# If $\Delta$ is the Laplace operator, what are $| \Delta |$ and $|\Delta +1|$

Assuming $\Delta : H^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ be the Laplace operator, then:

What is the exact definition of $| \Delta |$? What is $|\Delta +1|$ also?

This answer to another question of mine suggests, that $| \Delta | = -\Delta$. Is that true? And on which definition is that based?

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Recall that by the Borel functional calculus, $|A| := \sqrt{A^\ast A}$ for $A$ a densely-defined normal operator. There are then two possible answers to your question, depending on the convention at play:

1. It is very common in global analysis to define the Laplacian on $L^2(\mathbb{R}^n)$ so that it is positive semi-definite, i.e., $\Delta := -\sum_{k=1}^n \partial_k^2$, implying in particular that $\Delta + 1$ is a positive definite operator. It then follows immediately that $|\Delta| = \Delta$ and $|\Delta + 1| = \Delta + 1$.

2. If you use the other convention, so that $\Delta := \sum_{k=1}^n \partial_k^2$ is a negative semi-definite operator, then $\Delta + I$ is just some self-adjoint operator, and so you have to take $|\Delta| = \sqrt{\Delta^2} = -\Delta$ and $|\Delta + 1| = \sqrt{(\Delta+1)^2}$.

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Thank you! But if I consider the second case, namely $\Delta := \sum_{k=1}^n \partial_k^2$ and therefore $| \Delta | = \sqrt{ \Delta^2}$: Isn't $\sqrt{ \Delta^2}$ the unique positive operator $H^2 \to L^2$ such that $\sqrt{\Delta^2}\sqrt{\Delta^2} u = \Delta^2 u$ and therefore $\sqrt{\Delta^2}=(-\Delta)$? – mjb Jan 31 '13 at 9:06
Indeed it is, since $-\Delta$ would be positive semi-definite. Thanks! I've incorporated this into my answer. – Branimir Ćaćić Jan 31 '13 at 9:16
Thanks! Now that I think about it, the problem with that seams to be that if $\sqrt{\Delta^2} : H^2 \to L^2$ then $\sqrt{\Delta^2} \sqrt{\Delta^2} u$ doesn't make much sense, does it? – mjb Jan 31 '13 at 9:28
@mjb: yes and no. Note that the Laplace operator $\triangle$ can be extended to a map from $H^{k} \to H^{k-2}$ (you will need to know how to define the negative index Sobolev spaces; in Evans it is briefly discussed in section 5.9) for all $k\in \mathbb{Z}$. Then you can treat $\triangle:V\to V$ where $V = \oplus_{k\in \mathbb{Z}} H^k$ is the graded vector space corresponding to all integer Sobolev classes, but then self-adjoint becomes a problem... – Willie Wong Jan 31 '13 at 9:35

What you are looking at is called functional calculus. For self-adjoint operators on Hilbert spaces, usually the Borel flavour is used. You should of course consult a textbook on functional analysis, but here let me give an overview.

Before we start, one parting point: you should not consider functional calculus for $\triangle: H^2 \to L^2$. The idea of a functional calculus stems from the fact that for operators $T:X\to X$, it makes sense to compose $T$ with itself, since its range is a subset of its domain. Therefore "polynomials" of the operator would also make sense: things like $T^3 - 2T^2 + T - 5$ can be interpreted as an operator $X\to X$. But in your case you take the codomain to be $L^2 \not\subseteq H^2$. Essentially you want the space of operators to form an algebra, with addition being pointwise addition, and multiplication being operator composition, before you can speak of a calculus.

That said, let $T:X\to X$ be some operator. As described above by considering compositions, we can define polynomial functions of $T$, which are still all operators from $X\to X$. Now, by the Stone-Weierstrass theorem any continuous function on a compact interval can be well approximated by polynomials. So we ask whether something like this is true also for operators.

And the answer is yes, under certain assumptions.

