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In reading I came across the claim that the following is a metric.

For the space $X$ of all integrable functions on the interval [$0,1$] , for $f, g \in X$, the following equation defines a metric:

$\rho(f, g) = \int_0^1 |(f(x) - g(x))| dx$

It is clear to me that $\rho(f, g) = \rho(g, f)$ and also $\rho(f, g) = 0$ if and only $ f = g$. However, does the triangle inequality hold?

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Triangle inequality follow from triangle inequality for real numbers. In which sense do you mean integrable? Riemann? Lebesgue? (in the last case we can have $\rho(f,g)=0$ even if $f=g$). – Davide Giraudo Apr 19 '12 at 16:29

"$ρ(f,g)=0$ if and only $f=g$"

This isn't true if we're only working over the space of integrable functions: consider $f(x) \equiv 0$ and $g(x) = 0$ except at x = 0, when it equals 1 (say). $g - f$ is non-zero, but $\rho(f,g) = 0$.

For this to be a metric we usually work over the space of continuous functions.

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Isn't this exactly how we construct a Hilber space? – akkkk Apr 19 '12 at 16:43
The space C[0,1] equipped with the above norm isn't a Hilbert space: it fails on both counts. It isn't complete, consider $f_m(x) = 0 $ for $0 \le x \le \frac{1}{2} - m $ and $f_m(x) = 1 $ for $ \frac{1}{2} + m \le x \le 1$ and f being a straight line segment joining 0 and 1 between these two intervals. It's a Cauchy sequence, but it would necessarily converge to a function with a discontinuity at x= 1/2. So we could consider its completion, but this space cannot be an inner product space since it does not obey the – Conor Apr 19 '12 at 16:54

As Conor said, the space you should consider is that of continuous functions, not integrable functions. For the triangle inequality see:

Your case is $p=1$.

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Minkowski inequality is an overkill in this case, as we can simply apply the triangle inequality for absolute value on reals. – Damian Sobota Apr 19 '12 at 17:01

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