How is it possible for the L² norm of f − g to measure the area between the graphs of f and g?

Question

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

(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!

Answer 1

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 $g$.

Answer 2

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.