I have a question related to function normalization.
I'm dealing with a modulating function, which is a function that has some properties.
Let say that this modulating function is $\phi(t)$ and it is defined in an interval $[0, T]$. To $\phi(t)$ be a modulating function, it needs to satisfy the following properties:
$(P1) \quad \phi_{n}(t) = \begin{cases}sin^{n}\frac{(n\, \pi \, t)}{T}, \quad \text{for $t$} \in [0, T] \text{, $n \in \mathbb{N}$} \\ 0, \quad \text{otherwise}\end{cases}$
$(P2) \quad \phi_{n}^{(p)}(0) = 0 $
$(P2) \quad \phi_{n}^{(p)}(T) = 0$
$p$ in properties $P2$ and $P3$ is the derivative number. Also, the defined modulating function $\phi(t)$ is just a kind of modulating function called Shinbrot's Modulating Function. There are many other types of modulating functions.
Given the context, I would like to know how to normalize this function. When we normalize a vector we perform the following
\begin{equation} \vec{\hat{u}} = \frac{\vec{u}}{|| \vec{u} ||} \end{equation}
But when we work with functions, we have to compute the inner product of the function with itself? Like the following
\begin{equation}\label{a} <\phi(t), \phi(t)> \, = \int_{0}^{T}\phi(t) \bar{\phi}(t) dt \end{equation}
The $\bar{\phi}(t)$ indicates the conjugate of $\phi(t)$. But in this case, it is a real function, so we have
\begin{equation} <\phi(t), \phi(t)> \, = \int_{0}^{T}\phi^2 (t) dt \end{equation}
Then, due to the modulus operator in the denominator, we'll have the following
\begin{equation} \hat{\phi}(t) = \frac{\phi(t)}{<\phi(t), \phi(t)>} = \frac{\phi(t)}{\sqrt{\int_{0}^{T}\phi^2 (t) dt}} \end{equation}
So this last equation, with $\hat{\phi}(t)$ will be the normalized function? Are my assumptions correct?
Thanks in advance!
Given a non-zero element $v$ of a normed space $(X, \|\cdot\|)$, a normalized vector $\bar v$ is a vector in the same direction but with norm $1$. Namely $\bar v=\tfrac 1{\|v\|}v$. In particular, if $X$ is a real Hilbert space with an inner product $\langle\cdot,\cdot\rangle$ then $\|v\|=\sqrt{\langle v,v\rangle}$. So if $X$ is a space of a continuous functions on $[0,T]$ with $\langle\phi,\psi\rangle=\int_0^T \phi(t)\psi(t)dt$ for any functions $\phi,\psi\in X$, then a normalized non-zero function $\phi$ is $$\bar{\phi}(x)=\frac 1{\sqrt{\int_0^T (\phi(t))^2 dt}}\phi(x).$$