$A = \{ (1+\frac{d^2}{dx^2})f \;|\; f \in H^2(\mathbb R)\}$ is dense in $L^2(\mathbb R)$. Namely, I want to show that
For any $g \in L^2(\mathbb R)$, there exists $g_n \in A$ such that $\|g_n - g\|_{L^2(\mathbb R)} \to 0$.
How can I prove this?
Question on A Dense Subset in $L^2$
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real-analysis
functional-analysis
pde
1 Answers
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The operator $L=\frac{d^2}{dx^2}$ is closed, densely-defined and selfadjoint on $H^2(\mathbb{R})$. And $L$ has no eigenvalues. Therefore, $(L-\lambda I)$ always has dense range, regardless of the value of $\lambda$; this is because $$ \mathcal{R}(L-\lambda I)^{\perp} = \mathcal{N}(L^*-\overline{\lambda}I)=\mathcal{N}(L-\overline{\lambda}I)=\{0\}. $$ For $\lambda\notin\mathbb{R}$ the range is all of $L^2(\mathbb{R})$. However, every $\lambda\in(-\infty, 0]$ is in the spectrum of $L$; so the range of $L-\lambda I$ is dense but not closed for $\lambda\in(-\infty,0]$.