One particular case is that of the bounded self-adjoint operators. We start first from self-adjoint operators on a finite dimensional inner product space. By the spectral theorem, we can diagonalise $T = U^*DU$. And we have that $T^k = U^*D^k U$. This suggests that we can apply Stone-Weierstrass theorem entry-wise to the diagonal elements of $D$ and get a functional calculus (since the eigenvalues of $T$ are bounded!)

To be more precise, if $D = \mathrm{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n)$, we can define $f(D) = \mathrm{diag}(f(\lambda_1), f(\lambda_2),\ldots, f(\lambda_n))$ and define $f(T) = U^*f(D)U$.

Under suitable conditions (see the Wiki link for more detail) this can be done for self-adjoint operators on a Hilbert space. The method is the same: you "diagonalise" the operator by writing it in terms of its eigenvalues and eigenvectors (or more precisely, in terms of a spectral measure), act the function on the eigenvalues, and consider the new operator with the same eigenvectors but the new eigenvalues (this last step is often expressed in terms of an integral over the spectral measure).

The above is the general theory, but for functions of the Laplace operator there is a separate way of interpreting the functional calculus which has less dependence on the function spaces involved.

Let $\mathcal{S}$ denote the space of Schwartz functions on $\mathbb{R}^n$. We know that the Fourier transform is a map $\mathcal{S}\to\mathcal{S}$. One of its properties is that (possibly different constants depending on the normalisation of the transform) $$\widehat{\triangle f}(\xi) = - |\xi|^2 \hat{f}(\xi)$$ that it makes derivatives into coordinate multiplications. By using the Fourier transform to mimic the role of the spectral measure, we can define, for any reasonable function $h:\mathbb{R}\to\mathbb{R}$, the operator $h(\triangle)$ by using the Fourier transform: $$\widehat{h(\triangle)f}(\xi) = h(-|\xi|^2) \hat{f}(\xi)$$ One of the things to worry about is that we need to make sure $h(-|\xi|^2) \hat{f}(\xi)$ can have an inverse Fourier transform. For the most part this can be guaranteed by choosing $f$ to lie within a suitable function space and interpreting $h(\triangle)f$ in a suitable, possibly different, function space.

In your case, for functions $f\in H^2$, we have that $(1 + |\xi|^2)\hat{f} \in L^2$ by Plancherel. And so $|\triangle|f$ (or $|\xi|^2 \hat{f} = - \widehat{\triangle f}$) is also in $L^2$. Similarly we also have that $|\triangle + 1|$ can be expressed as the Fourier multiplier $$\hat{f} \mapsto \left|1 - |\xi|^2\right| \hat{f} \in L^2$$ so $|\triangle + 1|$ can be interpreted as an operator $H^2 \to L^2$.

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Wow, thank you! One question to the first part of your post: Why is $L^2(\mathbb{R}^n)$ not a subset of the sobolev space $H^2(\mathbb{R}^n)$? The latter is (at least in Evans, L.C.: Partial Differential Equations) defined as the space of all $L^2$-functions with first and second weak derivatives in $L^2$. – mjb Jan 31 '13 at 9:26
@mjb: write it out in symbols. $H^2 := \{f\in L^2 | \ldots\}$ so by definition $H^2 \subsetneq L^2$ and not the other way around. In words, an $H^2$ function is at least an $L^2$ function, but better. – Willie Wong Jan 31 '13 at 9:27
Oh sorry, that was a silly mistake :) Thank you. – mjb Jan 31 '13 at 9:30

It's abuse of notation, I think. The Laplace operator is negative semidefinite, i.e. is self-adjoint with $\left<\Delta f,f\right> \leq 0$ and so $|\Delta| = -\Delta$ is a positive operator.

I would call $|\Delta+1|$ nonsense, since $\Delta+1$ is indefinite: $\left<\Delta f + f, f\right>$ is positive for constant functions and negative for eigenvectors of the Laplacian with sufficiently high eigenvalue.

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