Parseval Equality says, Fourier transformation, as a linear map from one function space (function on x), to another function space (function on p), preserves 'norm'. Norm is just a fancy way of saying 'length of a vector'.

What do we mean by the length of a function?

Continuous Fourier transformation (OK, I switched to Boas convention) $f(x) = \int_\R F(p) e^{ipx} dp.$ $F(p) = (1/2\pi) \int_\R f(x) e^{-ipx} dx.$

Discrete Fourier transformation

Fix a positive integer $N$. $x,p$ are valued in the 'discretized circle' $\Z / N\Z \cong \{0,1,\cdots, N-1\}.$

$f(x) = \sum_{p \in \Z / N\Z} F(p) e^{2\pi i \cdot px/N}.$ $F(p) = (1/N) \sum_{x \in \Z / N\Z} f(x) e^{-2\pi i \cdot px/N}.$

Let $f(x)$ be a complex valued function on $x \in \R$, we define $\| f\|_x^2 := (1/2\pi) \int_\R |f(x)|^2 dx$

Let $F(p)$ be a complex valued function on $p \in \R$, we define $\| F\|_p^2 := \int_\R |F(p)|^2 dp$

$\| f\|_x^2 := (1/N) \sum_{x=0}^{N-1} |f(x)|^2$

Let $F(p)$ be a complex valued function on $p \in \R$, we define $\| F\|_p^2 := \sum_{p=0}^{N-1} |F(p)|^2$

If $F(p)$ is the Fourier transformation of $f(x)$, then $\|F\|^2_p = \|f\|^2_x.$ We proved in class the discrete case. The continuous case is similar in spirit, but harder to prove.

Consider two people, call them Alice and Bob, they each say an integer number, call it a and b. Suppose $a$ and $b$ both have equal probability of taking value within $\{1,2,\cdots, 6\}$, we can ask what is the probabity distribution of $a+b$?

We know $P(a=i) = 1/6$, $P(b=i) = 1/6$ for any $i=1,\cdots, 6$, otherwise the probabilit is 0. Then $P(a+b = k) = \sum_{i+j=k} P(a=i) P(b=j).$

This is an instance of convolution.

Convolution is usually denoted as $\star$.

If $f$ and $g$ are functions on the $x$ space, then we define $(f \star g)(x) = \int_{x_1} f(x_1) g(x-x_1) dx_1$ If $F$ and $G$ are functions on the $p$ space, then we define $(F \star G)(p) = \int_{p_1} F(p_1) G(p-p_1) dp_1$

Fourier transformation sends convolution of functions on one side to simply multiplication on the other side. $(1/2\pi) FT(f \star g) = F \cdot G.$ $FT(f \cdot g) = F \star G.$