a metric. – Jonas Meyer Sep 26 '14 at 19:43
  • There is a sense in which this is trivial. Obviously you must consider equivalence classes of functions (because of the measure 0 effect) or restrict the functions in some way to ensure you get at most one member of each equiv class. But then the result is almost trivial. – almagest Sep 26 '14 at 19:47
  • Of course one must take equivalence classes. Please forgive an analyst's hidden assumptions. :-) – Stromael Sep 29 '14 at 10:48
  • oops! yeah, it was supposed to say "The set of all integrable functions..." – Zachary F Sep 30 '14 at 03:49

2 Answers2

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No: you are taking the integral of the absolute value of the difference $$\displaystyle d(f,g)= \int_a^b \left|f(x)−g(x)\right|dx$$ while your counterexample would apply if you were taking the absolute value of the integral of the difference $$\displaystyle \left|\int_a^b (f(x)−g(x))\,dx\right|.$$

You may need to watch out for the possibility that $f(x)$ and $g(x)$ differ on a set of measure zero, but apart from that you will have $\left|f(x)−g(x)\right|$ as non-negative by definition, and positive if they differ, making the integral positive if they differ on a large enough subset of $[a,b]$.

Henry
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I think you are confusing that the limits of integration is arbitrary. The $a$ and $b$ are to be any number, you don't determine them, they are given to you.

ken
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  • That isn't relevant. The same properties will apply regardless of the interval. – Jonas Meyer Sep 26 '14 at 19:40
  • That's what I mean. It is regardless of the interval. – ken Sep 26 '14 at 19:41
  • I do not see how the point of possible confusion mentioned here is relevant to the question. There is no assumption that $a$ or $b$ have any particular value in the question. "you are confusing"--who is confusing this, where? – Jonas Meyer Sep 26 '14 at 19:47
  • @JonasMeyer, he said it might not work if $x = (a+b)/2$, but he doesn't choose the limits of integration. – ken Sep 26 '14 at 19:49
  • OP refers to the midpoint of the segment, which exists regardless of what the segment is. OP does not make any assumptions about what $a$ or $b$ is. The point of confusion illustrated there was that even if $f$ is odd about that midpoint, it does not mean that $|f|$ is. – Jonas Meyer Sep 26 '14 at 19:52
  • Oh okay, I misunderstood. – ken Sep 26 '14 at 19:53