# Friedrichs mollifiers and Sobolev spaces

$$\renewcommand{\epsilon}{\varepsilon}$$The following is from John Roe's book Elliptic operators, topology and asymptotic methods. $$S$$ is a vector bundle on a compact manifold $$M$$, but I think for my question it is sufficient to assume that $$S = M \times \mathbb{C}$$. I will probably be happy with an answer that only deals with the $$L^2$$ and Sobolev spaces of periodic functions on $$\mathbb{R}^n$$.

Later in the book the author claims that a Friedrichs' mollifier $$(F_\epsilon)$$, as well as $$([B, F_\epsilon])$$ for any first order differential operators $$B$$, are bounded families of operators on any Sobolev space $$W^k$$ (i.e. $$W^{k, 2}$$).

How could you prove this?

Everything that I could find online about this seems to talk about some "Friedrichs' Lemma", which I know as the statement that for a first-order smooth differential operator on an open subset of $$\mathbb{R}^n$$ and $$v \in L^2(\mathbb{R}^n)$$ $$[P, S_\varepsilon] v \to 0 \text{ in } L^2 \quad\text{for } \varepsilon \to 0$$ where $$S_\varepsilon$$ is a family of standard mollifiers. This certainly seems like it might be related to the question, but I don't really know how to use it.

• I've cross-posted this question from math.stackexchange (and deleted the question there). Feb 13, 2021 at 0:17

The claim that the $$F_\epsilon$$ map any Sobolev space to any other has nothing to do with the properties (i)-(iii), it is just a consequence of $$F_\epsilon$$ being a smoothing operator. In fact, in the literature the property of mapping any Sobolev space to any other is frequently used to define what it means to be a smoothing operator.

Recall that $$W^r=(-\Delta +1)^{-r/2}L^2(M)$$, where $$-\Delta$$ is the positive Laplacian on $$M$$ with respect to some Riemannian metric (the choice of which does not matter because $$M$$ is compact). Again because $$M$$ is compact, we can assume for our purposes that the volume density used to define $$L^2(M)$$ agrees with the Riemannian volume density of the metric we have chosen to define $$\Delta$$. For a smoothing operator $$A$$ with smooth kernel $$k$$ on $$M\times M$$, a smooth function $$f$$ on $$M$$, and $$r,r'\in \mathbb R$$ one has at $$x\in M$$ $$((-\Delta +1)^{r'/2}A(-\Delta +1)^{r/2}f)(x)=(-\Delta_x +1)^{r'/2}\int_{M}k(x,y)((-\Delta_y +1)^{r/2}f)(y)dvol(y)$$ $$=\int_{M}((-\Delta_x +1)^{r'/2}(-\Delta_y +1)^{r/2}k)(x,y)f(y)dvol(y)$$ because $$\Delta_y$$ is symmetric in $$L^2(M,dvol)$$ and we can interchange $$\Delta_x$$ with integration because $$M$$ is compact.

Thus, $$(-\Delta +1)^{r'/2}A(-\Delta +1)^{r/2}$$ is an integral operator with smooth kernel $$k_{r,r'}$$ on $$M\times M$$, given by $$k_{r,r'}(x,y):=((-\Delta_x +1)^{r'/2}(-\Delta_y +1)^{r/2}k)(x,y).$$ In particular, it defines a bounded operator on $$L^2(M)$$, which shows that $$A$$ is bounded from $$W^r$$ to $$W^{r'}$$.

A similar argument applies to $$B'\circ A\circ B$$ for any two differential operators $$B$$ and $$B'$$, with the only difference that if $$B$$ is not symmetric in $$L^2$$, then the operator that is applied to $$k$$ is the formal adjoint of $$B$$.

For the uniform boundedness of $$F_\epsilon$$ as an operator family on Sobolev spaces, one can use the following characterization of $$W^r$$ for integer $$r\geq 0$$: $$W^r=\{u\in \mathscr D'(M): Du\in L^2(M)\;\text{for every differential operator of order }\leq r \}.$$ Here $$\mathscr D'(M)$$ is the space of distributions on $$M$$ and the norm on $$W^r$$ can be defined (up to equivalence) using a partition of unity and the local norms $$\Vert u \Vert_r^2:=\sum_{|I|\leq r}\Vert \partial_I u \Vert_{L^2}^2,$$ where $$u\in C^\infty_c(U)$$, $$U\subset M$$ being a chart domain, and $$\partial_I:=\partial_{x_1}^{i_1}\cdots \partial_{x_n}^{i_n}$$, $$I=(i_1,\ldots,i_n)$$, $$n=\mathrm{dim}\,M$$, are the standard partial derivatives in $$U$$.

Now, given any $$r'\in \mathbb R$$ with $$r'\geq r$$ one has on $$U$$ $$\partial_I\circ F_\epsilon \circ(-\Delta +1)^{-r'/2}= [\partial_I, F_\epsilon] \circ(-\Delta +1)^{-r'/2}+F_\epsilon \circ\partial_I\circ (-\Delta +1)^{-r'/2},$$ where $$F_\epsilon$$ and $$[\partial_I, F_\epsilon]$$ are uniformly bounded on $$L^2$$ by (i) and (ii), and $$(-\Delta +1)^{-r'/2}$$ as well as $$\partial_I\circ (-\Delta +1)^{-r'/2}$$ are bounded on $$L^2$$ because they are pseudodifferential operators of orders $$\leq 0$$. Patching together these local estimates using a partition of unity, one obtains that $$F_\epsilon: W^{-r'}\to W^r$$ is uniformly bounded for every integer $$r\geq 0$$ and any real $$r'\geq r$$. For non-integer $$r\geq 0$$, one can use Sobolev interpolation to get the same result.

• So you're basically using that the operator $(1 - \Delta)^{r/2}$ (defined by some functional calculus I suppose) is an isometry between $W^r$ and $L^2$? Mar 1, 2021 at 19:40
• But I don't see how you get uniform boundedness in $\epsilon$... Mar 1, 2021 at 19:41
• @CarlosEsparza: $(1-\Delta)^{r/2}$ is defined using the functional calculus for unbounded symmetric operators. Here $-\Delta$ is non-negative and has a canonical self-adjoint extension called Friedrichs extension. For the uniform boundedness on the Sobolev spaces I'll add a pagragraph to my answer.
– B K
Mar 2, 2021 at 20:04