I came across the expression

$$\Vert\nabla f\Vert_{L^2(\Omega)}$$

for some function $f: \Omega \subset \mathbb R^2 \to \mathbb R$, but I couldn't find the definition. Can anyone tell me how the $L^2$ norm of a gradient is defined?

My best guess is

$$\sqrt{ \int_\Omega |\nabla f|^2 dx}$$

where $|\nabla f|^2 = (\partial_{x_1} f)^2 + (\partial_{x_2} f)^2$, but I'm not sure whether this is correct.

  • $\begingroup$ This is correct, but usually without the square root. $\endgroup$
    – user260822
    Feb 22, 2018 at 14:51
  • $\begingroup$ @user260822 What do you mean? Without the square root you wouldn't get a norm. $\endgroup$
    – Arnaud D.
    Feb 22, 2018 at 14:53
  • $\begingroup$ sorry, you are right! I was wrong coz I messed up the name. $\endgroup$
    – user260822
    Feb 22, 2018 at 14:56
  • $\begingroup$ Take the usual $L^2$ norm $\endgroup$ Feb 22, 2018 at 14:59

1 Answer 1


Usually it is done the way you have suggested, because that way $L^2(\Omega,\mathbb{R}^2)$ (the space that $\nabla f$ lives in, when the norm is finite) becomes a Hilbert space.

There would be more choices though: If $\Vert\cdot\Vert_0$ is any norm on $\mathbb{R}^m$, then $L^p(\Omega,\mathbb{R}^m)$ inherits a norm by putting $\Vert (f_1,...,f_m)\Vert=(\int\Vert (f_1(x),\dots,f_m(x))\Vert_0^pdx)^{1/p}$, quite often one takes for $\Vert\cdot\Vert_0$ the corresponding $p$-norm on $\mathbb{R}^m$, as you have done in the case $p=2$ intuitively as well. However all norms obtained in that way are equivalent.

  • $\begingroup$ Thanks for the explanation, that clears my confusion:) $\endgroup$
    – flawr
    Feb 22, 2018 at 16:18

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