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.

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    $\begingroup$ 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. $\endgroup$
    – Doug
    Commented Sep 19, 2022 at 16:00
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    $\begingroup$ That could be what it means, but both the meaning and the reason for normalizing depend on the context. $\endgroup$
    – aschepler
    Commented Sep 19, 2022 at 16:04

1 Answer 1


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.


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