I have to play a lot with the $L^2$-norm defined as $\|w\|=\sqrt{\int_a^b \langle f,f \rangle}$. However, I don't understand the interpretation of that norm.

We know that the Euclidean norm measure the length of a vector in the Euclidean space, but what does the $L^2$-norm?

Is there anyone could give me an intuitive (even a graph, if possible) explanation of that norm?

  • $\begingroup$ This might be of interest: math.stackexchange.com/questions/885998/… $\endgroup$
    – Tomas
    Commented May 31, 2016 at 17:18
  • $\begingroup$ I have seen that, but it is not what I am looking for $\endgroup$
    – user343429
    Commented May 31, 2016 at 17:19
  • $\begingroup$ I'd say that you might need some background in physics (specifically mechanics and waves) to appreciate the "energy" interpretation of the norm. Admittedly, it's quite hard to obtain intuition without that but it will come to you if you study Hilbert space long enough. $\endgroup$
    – BigbearZzz
    Commented May 31, 2016 at 17:26
  • $\begingroup$ @SpinningAtInfinity Then maybe you can give some thoughts about why the explanations in the link are not what you're looking for? $\endgroup$
    – Tomas
    Commented May 31, 2016 at 17:26
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    $\begingroup$ nobody said the $L^2([a,b])$ norm is exactly the same as the $\|.\|_2$ norm in $\mathbb{R}^n$ but with $n = \infty$ ? every (separable) Hilbert space have an orthonormal basis $(e_k)$ and its vectors are of the form $\sum_{k=1}^\infty c_ke_k$ for some constants $c_k$. This sequence $(c_k)$ is exactly a vector of $\mathbb{R}^\mathbb{N}$ and the norm is $\sqrt{\sum_{k=1}^\infty |c_k|^2}$. so everything stays the same: those separable Hilbert spaces are the limit when $n \to \infty$ of $\mathbb{R}^n$. An orthonormal basis of $L^2([0,1])$ example is $(e^{2i\pi kx})$ aka the Fourier series $\endgroup$
    – reuns
    Commented May 31, 2016 at 17:39

4 Answers 4


OK let's see if this helps you. Suppose you have two functions $f,g:[a,b]\to \mathbb{R}$. If someone asks you what is distance between $f(x)$ and $g(x)$ it is easy you would say $|f(x)-g(x)|$. But if I ask what is the distance between $f$ and $g$, this question is kind of absurd. But I can ask what is the distance between $f$ and $g$ on average? Then it is $$ \dfrac{1}{b-a}\int_a^b |f(x)-g(x)|dx=\dfrac{||f-g||_1}{b-a} $$ which gives the $L^1$-norm. But this is just one of the many different ways you can do the averaging: Another way would be related to the integral $$ \left[\int_a^b|f(x)-g(x)|^p dx\right]^{1/p}:=||f-g||_{p} $$ which is the $L^p$-norm in general.

Let us investigate the norm of $f(x)=x^n$ in $[0,1]$ for different $L_p$ norms. I suggest you draw the graphs of $x^{p}$ for a few $p$ to see how higher $p$ makes $x^{p}$ flatter near the origin and how the integral therefore favors the vicinity of $x=1$ more and more as $p$ becomes bigger. $$ ||x||_p=\left[\int_0^1 x^{p}dx\right]^{1/p}=\frac{1}{(p+1)^{1/p}} $$ The $L^p$ norm is smaller than $L^m$ norm if $m>p$ because the behavior near more points is downplayed in $m$ in comparison to $p$. So depending on what you want to capture in your averaging and how you want to define `the distance' between functions, you utilize different $L^p$ norms.

This also motivates why the $L^\infty$ norm is nothing but the essential supremum of $f$; i.e. you filter everything out other than the highest values of $f(x)$ as you let $p\to \infty$.

  • $\begingroup$ Remember at least one thing in your comment. When you insert examples like you did, it really helps the person questioned to understand a problem. Your example has helped me greatly, thank you. $\endgroup$
    – user343429
    Commented May 31, 2016 at 18:15
  • $\begingroup$ $[\int_0^1 x^p dx]^{1/p}=(\frac{x^{p+1}}{p+1}|_0^1)^{1/p}=(\frac{1}{p+1})^{1/p}$ $\endgroup$
    – Frank
    Commented Mar 20, 2019 at 23:23

There are several good answers here, one accepted. Nevertheless I'm surprised not to see the $L^2$ norm described as the infinite dimensional analogue of Euclidean distance.

In the plane, the length of the vector $(x,y)$ - that is, the distance between $(x,y)$ and the origin - is $\sqrt{x^2 + y^2}$. In $n$-space it's the square root of the sum of the squares of the components.

Now think of a function as a vector with infinitely many components (its value at each point in the domain) and replace summation by integration to get the $L^2$ norm of a function.

Finally, tack on the end of last sentence of @levap 's answer:

... the $L^2$ norm has the advantage that it comes from an inner product and so all the techniques from inner product spaces (orthogonal projections, etc) can be applied when we use the $L^2$ norm.

  • $\begingroup$ Your third paragraph is exactly what made it "click" for me when I was learning these things. $\endgroup$
    – Neal
    Commented Jun 4, 2016 at 19:39

If you have some physics background, then $L^2$ norm can often be interpreted as the "energy" of the wave functions. Physical interpretation of L1 Norm and L2 Norm

In quantum physics, the $L^2$ norm represents the probability of detecting a particular pure state amount many mixed states.

In statistic, minimizing the $L^2$ norm of the difference between 2 functions is equivalent to the process called "least square method". Differences between the L1-norm and the L2-norm

In mathematics, we prefer it over many other possible norm because it induces the Hilbert Spaces structure on the functions spaces. There are many beautiful theory available once we embrace the $L^2$ norm. I'd say that you'll gain intuition along the way of studying it. It may not seem why we prefer it now but, eventually, you'll see.

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    $\begingroup$ You can almost guess that $L^2$ is the correct norm in physics because space and momentum are Fourier Transforms of each other and Fourier transforms preserve only the $L^2$ norm. $\endgroup$
    – Alex R.
    Commented May 31, 2016 at 18:20

I'm not sure how intuitive is this, but the norm $||f||_{L^2([a,b])}$ of an integrable function defined on $[a,b]$ is the square root of the "area under the graph" of $|f|^2$ on the interval $[a,b]$. For example, if $f \equiv C$ is a constant function, then the area under the graph of $f^2 = C^2$ is the area of a rectangle given by $(b - a)C^2$ and so $||f||_{L^2([a,b])} = |C| \sqrt{b - a}$.

There are various other norms one can define for functions, the most intuitive of which is probably the $L^1$ norm given by $||f||_{L^1([a,b])} = \int_a^b |f(x)| \, dx$ which simply gives you the area under the graph of $|f|$. Unlike the $L^1$ norm, the $L^2$ norm has the advantage that it comes from an inner product and so all the techniques from inner product spaces (orthogonal projections, etc) can be applied when we use the $L^2$ norm.


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