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Here is the definition of a norm given by my textbook;

enter image description here

(This is from Fourier Series and Boundary Value Problems by James Ward Brown and Ruel V. Churchill, Chapter 7)

I'm confused by what authors say after (9). I was under the impression that the area between two curves on an interval [a,b] is just the difference of the integral on [a,b] of those functions. That is,

Area between $f(x)$ and $g(x)$ on $[a,b] = \int_a^b f(x) - g(x) dx =\int_a^b f(x) dx - \int_a^b g(x) dx$

How does the norm represent the same thing? Or am I simply missing that somehow,

$\left(\int_a^b [f(x) - g(x)]^2dx \right)^{\frac{1}{2}} = \int_a^b f(x) - g(x) dx$

for all functions in a function space?

Sorry if this is simple, thanks in advance!

user3840170
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ant
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    They don't. I think this is only a vocabulary problem, it is in fact a way to measure the difference of area but in a nonlinear way, you don't want to count every distance the same way, this one is more circular and the other one in straight lines. It is only written to give some intuition. If you have done some probability theory it may help, or even convex optimisation? – julio_es_sui_glace Mar 06 '24 at 19:43
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    Please do not use images to convey information not otherwise present in your post, and please do not use mathjax to format text. – Arturo Magidin Mar 06 '24 at 19:52
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    Yeah, the sentence after $(9)$ is just wrong. That sentence describes a norm, the $L^1$ norm, if you take a strict sense of positive area, but it is different from the norm defined here – Thomas Andrews Mar 06 '24 at 19:52
  • Please don’t write “Confusion on” titles next time. – user3840170 Mar 07 '24 at 17:11

2 Answers2

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Firstly, the area between the graph of two functions $f,g: [a,b]\to\mathbb{R}$ is given by $$\displaystyle\int_{a}^{b}|f(x)-g(x) |dx,$$

because area cannot be negative and we don't know, in general, which one of the functions attains bigger values.

The author says $$|| f-g||=\left(\displaystyle\int_{a}^{b}|f(x)-g(x) |^2dx \right)^{1/2}$$

is a measure of the area, not exactly what that area is equal. They continue to explicitly mention the reasoning behind this measure: it is the mean square deviation between $f$ and $g$.

Julio Puerta
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    Nice! I was hoping that it was more confusion about the writing than something else. I think I just disagree with the use of the word measure here, but it might be I just don't have a concrete enough understanding of the word. This cleared up my confusion, thank you! – ant Mar 06 '24 at 19:56
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    @ant The author says it is a measure of the area, you are more used to a different measure. In every day parlance there is only one reasonable measure for area (up to units) but in maths there can be multiple different measures. Similar to how you can measure for example the wealth of a country in more than one way, so you could say GDP is a measure of wealth but there are other measures for that as well. – quarague Mar 07 '24 at 10:00
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You are right that the integral of the difference is the area between the graphs. The book is using the integral of the square of the difference. That integral divided by the length $b-a$ tells you the average square distance between the graphs.

If you work out the finite analogue for a pair of sequences of length $n$ you see the well known mean squared error.

Ethan Bolker
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