Recently I came across the following inequality. Let $f:\,\mathbb{R}^n\rightarrow\mathbb{R}$ be smooth, then the inequality $$\left\lVert f\right\lVert_{\dot H^2(\mathbb{R}^n)}\leq\left\lVert\Delta f\right\lVert_{L^2(\mathbb{R}^n)}$$ holds, where $\dot H^2(\mathbb R^n)$ denotes the homogeneous Sobolev norm on $\mathbb R^n$.
I tried to prove this brute-force, but I did not see where this was going. Is there an elegant and quick way of obtaining the estimate?
$$ \|D^2 v \|_2^2 = \int \sum_{i,j=1}^N v_{x_ix_j} v_{x_i x_j} = - \int \sum_{i,j=1}^N v_{x_i x_j x_i} v_{x_j} = \int \sum_{i=1}^N v_{x_i x_i} \sum_{j=1}^N v_{x_j x_j} = \|\Delta v\|_2^2 $$ for every $v \in C_0^3(\mathbb{R}^N)$. If you use Fourier analysis to define homogeneous Sobolev spaces, then the proof is even easier. The identity might become an inequality because people may use equivalent Sobolev constants.