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Corrected the statement as to what the domain of $\sqrt{-L}$ is.
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Jeff Schenker
Jeff Schenker
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In short, "yes, probably, but you should be careful about boundary conditions."

The long version:

First a cautionary result. Let the Hilbert space be $L^2(0,1)$ and let $Lf =f''$ on the domain of functions $f\in L^2(0,1)$ with two derivatives in $L^2$ and such that $f(0)=-f'(1)$ and $f(1)=f'(0)$. This is a self-adjoint operator and by integration by parts

$$(f,(-L)f)= 2 \mathrm{Re}(\overline{f(1)}f(0))+ \int_0^1 |f'(x)|^2 dx.$$

However, this counter example is crazy, since it is not possible to interpret $L$ as $- D^\dagger D$ for an operator $D$ which takes one derivative. Also the boundary conditions I used would be very unlikely to arise in practice. Nevertheless it shows that some caution is needed.

So, let me interpret your question as follows:

"Let $L$ be the (unique!) operator defined on the domain $\mathcal{D}(L)\subset > \mathcal{D}(\nabla)$ of functions $f$ such that $A(x)\nabla f \in > \mathcal{D}(\nabla^\dagger)$, with $Lf=-\nabla^\dagger A(x) \nabla f$. Does the identity $$((-L)f,f)=\int \nabla f \cdot A(x)\nabla f d\pi$$ hold for $f\in \mathcal{D}(L)$?"

To begin with it may not be obvious that such an operator exists, or that it is self-adjoint, however this is the case and further more your integration by parts identity always holds. In fact, under the standard construction -- originally due to Friedrichs I think -- the answer is trivially yes since integration by parts is essentially the definition of $L$!

The Friedrichs construction is based on a theorem from functional analysis that says that any closed, positive quadratic form $q$ on a Hilbert space is the quadratic form of a positive self-adjoint operator. (See, for example, Thm. VIII.15 in Reed and Simon Vol. I.) In the present case we would define the quadratic form

$$q(g,f)= \int \overline{\nabla g(x)} \cdot A(x) \nabla f(x) d\pi(x) $$

which is easily shown to be closed and positive so long as $\nabla$ is a closed operator and $A(x)$ is symmetric positive definite as you assume. The proof of the theorem goes by showing that the domain $\mathcal{D}$ of functions $f$ such that $|q(g,f)| \le C \|g\| $ is dense so that it makes sense (by the Riesz thm. on linear functionals) to define an operator $L$ on this domain by the identity

$$q(g,f)= (g,(-L)f).$$

Note that the identity you want is a special case of this defintion!

It is easy to see now that the domain of $L$ consists of all functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$ and that thisthe quadratic form domain agrees with the domain of $\sqrt{-L}$ so that we have

$$\|\sqrt{-L}f\|^2 =q(f,f).$$

Note that, none of what was done above relied on the derivatives being implemented as anti-self-adjoint operators as you asked for. Returning to the one-dimensional case with Hilbert space $L^2(0,1)$ and $A=1$, first let $\nabla = d/dx$ on the domain of functions in $L^2(0,1)$ with one derivative in $L^2$. The resulting operator $L_N$ is the Neumann second derivative defined on the domain $\mathcal{D}(L_N)$ of twice differentiable functions with derivatives that vanish at $0$ and $1$ and the identity holds, however $\nabla$ is not anti-self-adjoint.

In short, "yes, probably, but you should be careful about boundary conditions."

The long version:

First a cautionary result. Let the Hilbert space be $L^2(0,1)$ and let $Lf =f''$ on the domain of functions $f\in L^2(0,1)$ with two derivatives in $L^2$ and such that $f(0)=-f'(1)$ and $f(1)=f'(0)$. This is a self-adjoint operator and by integration by parts

$$(f,(-L)f)= 2 \mathrm{Re}(\overline{f(1)}f(0))+ \int_0^1 |f'(x)|^2 dx.$$

However, this counter example is crazy, since it is not possible to interpret $L$ as $- D^\dagger D$ for an operator $D$ which takes one derivative. Also the boundary conditions I used would be very unlikely to arise in practice. Nevertheless it shows that some caution is needed.

So, let me interpret your question as follows:

"Let $L$ be the (unique!) operator defined on the domain $\mathcal{D}(L)\subset > \mathcal{D}(\nabla)$ of functions $f$ such that $A(x)\nabla f \in > \mathcal{D}(\nabla^\dagger)$, with $Lf=-\nabla^\dagger A(x) \nabla f$. Does the identity $$((-L)f,f)=\int \nabla f \cdot A(x)\nabla f d\pi$$ hold for $f\in \mathcal{D}(L)$?"

