$\renewcommand{\epsilon}{\varepsilon}$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 $\mathbb{R}^n$.

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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.

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

1 Answer 1


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.

  • $\begingroup$ 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 $L^2$? $\endgroup$ Mar 1, 2021 at 19:40
  • $\begingroup$ But I don't see how you get uniform boundedness in $\epsilon$... $\endgroup$ Mar 1, 2021 at 19:41
  • $\begingroup$ @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. $\endgroup$
    – B K
    Mar 2, 2021 at 20:04

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