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 March 1st, 2017, 11:58 PM #1 Newbie   Joined: Mar 2017 From: tbilisi Posts: 1 Thanks: 0 metric space How to showed that $\displaystyle \rho(f,g)=\int_{a}^{b}|f(x)-g(x)|^p dx$ defined metric on $\displaystyle L^p[a,b]$?
 March 2nd, 2017, 07:44 AM #2 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,537 Thanks: 108 Step 1: Define $\displaystyle L^{p} [a,b]$ Thanks from topsquark
March 2nd, 2017, 11:58 AM   #3
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Quote:
 Originally Posted by zylo Step 1: Define $\displaystyle L^{p} [a,b]$
Zylo I see that you're starting to channel my own pickiness. Happy to see that In this case there is a great subtlety lurking.

We typically think of $\displaystyle L^p[a,b]$ as the set of functions $\displaystyle f : [a,b] \to \mathbb R$ such that $\displaystyle \int_{[a,b]} f(x) \ \mathrm dx < \infty$. I'm notating the integral that way to indicate that this is the Lebesgue and not the Riemann integral.

To show this induces a metric we need to prove (among other things) that $\displaystyle d(f, g) = 0$ if and only if $\displaystyle f = g$.

But now consider $\displaystyle f(x) = 1$ and $\displaystyle g(x) = 1$ if $x$ is irrational and $\displaystyle g(x) = 0$ if $\displaystyle x$ is rational.

Clearly $\displaystyle f \neq g$ since these two functions differ on the rationals, but $\displaystyle d(f,g) = 0$. What happened?

The answer is that when we put on our picky hats, $\displaystyle L^p[a,b]$ is the set of equivalence classes of functions where we say that $\displaystyle f$ and $\displaystyle g$ are equivalent if $\displaystyle f$ and $\displaystyle g$ differ on a set of measure zero.

With that refinement (which we typically never think about) we can show that $\displaystyle d$ is a metric.

This is how to get the proof started.

Last edited by Maschke; March 2nd, 2017 at 12:07 PM.

March 2nd, 2017, 03:38 PM   #4
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Quote:
 Originally Posted by giokakocha How to showed that $\displaystyle \rho(f,g)=\int_{a}^{b}|f(x)-g(x)|^p dx$ defined metric on $\displaystyle L^p[a,b]$?
Usually the pth root of the integral defines the metric.

 March 3rd, 2017, 09:27 AM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,537 Thanks: 108 Function Space: Function considered as infinite dimensional vector. What is size (Norm) of a vector? What is distance d between vectors? In Euclidean space, N=|a| and d=|b-a|. Generalization to function space, L$\displaystyle ^{p}$[a,b]: Definition, p-integrable: Lim $\displaystyle \Sigma |f(x_{i})|^{p}\Delta x_{i}$ exists and denoted by $\displaystyle \int |f(x)|^{p}dx$ $\displaystyle \int$ is Riemannian, or other if dx is d$\displaystyle \mu$, where $\displaystyle \mu$ is a defined measure. Norm: N = ||f|| = $\displaystyle \left [ \int |f(x)|^{p}dx \right ]^{ 1/p}$ Application: Express arbitrary functions in terms of basis functions. Fourier series for example. Which brings us to OP: Given: p-integrable (Riemannian) function space. Does $\displaystyle ||f_{1}-f_{2}||$ satisfy requirements of distance function? It does if it satisfies triangle inequality: $\displaystyle ||f_{1}-f_{3}||\leq ||f_{1}-f_{2}||+||f_{2}-f_{3}||$: https://press.princeton.edu/chapters/s9627.pdf Ttheorem 1.2 Thanks from topsquark

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