To begin with it may not be obvious that such an operator exists, or that it is self-adjoint, however this is the case and further more your integration by parts identity always holds. In fact, under the standard construction -- originally due to Friedrichs I think -- the answer is trivially yes since integration by parts is essentially the definition of $L$!

The Friedrichs construction is based on a theorem from functional analysis that says that any closed, positive quadratic form $q$ on a Hilbert space is the quadratic form of a positive self-adjoint operator. (See, for example, Thm. VIII.15 in Reed and Simon Vol. I.) In the present case we would define the quadratic form

$$q(g,f)= \int \overline{\nabla g(x)} \cdot A(x) \nabla f(x) d\pi(x) $$

which is easily shown to be closed and positive so long as $\nabla$ is a closed operator and $A(x)$ is symmetric positive definite as you assume. The proof of the theorem goes by showing that the domain $\mathcal{D}$ of functions $f$ such that $|q(g,f)| \le C \|g\| $ is dense so that it makes sense (by the Riesz thm. on linear functionals) to define an operator $L$ on this domain by the identity

$$q(g,f)= (g,(-L)f).$$

Note that the identity you want is a special case of this defintion!

It is easy to see now that the domain of $L$ consists of all functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$ and that this domain agrees with the domain of $\sqrt{-L}$ so that we have

$$\|\sqrt{-L}f\|^2 =q(f,f).$$

Note that, none of what was done above relied on the derivatives being implemented as anti-self-adjoint operators as you asked for. Returning to the one-dimensional case with Hilbert space $L^2(0,1)$ and $A=1$, first let $\nabla = d/dx$ on the domain of functions in $L^2(0,1)$ with one derivative in $L^2$. The resulting operator $L_N$ is the Neumann second derivative defined on the domain $\mathcal{D}(L_N)$ of twice differentiable functions with derivatives that vanish at $0$ and $1$ and the identity holds, however $\nabla$ is not anti-self-adjoint.

In short, "yes, probably, but you should be careful about boundary conditions."

The long version:

First a cautionary result. Let the Hilbert space be $L^2(0,1)$ and let $Lf =f''$ on the domain of functions $f\in L^2(0,1)$ with two derivatives in $L^2$ and such that $f(0)=-f'(1)$ and $f(1)=f'(0)$. This is a self-adjoint operator and by integration by parts

$$(f,(-L)f)= 2 \mathrm{Re}(\overline{f(1)}f(0))+ \int_0^1 |f'(x)|^2 dx.$$

However, this counter example is crazy, since it is not possible to interpret $L$ as $- D^\dagger D$ for an operator $D$ which takes one derivative. Also the boundary conditions I used would be very unlikely to arise in practice. Nevertheless it shows that some caution is needed.

So, let me interpret your question as follows:

"Let $L$ be the (unique!) operator defined on the domain $\mathcal{D}(L)\subset > \mathcal{D}(\nabla)$ of functions $f$ such that $A(x)\nabla f \in > \mathcal{D}(\nabla^\dagger)$, with $Lf=-\nabla^\dagger A(x) \nabla f$. Does the identity $$((-L)f,f)=\int \nabla f \cdot A(x)\nabla f d\pi$$ hold for $f\in \mathcal{D}(L)$?"

To begin with it may not be obvious that such an operator exists, or that it is self-adjoint, however this is the case and further more your integration by parts identity always holds. In fact, under the standard construction -- originally due to Friedrichs I think -- the answer is trivially yes since integration by parts is essentially the definition of $L$!

The Friedrichs construction is based on a theorem from functional analysis that says that any closed, positive quadratic form $q$ on a Hilbert space is the quadratic form of a positive self-adjoint operator. (See, for example, Thm. VIII.15 in Reed and Simon Vol. I.) In the present case we would define the quadratic form

$$q(g,f)= \int \overline{\nabla g(x)} \cdot A(x) \nabla f(x) d\pi(x) $$

which is easily shown to be closed and positive so long as $\nabla$ is a closed operator and $A(x)$ is symmetric positive definite as you assume. The proof of the theorem goes by showing that the domain $\mathcal{D}$ of functions $f$ such that $|q(g,f)| \le C \|g\| $ is dense so that it makes sense (by the Riesz thm. on linear functionals) to define an operator $L$ on this domain by the identity

$$q(g,f)= (g,(-L)f).$$

Note that the identity you want is a special case of this defintion!

