# What does normalizing a real valued function mean?

I might be wrong because I didn't find a proper definition in a google search but what I have found is that if we have a real-valued function $$f:[0,1]\to\mathbb{R}$$ defined by $$f(x)=x^2$$ then normalizing this function means the integral value should be $$1$$.Like here if I integrate then we have $$\int_{0}^{1}x^2dx=\frac{1}{3}$$ so normalizing this function means we want another function say $$g$$ which is isomorphic to $$g$$ such that $$\int_{0}^{1}g(x)=1$$ . If whatever I have written is correct then I want to know why we are doing this. If whatever I have written is not correct then please if someone gives the correct meaning or definition of normalizing a function and also how to normalize such function, that will be a great help, Thanks.

• I typically think of normalizing a data set (or a function) as dividing by the largest value (largest data point, or sup norm in case of a function). The integral of a real-valued function is not necessarily a norm.
– Doug
Commented Sep 19, 2022 at 16:00
• That could be what it means, but both the meaning and the reason for normalizing depend on the context. Commented Sep 19, 2022 at 16:04

This depends on which norm you're using. Every continuous real function with compact domain assumes both its maximum and minima at least once, so on a compact domain $$D$$ it's natural to define the norm of a real function $$f$$ as being:
$$\|f\| = \sup_{x \in D} |f(x)|$$
In your case, $$\|[0, 1] \ni x \mapsto x^2\| = 1$$
and so the normalization of $$f$$, $$\frac{f}{\|f\|}$$, coincides with $$f$$. There are, of course, other norms one might consider, where $$f$$ might not necessarily be already normalized.