(Revised version) How to show $A:u(t) \rightarrow -u''(t)$ is closed linear operator?

Question

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

$H^2[0,1]$ is Sobolev space, namely, composed by functions which's second order weak derivative belongs to $L^2[0,1]$.

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 $Ax=y$.

Yesterday I asked almost the same question. Due to my personal insufficiency and the wrong guidance of the book, the question I asked doesn't hold true. Severin Schraven has constructed a counterexample. Based on yesterday's discussion, I modified the problem. Although I know what $L^2$ and $H^2$ spaces are, I know very little about their properties (I know that they are complete and what is weak derivative). So, I still don't know how to prove that $A$ is a closed linear operator.

If there are any mistakes in the problem, tell me and I will correct them. Thanks.

Answer 1

The Fourier series expansion $f(t)=\sum_{n\in\mathbb Z}\hat{f}_ne^{2\pi i n t}$ establishes an isomorphism between the two Hilbert spaces $L^2[0, 1]$ and $\ell^2(\mathbb Z)$. Also, the weak derivative of $f$ if in $L^2$ must have Fourier expansion $\sum_{n\in\mathbb Z}\hat{f}_n(2\pi i n) e^{2\pi i nt}$(See e.g. this post). Therefore, up to a positive constant, $f\mapsto -f''$ is unitarily equivalent to the multiplication operator, $$\begin{align}B: D(B)\subset \ell^2(\mathbb Z) &\rightarrow \ell^2(\mathbb Z) \\ (a_n)&\mapsto (n^2a_n)\end{align}$$

And $D(B)=\{(a_n): (n^2a_n)\in\ell^2(\mathbb Z) \}$ is precisely the image of $D(A)$ under the Fourier expansion.

The operator $B$ is clearly closed. If $(a_n)_m\xrightarrow{m\rightarrow\infty} (b_n)\in \ell^2(\mathbb Z)$ and $(n^2a_n)_m\xrightarrow{m\rightarrow\infty} (c_n)\in \ell^2(\mathbb Z)$, then componentwisely we must have $a_n^{(m)}\rightarrow b_n, n^2a_n^{(m)}\rightarrow n^2b_n$, hence $c_n = n^2b_n$. And since $(c_n)\in\ell^2(\mathbb Z)$, this proves $(b_n)\in D(B)$ and $B((b_n))=(c_n)$.

In the false claim from the book, $D(A)=\{f\in C^2[0, 1]: f(0)=f(1), f'(0)=f'(1)\}$ guarantees that $f', f''$ can be obtained by differentiating the Fourier expansion of $f$ term by term, but the conditions are only sufficient, not necessary, hence $A$ is closable but not closed.

Answer 2

• First of all, your operator $A$ is symmetric with respect to the $L^2([a,b])$ inner product, because if $f,g\in D(A)$, then using integration by parts twice, we have \begin{align} \langle A(f),g\rangle_{L^2([a,b])}&:=\int_a^b(-f’’)\overline{g}\,dt\\ &=[-f’\overline{g}]_a^b+\int_a^bf’\overline{g’}\,dt\\ &=0+ [f\overline{g’}]_a^b-\int_a^bf\overline{g’’}\,dt\\ &=0+0+\int_a^bf\overline{A(g)}\,dt\\ &=\langle f,A(g)\rangle. \end{align} Note that the boundary terms cancel each other out in pairs precisely because of the definition of $f,g\in D(A)$. The justification of the integration by parts for these Sobolev functions is a standard exercise which I leave to you (one way is to first prove this for smooth $f,g$ in $D(A)$, then argue by density).

• Next, it is a general fact that if $H$ is a Hilbert space and $A:D(A)\subset H\to H$ is a densely-defined symmetric linear map, then it is closable. So, atleast we have that going for us (at this stage this is also true for your previous question with the domain consisting of $C^2$ functions. It is also true if we allow $C^{\infty}$ functions… you see, symmetry is just a matter of ensuring the boundary terms vanish, and there are lots of possible dense domains which allow for this possibility).

• Since $A$ is densely-defined symmetric, it follows $A\subset A^*$ (i.e the adjoint is an extension, or said another way, the graph of $A$ is contained in the graph of $A^*$). I claim that the opposite inclusion is also true. If we manage to prove this then it would then tell us that $A=A^*$ is actually self-adjoint; in particular $A$ is the adjoint of some densely-defined linear map; and it is a general fact that adjoints are always closed operators (because the graph of an adjoint is related to the annihilator subspace, or in the Hilbert space case, to orthogonal complements, and these are always closed subspaces).