It is easy to see now that the domain of $L$ consists of all functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$ and that the quadratic form domain agrees with the domain of $\sqrt{-L}$ so that we have

$$\|\sqrt{-L}f\|^2 =q(f,f).$$

Note that, none of what was done above relied on the derivatives being implemented as anti-self-adjoint operators as you asked for. Returning to the one-dimensional case with Hilbert space $L^2(0,1)$ and $A=1$, first let $\nabla = d/dx$ on the domain of functions in $L^2(0,1)$ with one derivative in $L^2$. The resulting operator $L_N$ is the Neumann second derivative defined on the domain $\mathcal{D}(L_N)$ of twice differentiable functions with derivatives that vanish at $0$ and $1$ and the identity holds, however $\nabla$ is not anti-self-adjoint.

Source Link
Jeff Schenker
Jeff Schenker
  • 1.5k
  • 10
  • 16

In short, "yes, probably, but you should be careful about boundary conditions."

The long version:

First a cautionary result. Let the Hilbert space be $L^2(0,1)$ and let $Lf =f''$ on the domain of functions $f\in L^2(0,1)$ with two derivatives in $L^2$ and such that $f(0)=-f'(1)$ and $f(1)=f'(0)$. This is a self-adjoint operator and by integration by parts

$$(f,(-L)f)= 2 \mathrm{Re}(\overline{f(1)}f(0))+ \int_0^1 |f'(x)|^2 dx.$$

However, this counter example is crazy, since it is not possible to interpret $L$ as $- D^\dagger D$ for an operator $D$ which takes one derivative. Also the boundary conditions I used would be very unlikely to arise in practice. Nevertheless it shows that some caution is needed.

So, let me interpret your question as follows:

"Let $L$ be the (unique!) operator defined on the domain $\mathcal{D}(L)\subset > \mathcal{D}(\nabla)$ of functions $f$ such that $A(x)\nabla f \in > \mathcal{D}(\nabla^\dagger)$, with $Lf=-\nabla^\dagger A(x) \nabla f$. Does the identity $$((-L)f,f)=\int \nabla f \cdot A(x)\nabla f d\pi$$ hold for $f\in \mathcal{D}(L)$?"

To begin with it may not be obvious that such an operator exists, or that it is self-adjoint, however this is the case and further more your integration by parts identity always holds. In fact, under the standard construction -- originally due to Friedrichs I think -- the answer is trivially yes since integration by parts is essentially the definition of $L$!

The Friedrichs construction is based on a theorem from functional analysis that says that any closed, positive quadratic form $q$ on a Hilbert space is the quadratic form of a positive self-adjoint operator. (See, for example, Thm. VIII.15 in Reed and Simon Vol. I.) In the present case we would define the quadratic form

$$q(g,f)= \int \overline{\nabla g(x)} \cdot A(x) \nabla f(x) d\pi(x) $$

which is easily shown to be closed and positive so long as $\nabla$ is a closed operator and $A(x)$ is symmetric positive definite as you assume. The proof of the theorem goes by showing that the domain $\mathcal{D}$ of functions $f$ such that $|q(g,f)| \le C \|g\| $ is dense so that it makes sense (by the Riesz thm. on linear functionals) to define an operator $L$ on this domain by the identity

$$q(g,f)= (g,(-L)f).$$

Note that the identity you want is a special case of this defintion!

It is easy to see now that the domain of $L$ consists of all functions $f$ such that $A(x)\nabla f \in \mathcal{D}(\nabla^\dagger)$ and that this domain agrees with the domain of $\sqrt{-L}$ so that we have

$$\|\sqrt{-L}f\|^2 =q(f,f).$$

Note that, none of what was done above relied on the derivatives being implemented as anti-self-adjoint operators as you asked for. Returning to the one-dimensional case with Hilbert space $L^2(0,1)$ and $A=1$, first let $\nabla = d/dx$ on the domain of functions in $L^2(0,1)$ with one derivative in $L^2$. The resulting operator $L_N$ is the Neumann second derivative defined on the domain $\mathcal{D}(L_N)$ of twice differentiable functions with derivatives that vanish at $0$ and $1$ and the identity holds, however $\nabla$ is not anti-self-adjoint.