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Let $Af(x)=xf(x)$ be the unbounded operator defined on $L^2(\mathbb R)$ with as its domain the space of infinitely differentiable functions with compact support on $\mathbb R$ (denoted by $C_c^{\infty}(\mathbb R)$). I remember my lecturer once when I asked him about the closedness of $A$, he told me that $A$ is not closed for its domain is too "small". Now, I wanted to write down details but I got stack and this why I am seeking your help. I would like a proof of the non closedness of this operator $A$ on $C_c^{\infty}(\mathbb R)$.

Many thanks in advance for your help.

Math

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Define $f:=\chi_{[-1,1]}$. Define $g(x):=xf(x)$. Define $\eta_\varepsilon$ as the standard mollifier. Define $\forall k \in \mathbb{N}, f_k:=f*\eta_\frac{1}{k}$, where $*$ is the convolution. Then $$\forall k \in \mathbb{N}, f_k\in C_c^\infty(\mathbb{R})$$ and $$\|f_k-f\|_{L^2(\mathbb{R})}\rightarrow0, k\rightarrow\infty$$ and $$\|Af_k-g\|_{L^2(\mathbb{R})}\rightarrow0, k\rightarrow\infty.$$ Then $$\|(f_k,Af_k)-(f,g)\|_{L^2(\mathbb{R})\times L^2(\mathbb{R})}\rightarrow0, k\rightarrow\infty,$$ and so $(f,g)$ is in the closure of $\operatorname{graf}(A)$. However $f$ is not in the domain of $A$ and then $(f,g)\notin\operatorname{graf}(A)$. So $\operatorname{graf}(A)$ isn't closed in $L^2(\mathbb{R})\times L^2(\mathbb{R})$ which, by definition, means that $A$ is not closed.

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  • $\begingroup$ Dear Bob. This is great. I kind of started it like you did but your proof is very clear. Thanks again. $\endgroup$
    – Math
    Jul 9 '18 at 21:52
  • $\begingroup$ One last point though Bob, is it so clear that $Af_k$ goes to $g$ in the $L^2$ norm? $\endgroup$
    – Math
    Jul 9 '18 at 21:58
  • $\begingroup$ $Af_k\rightarrow g$ pointwise a.e. and the each element of the sequence has a square that is dominated by a common bounded function with compact support, say $\chi_{[-2,2]}$. Then by dominated convergence theorem $\int |Af_k|^2\rightarrow \int |g|^2$ and by Scheffe's lemma you get the conclusion. $\endgroup$
    – Bob
    Jul 9 '18 at 22:05
  • $\begingroup$ Thanks a lot Bob and you've been of great help! I appreciate it. Math $\endgroup$
    – Math
    Jul 9 '18 at 22:08
  • $\begingroup$ Nice to help you. If you're satisfied with the answer, I would be grateful if you'll accept the answer. $\endgroup$
    – Bob
    Jul 9 '18 at 22:12

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