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Is there a direct relationship between the Dirichlet energy of a function:

$$E(f)=\int_{\Omega}\lvert\nabla f(\mathbf{x})\rvert^2\mathrm{d}V$$

and its Fourier transform

$$\hat{f}(\mathbf{k})=\int_{\Omega}f(\mathbf{x})e^{-2\pi i\mathbf{k}\cdot\mathbf{x}}\mathrm{d}V$$

Since the Dirichlet energy measures the variability of a function in some region, and the Fourier transform measures the amplitude of its frequencies, I think some expression involving the Fourier transform at high frequencies should yield the Dirichlet energy.

Is it possible to connect these two expressions? If so, how?

I think the functional relationships of the Fourier transform listed here might be relevant.

jd27
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user76284
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1 Answers1

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There is indeed a connection, and it is a very useful one. Namely, the Parseval identity states that the $L^2$ energy (squared norm) of $f$ is the same as that of $\hat f$: $$ \int|f(x)|^2=\int|\hat f(k)|^2. $$

If you want to study the $L^2$ norm of the gradient instead, using $\widehat{\nabla f}(k)=2\pi ik\hat f(k)$ with the Parseval identity tells you that $$ \int|\nabla f(x)|^2=(2\pi)^2\int|k|^2|\hat f(k)|^2 $$ and $$ \int|\nabla\hat f(k)|^2=(2\pi)^2\int|x|^2|f(x)|^2. $$ The Dirichlet energy of $f$ can be calculated from $\hat f$ in this explicit way, and it depends more on $\hat f$ for high than for low frequencies.

This makes sense intuitively: The Dirichlet energy does not see the "total mass" of the function but its "total oscillation" in some sense, and the amount of oscillation corresponds to the size of the Fourier variable $k$. High frequencies contribute more to the energy.

If you use different conventions for the Fourier transform, you may have different multiplicative constants in the identities.

Joonas Ilmavirta
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    Thank you! This makes intuitive sense as well since greater values of $x$ correspond to higher frequencies and should therefore contribute more to the energy. – user76284 Jul 15 '15 at 00:00
  • This answer is only correct in the 1-dimensional setting. The problem is that the gradient $\nabla f(x)$ is a vector field and not a function when dim > 1, and so it doesn't make sense to say $\hat{\nabla f(k)}=2\pi i k \hat{f(k)}$. I would be very curious to see if there is a correct version of this that says something along the lines of "If we decompose $\nabla f(x)$ with respect some appropriate "Fourier basis" then the Dirichlet energy comes primarily from "high frequency" terms". Here when I say Fourier basis, I think I means an eigenbasis of some appropriate Laplacian on vector fields. – Sam Ballas Apr 09 '24 at 12:57
  • @SamBallas No, it does still make sense in all dimensions. The gradient is a vector, but so is $k$, so both sides are vectors. The equation is still true, with the absolute value standing for the norm. – Joonas Ilmavirta Apr 09 '24 at 13:44
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    @JoonasIlmavirta. Apologies for my confusion. I retract my previous complaint about the solution – Sam Ballas Apr 09 '24 at 14:01