Global Analysis — Proof of local elliptic regularity

# A Proof of local elliptic regularity

We fill in the details of the proof of local elliptic regularity. We fix $L$ a differential operator of order $k \geq 1$ on $\mathbb {R}^ n$, $U \subseteq \mathbb {R}^ n$ precompact and $L$ elliptic over $\bar{U}$.

Lemma

If $L$ has constant coefficients, i.e.

$L = \sum _{|\alpha | \leq k} a^\alpha \mathrm{D}_\alpha ,$

then there is an $A$ such that for all $u \in C_ c^\infty (U)$,

$\| u\| _{s + k} \leq A(\| u\| _ s + \| Lu\| _{s}).$

Proof
Observe that we have $\widehat{Lu}(\xi ) = p(\xi ) \hat{u}(\xi ).$ for some polynomial $p$ of degree at most $k$. By ellipticity, for some $R \gg 0$ and constant $A > 0$, we have $A |p(\xi )| \geq (1 + |\xi |^2)^{k/2}.$ So in the decomposition $\int |\hat{u}(\xi )|^2 (1 + |\xi |)^{s + k} \; \mathrm{d}\xi = \left(\int _{|\xi | \leq R} + \int _{|\xi | \geq R}\right) |\hat{u}(\xi )|^2 (1 + |\xi |)^{s + k} \; \mathrm{d}\xi ,$ we can bound the first term by $(1 + R^2)^ k \| u\| _ s$, and we can bound the second term by $\int _{|\xi |\geq R} |\widehat{Lu}(\xi )|^2 (1 + |\xi |^2)^ s\; \mathrm{d}\xi \leq \| Lu\| _ s.\qedhere$

Lemma

For any fixed $L$ and $x_0 \in U$, there is some neighbourhood $V \subseteq U$ of $x$ and $A > 0$ such that for all $u \in C_ c^\infty (V)$, we have

$\| u\| _{s + k} \leq A(\| u\| _ s + \| Lu\| _{s}).$

Proof
Let $L_0$ be the differential operator with constant coefficients that agree with $L$ at $x_0$. Then $L_0$ is also an elliptic operator, and the above applies. So for any $u$, we have $\| u\| _{s + k} \leq A_1(\| u\| _ s + \| L_0u\| _ s) \leq A'(\| u\| _ s + \| (L - L_0) u\| _ s + \| Lu\| _ s).$ So we have to control the term $\| (L - L_0)u\| _ s$. For a tiny $\delta \ll 1$, pick a neighbourhood $V$ of $x_0$ such that the coefficients of $L - L_0$ are bounded by $\delta$. Then $\| (L - L_0)u\| _ s \leq \delta A_2 \| u\| _{s + k} + A_3 \| u\| _{s + k- 1},$ where $A_2, A_3$ are fixed, independent of $u$ and $V$. By the next lemma, for any $\varepsilon > 0$, we can bound $A_1 A_3 \| u\| _{s + k - 1} \leq \varepsilon \| u\| _{s + k} + A_4(\varepsilon ) \| u\| _ s.$ We then deduce that $\| u\| _{s + k} \leq A_1 \| Lu\| _ s + (\delta A_1 A_2 + \varepsilon ) \| u\| _{s + k} + (A_1 + A_4(\varepsilon )) \| u\| _ s.$ Picking $\delta$ and $\varepsilon$ to be small enough, we are done.

Lemma

For any $r < s < t$ and $\varepsilon > 0$, there exists $C(\varepsilon )$ such that

$(1 + |\xi |^2)^ s \leq (1 + |\xi |^2)^ t \varepsilon + (1 + |\xi |^2)^ r C(\varepsilon )$

for all $\xi$. Hence

$\| u\| _ s \leq \varepsilon \| u\| _ t + C(\varepsilon ) \| u\| _ r.$

Proof
The claim is the same as $1 \leq (1 + |\xi |^2)^{t - s} \varepsilon + (1 + |\xi |^2)^{r - s} C(\varepsilon ).$ Observe that for any $y$, we always have $1 \leq y^{t - s} + (1/y)^{s - r}.$ Then take $y = (1 + |\xi |^2) \varepsilon ^{1/(t - s)}$.

Theorem

For any $L$, there exists $A$ such that

$\| u\| _{s + k} \leq A (\| u\| _ s + \| Lu\| _ s).$

Proof
Pick $W \supseteq \bar{U}$ such that $L$ is elliptic on $W$, and cover $W$ (and hence $\bar{U}$) with finitely $V_ i$ where we have a bound as above, say $\| u\| _{s + k} \leq A (\| u\| _ s + \| Lu\| _{s - k})$ for any $u$ supported in the $V_ i$'s. Now pick a partition of unity $\{ \mu _ i\}$ subordinate to $\{ V_ i\}$. Then We can bound the first two by a constant multiple of $\| u\| _ s$ and $\| Lu\| _ s$. To bound the last term, we use that $[L, \mu _ i]$ is a differential operator of order $k - 1$, and hence $\| [L, \mu _ i] u\| _ s \leq C \| u\| _{s + k - 1} \leq \varepsilon \| u\| _{s + k} + C(\varepsilon ) \| u\| _ s.\qedhere$