Dirichlet energy and Fourier transform 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.
 A: 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.
