Reason for Defining the Inner Product of Functions via Definite Integration
Buildup
The general definition of the inner product is as follows:
Let $H$ be a vector space. For $x,y,z \in H$ and $\alpha, \beta \in \mathbb{C}$, a function that satisfies the following conditions
$$ \langle \cdot , \cdot \rangle \ : \ H \times H \to \mathbb{C} $$
is defined as an inner product, and $\left( H, \langle \cdot ,\cdot \rangle \right)$ is called an inner product space.
- Linearity: $\langle \alpha x + \beta y ,z \rangle =\alpha \langle x,z\rangle + \beta \langle y,z\rangle$
- Conjugate Symmetry: $\langle x,y \rangle = \overline{ \langle y,x \rangle}$
- Positive-definiteness: $\langle x,x \rangle \ge 0 \quad \text{and} \quad \langle x,x \rangle = 0\iff x=0$
Especially, the inner product in the function space is defined using a definite integral as follows:
$$ \langle f, g \rangle := \int_{a}^{b} f(x) g(x) dx $$
It is easy to show that $\langle , \rangle$ becomes an inner product by this definition, but why it is defined in this way might be hard to understand. It feels quite detached from the definition of the inner product in Euclidean space, which is the sum of the products of corresponding elements, and it also doesn’t seem practical at all. However, these definitions are not only natural but also beautifully fit together as one learns more about functional analysis.
Example
Let’s understand through an example:
Consider two vectors $\mathbf{f} = ( {\color{blue} 1} , {\color{orange} 5} , 0 , {\color{purple} 4} , {\color{red} 2} , {\color{Green} 1} )$ and $\mathbf{g} = ( {\color{blue} 9} , {\color{orange} 6} , 0 , {\color{purple} 1} , {\color{red} 2} , {\color{Green} 5} )$. Calculating the inner product gives
$$ \mathbf{f} \cdot \mathbf{g} = {\color{blue} 1 \cdot 9 } + {\color{orange} 5 \cdot 6} + 0 \cdot 0+ {\color{purple} 4 \cdot 1 } + {\color{red} 2 \cdot 2 } + {\color{Green} 1 \cdot 5} = 52 $$
If we represent the size of each component of the vector with a bar graph, it looks like this.
Considering two functions defined to make the shapes in the bar graph above from $[-3,3]$
$$ f(x) := \begin{cases} 1 & , -3 \le x \le -2 \\ 5 & , -2 \le x < -1 \\ 0 & , -1 \le x \le 0 \\ 4 & , 0 \le x < 1 \\ 2 & , 1 \le x < 2 \\ 1 & , 2 \le x \le 3 \end{cases} $$
$$ g(x) := \begin{cases} 9 & , -3 \le x \le -2 \\ 6 & , -2 \le x < -1 \\ 0 & , -1 \le x \le 0 \\ 1 & , 0 \le x < 1 \\ 2 & , 1 \le x < 2 \\ 5 & , 2 \le x \le 3 \end{cases} $$
Then,
$$ f(x) g(x) = \begin{cases} 9 & , -3 \le x \le -2 \\ 30 & , -2 \le x < -1 \\ 0 & , -1 \le x \le 0 \\ 4 & , 0 \le x < 1 \\ 4 & , 1 \le x < 2 \\ 5 & , 2 \le x \le 3 \end{cases} $$
therefore, it is $\displaystyle \int_{-3}^{3} f(x) g(x) dx = 52$ and amazingly matches with $\mathbf{f} \cdot \mathbf{g}$.
Of course, not all functions are conveniently like this, but if it’s an integrable function, the idea of Riemann sums can be applied. Since the definite integral itself includes dividing, multiplying, and adding, there is nothing lacking in calling it an ‘inner product’. The inner product of functions can be seen as a generalization of the inner product of finite-dimensional vectors to infinite dimensions, emphatically covered by the conventional concept of inner product.