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Let $X= L^2[0,1]$, and $A:u(t)\mapsto-u''(t)$. The domain of $A$ is $$ D(A)=\{ u\in C^2[0,1]: u(0)=u(1), u'(0)= u'(1) \} $$ Then, how to show $A$ is a closed linear operator?

Definition: $A$ is a closed linear operator, if for any $\{x_n\}\subset D(A)$ satisfying $x_n\to x \in X$ and $Ax_n \rightarrow y$, we have $x\in D(A)$ and $Ax=y$.

I get the problem in a Chinese functional analysis book.

Source of the problem.(translated) Perhaps there is indeed a mistake in this example. If so, could you tell me which book has a similar example?

Example 2.6.2 Suppose that $\mathscr{X}=L^2[0,1], A: u(t)\mapsto -\frac{d^2}{dt^2}u(t)$, where $$D(A)=\{u\in C^2[0,1]\vert u(0)=u(1), u'(0)=u'(1)\},$$ then $A$ is a closed linear operator and $$\sigma(A)=\sigma_p(A)=\{(2n\pi)^2\vert n=0,1,\cdots\}.$$

Proof: On the one hand, we have $$-\frac{d^2}{dt^2}\sin(2n\pi t)=(2n\pi)^2\sin(2n\pi t)\quad \text{and}\quad -\frac{d^2}{dt^2}\cos(2n\pi t)=(2n\pi)^2\cos(2n\pi t), \forall\ n\in\mathbb{N}.$$ On the other hand, when $\lambda\neq (2n\pi)^2$, $\forall\ f\in L^2[0,1]$, the equation $$\left(-\frac{d^2}{dt^2}-\lambda\right)u(t)=f(t)$$ has a unique solution $$u(t)=\sum_{n=-\infty}^\infty\frac{C_n}{(2n\pi)^2-\lambda}e^{2\pi int},$$ where $$C_n=\int_0^1f(t)e^{-2\pi int}dt\ (n\in\mathbb{Z}).$$ It's easy to check that $u\in D(A)$ and $$\vert\vert u\vert\vert^2=\sum_{n=-\infty}^\infty\frac{\vert C_n\vert}{\vert (2n\pi)^2-\lambda\vert^2}\leq M_\lambda^2\sum_{n==\infty}^\infty\vert C_n\vert^2=M_\lambda^2\vert\vert f\vert\vert^2,$$ where $$M_\lambda=\sup_{n\in\mathbb{Z}}\frac{1}{\vert (2n\pi)^2-\lambda^2 \vert}<\infty.$$

Bowei Tang
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Enhao Lan
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    $D(A)$ is not closed. It is dense In $X$. – Kavi Rama Murthy Jun 15 '25 at 09:01
  • @KaviRamaMurthy My expression is wrong. I revise it. Thank you. – Enhao Lan Jun 15 '25 at 09:17
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    The definition is not very clear to me. $x$ was defined previously as the limit of the sequence $x_n$, so why, after this, you say "there are $x\in D(A)$"? – Zima Jun 15 '25 at 09:20
  • @Zima I mean that the sequence of convergence points of $D(A)$ must converge to $D(A)$. – Enhao Lan Jun 15 '25 at 09:27
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    The Wikipedia page on uniform convergence gives the example $f_n = \frac{1}{\sqrt n} \sin(nx)$ for a sequence of functions which uniformly converges to $0$, but whose derivatives do not converge to anything. I think the second derivatives also do not converge to anything. Would that be a counterexample to your statement? – Jayanth R Varma Jun 15 '25 at 10:10
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    @JayanthRVarma Not quite. The statement the OP tries to prove is that the linear operator has a closed graph. This means, for a counterexample one needs to find a sequence $(x_n, Tx_n)_n$ which converges $L^2\times L^2$ to $(x,y)$ and $y\neq Tx$ or $x\notin D(T)$ (i.e. the limit is not in the graph of $T$). – Severin Schraven Jun 15 '25 at 10:22
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    @EnhaoLan I don't think the statement is true. Is it possible that you are meant to prove that the operator is closable (i.e. admits a closed extension)? – Severin Schraven Jun 15 '25 at 10:26
  • @SeverinSchraven I also think it is wrong, since the example (above picture) has more than just this one problem. Do you know any book talk about the spectral of the above operator $A$ ? – Enhao Lan Jun 15 '25 at 10:47
  • Essentially you are looking at the Laplacian on the circle. Pretty much any standard book on Schrödinger operators should treat that. – Severin Schraven Jun 15 '25 at 10:53
  • However, if you really know how to show $x\in D(A)$, then the statement should be true. If you are familiar with Sobolev spaces, then you get that $(x_n)_n$ is a Cauchy sequence in $H^2$. Thus, for the limit, we have $y=-x''$ (with distributional derivatives). However, if $x$ is $C^2$, then those the usual derivatives. – Severin Schraven Jun 15 '25 at 10:57
  • @SeverinSchraven I can't show $x\in D(A)$. My previous idea was wrong. It seems that everything would make sense if let $D(A)={ u\in H^2[0,1]: u(0)=u(1), u'(0)= u'(1) }$. And another problem (https://math.stackexchange.com/questions/5061992/unique-solution-of-left-fracd2dt2-lambda-right-ut-ft-w) also imply it should be $H^2$. – Enhao Lan Jun 15 '25 at 11:37
  • @SeverinSchraven Thank you very much for your patience and discussion. I have been stuck with this example problem for a long time. It seems that the key point has been found now. Although I still don't know how to do it, the direction seems to be right. – Enhao Lan Jun 15 '25 at 11:40
  • @EnhaoLan Indeed, if we had $H^2$, we would be good to go. For a counterexample one should be able to approximate some $x\in H^2(S^1)\setminus C^2(S^1)$ to get a counterexample. – Severin Schraven Jun 15 '25 at 12:22
  • For $Ax_n\to y$ to make sense, you should edit your post to specify the codomain of $A$, and its topology. – Anne Bauval Jun 15 '25 at 16:08

1 Answers1

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The operator is not closed. Indeed, after shifting, we can assume that our domain is $$D(A)=\{ f\in C^2([-1/2,1/2]) \ : \ f(-1/2)=f(1/2), f'(-1/2)=f'(1/2)\}.$$ Now pick a smooth cut-off function $\chi\in C^\infty(\mathbb{R})$ with $\chi(x)= 0$ for $\vert x\vert>1/4$ and $\chi(x)=1$ for $\vert x\vert\leq 1/5$. Define $f(x)=\chi(x) x^2\ln(\vert x \vert).$ A direct computation shows that $f\in H^2$, and satisfies all conditions of $D(A)$ except that it is not $C^2$ at the origin.

Now pick a suitable mollifier, see Approximate $L^2$ function by convolving with mollifiers, we can find $f_n\in C^\infty([-1/2, 1/2])$, $f_n(x)=0$ for $1/3\leq \vert x\vert \leq 1/2$ (in particular $f_n \in D(A)$) and such that $$ \Vert f-f_n\Vert_{L^2}, \Vert f_n'-f'\Vert_{L^2}, \Vert f_n''-f''\Vert_{L^2} \rightarrow 0, n\rightarrow \infty.$$ Thus, $$(f_n, Af_n)=(f_n, -f_n'')\rightarrow (f, -f'')$$ where the convergence is in $L^2\times L^2.$ As $f\notin D(A)$, we get that the graph of $A$ is not closed and hence, $A$ is not closed.

Severin Schraven
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