Parseval's theorem/Identity - Definition seems wrong

Question

My book shows with some steps that $$\int_{-L}^L {f(x)}^2dx=\int_{-L}^L\left\{\frac{1}{2}a_0+\sum_{n=1}^\infty\left[a_n\cos{\left(\frac{n\pi x}{L}\right)}+b_n\sin{\left(\frac{n\pi x}{L}\right)}\right]\right\}^2=L\left[\frac{1}{2}a_0^2+\sum_{n=1}^\infty a_n^2+b_n^2\right]$$

and hence it says

This can be restated as $$\lVert f \rVert^2=\sum\text{squares of componentsof $f$ on basis vectors}$$ the norm of $f$ is often called the power contained in the function.

However it doesn't make any sense to me (not the passages). What is the norm of the function? And what does this identity or theorem actually say?

Looking here and here I could only find different formulations. However in all these sources they seem to forget that $\sum\text{squares of componentsof $f$ on basis vectors}$ is multiplied by $L$.

Answer 1

It's easier to understand what's going on here if we talk in more generality, so as to avoid getting bamboozled by all the trig functions (plus, it's useful to know about these ideas for other bases).

The norm (or at least, the one it means) is defined as $$ \lVert f \rVert_2 = \sqrt{\int_{-L}^L f(x)^2 \, dx}, $$ and there is an associated inner product $\langle f,g \rangle = \int_{-L}^L f(x) g(x) \, dx$, so $\langle f,f \rangle = \lVert f \rVert_2^2 $.

Suppose we want to have an expansion of $f$ in terms of an orthogonal basis of functions $ (e_n)_{n=0}^{\infty}$ (orthogonal means $\langle e_m,e_n \rangle=0$ if $n \neq m$), $$f(x) = \sum_{n=0}^{\infty} A_n e_n(x)$$ (this is what a Fourier series is, if we choose $e_n$ to include $\frac{1}{2}$, $\cos{(k\pi x/L)}$ and $\sin{(k\pi x/L)}$). We find $A_n$ using the orthogonality: $$ \langle f, e_n \rangle = \sum_{m=0}^{\infty} A_m\langle e_m,e_n \rangle = A_n \langle e_n , e_n \rangle = A_n \lVert e_n \rVert_2^2 $$ Thus $$f = \sum_{n=0}^{\infty} \frac{\langle f, e_n \rangle}{\lVert e_n \rVert_2^2} e_n. $$

Parseval's identity gives the norm of $f$ in terms of the coefficients $A_n$: \begin{align} \lVert f \rVert_2^2 &= \langle f , f \rangle \\ &= \left\langle \sum_{n=0}^{\infty} A_n e_n , \sum_{m=0}^{\infty} A_m e_m \right\rangle \\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} A_n A_m \langle e_n , e_m \rangle \\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} A_n^2 \lVert e_n \rVert^2. \end{align} To get back to your problem, the important thing is that this includes the norm of the basis functions $e_n$. Therefore if the basis functions are not normalised to have $\lVert e_n \rVert^2=1$, you'll get an extra factor. Going back to Fourier series explicitly, the basis functions are not normalised in the usual formulation: $$ \int_{-L}^L \cos^2{\left( \frac{n\pi x}{L} \right)} \, dx = \int_{-L}^L \sin^2{\left( \frac{n\pi x}{L} \right)} \, dx = L, \\ \int_{-L}^L \frac{1}{2} \, dx = L. $$ Hence the extra factor of $L$.

Answer 2

Fourier transforms are often a bit tough to digest. There are many subjects and concepts coming together.

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You first have an inner product space. Not just vectors containing a bunch of numbers like vectors in $\mathbb R^n$ having some matrix defining lengths, but functions are the vectors.

The meaning of a function being orthogonal to another is just as in linear algebra : inner product between two vectors is 0. Although the inner product in our case is an integral.

So one $f$ (which is linear combination of sines and cosines) times itself and then integrated.

$$\begin{array}{|c|c|}\hline \text {Subject}&\text{Linear Algebra}& \text{Fourier Analysis} \\\hline \text{vector } f& {\bf f} = [f_1,f_2,\cdots,f_n]^T&x\to f(x)\\\hline \text{Inner Product } \langle f,g \rangle&{\bf f}^T{\bf Gg}&\int f(x)g(x)dx\\\hline\text{Squared norm: }\|f\|^2&{\bf f}^T{\bf Gf}&\int f(x)^2dx\\\hline\end{array}$$

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Parseval's identity basically say that after change from trivial to Fourier basis, the "lengths of the vectors" or the "energy of the functions" are the same as before.