Understanding dot product of continuous functions

Question

I'm reading about Fourier analysis and in my book the author speaks about dot product for continuous functions $f, g\in L^2(a,b)$(the set of functions which are square-integrable on the interval $[a,b]$), which is defined as:

$$\langle f, g\rangle = \int_a^b f(x)\overline{g(x)}\;dx.$$

The author mentions that the reader should think the functions as vectors with the components of the vectors corresponding to the function values. This notation confuses me. If I had discrete $n$-dimensional vectors $\textbf{x}$ and $\textbf{y}$ the dot product would be:

$$\langle \textbf{x}, \textbf{y}\rangle = \sum_{i=1}^n \textbf{x}(i)\overline{\textbf{y}(i)},$$

where $\textbf{x}(i)$ corresponds to the $i$th component of vector $\textbf{x}$. To me the integral

$$\langle f, g\rangle = \int_a^b f(x)\overline{g(x)}\;dx$$

seems like area under the curve, not a dot product. I guess the $dx$ is confusing me...could someone make this formula a bit more intuitive for me? Please let me know if my question is unclear.

Thank you! =)

Answer 1

The formulas are actually exactly the same. An integral is just a sum of infinitesimally short areas over the interval. The essential part is the $f(x)\overline{g(x)}$ which ensures that this is in fact a product with all the properties of the dot product.

You can even write a Riemann sum for the integral:

$$\int_a^b f(x)\overline{g(x)}dx=\lim_{N\to \infty}\sum_{i=0}^N f(a+(b-a)i/N)\overline{g(a+(b-a)i/N)}\frac{b-a}{N}$$

Now both are the same.

The "area under the curve" interpretation is confusing you here. Yes, you can interpret it like that, but no less than the sum can be represented as a total number of apples... the function that you are integrating is of the form that ensures that the functional is a positive semidefinite square form (linear in both terms), which is what dot product is all about.

Answer 2

For introductory purposes, i would switch to

$$\langle f, g\rangle = \frac{1}{2 \pi} \; \;\int_0^{2 \pi} f(x)\overline{g(x)}\;dx.$$

Now we can take an " orthonormal basis" made up of functions $$ \{ 1, e^{inx}, e^{-inx} \} $$ Note that the integral is over the real variable we are calling $x.$ We are integrating some function of $x,$ possibly constant, which has real and imaginary parts; these can just be integrated separately and the results combined later with an $i$ in the correct place.

It is worth your effort to write out the integrals showing the orthonormal part, $f = e^{inx} $ and $g = e^{\pm imx},$ once for $m = n $ and once for $m \neq n,$ so that makes four integrals because of the $\pm$ signs. It really will help.

Answer 3

A useful heuristic analogy to keep in mind: $$ \text{integration is for continuous $X$ as summation is for discrete $X$.} $$

The feature of the inner product for elements of a vector space with discrete basis that carries over to the continuous world is that sum samples each vector in each of its components. The analogous quality for the function space inner product is to integrate over an interval: all values of each function (on the interval) are relevant.