• Finally, to show that $A^*\subset A$, we really just need to prove the statement about domains, i.e that if $\xi\in D(A^*)$, then $\xi\in D(A)$. Ok, so let $\xi\in D(A^*)$ be arbitrary. By definition, for all $x\in D(A)$, we have $\langle A^*(\xi),x\rangle=\langle\xi,A(x)\rangle$, i.e $\int_a^bA^*(\xi)\overline{x}\,dt=\int_a^b\xi\cdot (-\overline{x’’})\,dt$. Since this is in particular true for all $x\in C^{\infty}_c((a,b))$, it follows by definition that $\xi$ has two weak derivatives in $L^2$, i.e $\xi\in H^2([a,b])$. This is one step in the right direction. Next, we fully exploit this adjoint formula: for all $x\in D(A)$, \begin{align} 0&= \int_a^bA^*(\xi)\overline{x}\,dt-\int_a^b\xi\cdot (-\overline{x’’})\,dt\\ &=\int_a^b(-\xi’’)\overline{x}\,dt+\int_a^b\xi\overline{x’’}\,dt\\ &=[-\xi’\overline{x}]_a^b+[\xi\overline{x’}]_a^b\tag{IBP}\\ &=\overline{x(a)}[\xi’(a)-\xi’(b)]+ \overline{x’(a)}[\xi(b)-\xi(a)]. \end{align} Notice that I did integration by parts on each term, which leaves us with the boundary terms, while the integral term $\int_a^b\xi’\overline{x’}\,dt$ appears twice with opposite sign hence cancels out. In the final step I used explicitly the definition of $D(A)$ to simplify the boundary terms to write $x(b)=x(a)$ and $x’(b)=x’(a)$. Now, notice that $x$ is an arbitrary $H^2([a,b])$ function such that $x,x’$ agree at the endpoints; but other than that, $x(a)$ and $x’(a)$ are arbitrary. So, this implies that each of the square brackets vanishes, i.e $\xi(a)=\xi(b)$ and $\xi’(a)=\xi’(b)$. Thus, by definition, we have $\xi\in D(A)$, and hence the proof that $A=A^*$ is complete. As I said above, this self-adjointness trivially implies $A$ is closed.

If you now mess around with the domain of $A$ slightly, then you’ll see that the computation of the adjoint remains mostly the same. It’s just at this final step or two that you have to carefully look at what it means for $x$ to be in $D(A)$, and then conclude accordingly. For example, if $A_0$ is the operator with domain $D_{A_0}=\{u\in H^2([a,b])\,:\, u^{(j)}(a)= u^{(j)}(b)=0, \quad\text{for $j=0,1$}\}$, then the argument carries through till we reach the step (IBP) above. But now both these boundary terms vanish; this means there are no further conditions on $\xi$, i.e $D(A_0^*)=H^2([a,b])$, so this $A_0$ is not self-adjoint. See Rudin’s functional analysis book, example 13.4 for more details on the first-order derivative $f\mapsto if’$ case.

This type of integration-by-parts argument is also very closely related to the Fourier series description of the $L^2$-based Sobolev spaces.

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A slightly more direct outline for closedness of $A$

Consider any $\xi,\eta\in L^2([a,b])$ such that there exists a sequence $(x_n)_{n=1}^{\infty}\subset D(A)$ such that $x_n\to \xi$ and $A(x_n)=-x_n’’\to \eta$ in $L^2([a,b])$ (i.e $(\xi,\eta)$ is in the closure of the graph of $A$). We have to show $\xi\in D(A)$ and $A(\xi)=\eta$.

• Sobolev-embedding theorem tells us that $H^2([a,b])\to C^0([a,b])\times C^0([a,b])$, $f\mapsto (f,f’)$ is linear and continuous (when the domain is given the $H^2$-norm), so in particular, for each $t\in [a,b]$, the evaluation map $\epsilon_t:H^2([a,b])\to \Bbb{C}^2$, $\epsilon_t(f):= (f(t),f’(t))$ is continuous. Hence, $D(A)$, which is the kernel of $\epsilon_a-\epsilon_b$, is a closed subspace of $H^2([a,b])$ relative to $H^2$-norm, i.e $\mathscr{H}:=(D(A), \|\cdot\|_{H^2([a,b])})$ is a Hilbert space in its own right.
• Show from the embedding theorem and the assumptions that $\sup\limits_{n\geq 1}\|x_n\|_{\mathscr{H}}<\infty$. This shouldn’t be too hard (in higher dimensions, this requires us to use elliptic estimates, but in this 1D case, it’s much easier).
• Hilbert-space theory (a special case of Banach-Alaoglu) tells us that there every norm-bounded sequence has a weakly-convergent subsequence, i.e exists a $\zeta\in \mathscr{H}$ such that (after passing to a subsequence) $x_n\to \zeta$ weakly in $\mathscr{H}$. Hence, $x_n\to \zeta$ weakly in $L^2$ as well. But by hypothesis $x_n\to \xi$ weakly in $L^2$, so by uniqueness of weak limits in $L^2$, we have $\xi=\zeta$, but by definition of $\zeta$, we thus have $\xi\in \mathscr{H}$.
• The operator $A: \mathscr{H}\to L^2$, is bounded so it preserves weak limits, so $A(x_n)\to A(\zeta)=A(\xi)$ weakly in $L^2([a,b])$. But we also have by hypothesis $A(x_n)\to \eta$ strongly hence weakly in $L^2$, so by uniqueness of weak limits, we have $A(\xi)=\eta$.

