Are these $ L^2 $ norms equivalent?

Question

Could anyone help me to prove that these two norms are equivalent? The norms are $$\Vert \nabla^{2} f\Vert_{L^{2}} $$ and $$ \Vert\Delta f\Vert_{L^{2}}.$$ Maybe it is possible to prove using the Fourier Transform?

Answer 1

If $\nabla^2$ is just another notation for div grad, then the two are equivalent because $\Delta f=\pm\nabla^2 f$, sign convention due to author's choice.

On the other hand, if $\nabla^2$ means the Hessian, then taking Fourier transform, $$ \lVert \partial_{ij}f\rVert_{L^2(\mathbb{R}^n)}^2=\lVert \omega_i\omega_j(\mathcal{F}f)(\omega)\rVert_{L^2(\mathbb{R}^n)}^2 $$ and so \begin{align*} \lVert\nabla^2 f\rVert_{L^2(\mathbb{R}^n)}^2 &= \sum_{i,j}\lVert (\omega_i\omega_j)(\mathcal{F}f)(\omega)\rVert_{L^2(\mathbb{R}^n)}^2\\ &\leq\sum_{i,j}\frac12\lVert(\omega_i^2+\omega_j^2)(\mathcal{F}f)(\omega)\rVert_{L^2(\mathbb{R}^n)}^2\\ &=n\sum_i\lVert\omega_i^2(\mathcal{F}f)(\omega)\rVert_{L^2(\mathbb{R}^n)}^2\\ &=n\lVert\Delta f\rVert_{L^2(\mathbb{R}^n)}^2 \end{align*} and the other bound is clear.

Answer 2

This is a notational difference, with the definition being $$ \nabla^2 f = \Delta f = \sum_{i} \frac{\partial}{\partial x_i}f $$

Therefore the norms are the same.

I will say that the $\nabla^2 $ notation seems to appear much more in physics heavy areas of mathematics, while the $\Delta f$ notation is much more common in analysis and other less physical areas of mathematics. But that's not a hard-and-fast rule, just a slight cultural observation. The former notation comes from a combination of the gradient and the divergence notation you see pop up in vector calculus.