So far I know, the Schwartz space $S(\mathbb{R}^n)$ is dense in $L^2(\mathbb{R}^n)$ with respect to the usual $\lVert \cdot \rVert_{L^2}$-norm, i.e., for every function $f \in L^2(\mathbb{R}^n)$ there is a sequence $(\varphi_n)$ such that $ \lim_{n \to \infty} \lVert \varphi_n - f \rVert_{L^2} \rightarrow 0$. The first Sobolev space $H^1(\mathbb{R}^n)$ is a subspace of $L^2(\mathbb{R}^n)$ and can be made into a Hilbert space via the scalar product $\langle g,f \rangle_{H^1} = \langle g,f \rangle_{L^2} + \intop_{\mathbb{R}^n}\overline{\nabla g} \cdot \nabla f dx^n$. So, my question is if $S(\mathbb{R}^n)$ is also dense in $H^1(\mathbb{R}^n)$ with respect to the $\lVert \cdot \rVert_{H^1}$-norm. With respect to the $\lVert \cdot \rVert_{L^2}$-norm the answer is yes and this is trivial, but my problem is that the $\lVert \cdot \rVert_{H^1}$-norm is always bigger or equal than the $\lVert \cdot \rVert_{L^2}$-norm.
dense subspaces in Hilbert spaces
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functional-analysis
sobolev-spaces