This completes the proof that the graph of $A$ is closed in $L^2([a,b])\times L^2([a,b])$. This type of argument also works for dealing with Laplace-type or Dirac-type operators acting on suitable sections of a vector bundle over compact boundaryless Riemannian manifolds. The key is to establish suitable elliptic estimates. You could avoid phrasing things so abstractly in terms of weak limits if you instead look in terms of the formal adjoint (hence why I presented the integration-by-parts arguments first). See Roe’s book on elliptic operators (the chapter on analysis of Dirac operators) for these more general statements.

Answer 3

Henceforth let $U=(0,1)$.

IN SUMMARY: You need to be careful about the spaces which you are talking about, and what norms they are equipped with. Letting $X$ and $Y$ be normed spaces each with two different norms $\Vert\cdot\Vert _{X},\Vert\cdot \Vert'_X$ and $\Vert\cdot\Vert _{Y},\Vert\cdot \Vert'_Y$ , an arbitrary mapping $T:X\to Y$ may be closed when considered as an operator from $(X,\Vert \cdot \Vert_X)\to (Y,\Vert \cdot \Vert_Y) $, but may not be closed when considered as an operator from $(X,\Vert \cdot \Vert'_X)\to (Y,\Vert \cdot \Vert'_Y)$, when these two pairs of norms are not equivalent.

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Understanding $A$ as an operator from $(H^2(U),\Vert \cdot \Vert_{H^2(U)})\to (L^2(U),\Vert \cdot\Vert_{L^2(U)})$, $A$ is quite obviously closed. The reasoning is as follows: Clearly, $A$ is linear. Therefore, via the closed graph theorem, $A$ being continuous implies $A$ is closed. So it suffices to show $A$ is continuous, or equivalently, $A$ is bounded.

In other words we would like to show that $\exists C>0$ such that $$\Vert Au\Vert_{L^2(U)}\leq C\Vert u \Vert_{H^2(U)}$$ But $A=-D^2$, in other words $$\left\Vert -D^2 u\right\Vert_{L^2(U)}\leq C\left(\Vert u\Vert^2_{L^2 (U)}+\left\Vert D^1u \right\Vert_{L^2(U)}^2+\left\Vert D^2 u \right\Vert^2_{L^2(U)} \right)^{1/2}$$

Which clearly holds - just take $C=1$.

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(As in the other question, we now take a symmetric interval $U=(-1/2,1/2)$

As shown in the answer to your other question, this is not the case when we consider $A$ as an operator from $(\text{domain},\Vert \cdot\Vert_{L^2(U)})\to (L^2(U),\Vert \cdot \Vert_{L^2(U)})$. Here, $$\text{domain}=\{u:U\to \mathbb R~\text{such that}~u\in C^2(U)\cap H^2(U),~u(-1/2)=u(1/2),~Du(-1/2)=Du(1/2)\}$$

Again, $A$ is linear, so in order for $A=-D^2$ to be closed it must be bounded. Therefore, there would need to exist $C>0$ such that $$\left\Vert D^2 u \right\Vert_{L^2(U)}\leq C\Vert u\Vert_{L^2(U)}$$ $\forall u\in \text{domain}$. But, for instance consider the sequence $(u_k)_{k\in \mathbb N_0}\in\text{domain}$ $$u_k(x):=x(1/2+x)(1/2-x)\exp(-k^2x^2)$$ You can check $\Vert u_k \Vert_{L^2(U)}\to 0$ as $k\to\infty$, however $\left\Vert D^2 u_k \right\Vert_{L^2(U)}\to\infty $ as $k\to\infty$ (which takes some work, but it is doable) .

Thus no such constant $C$ can exist.