Global AnalysisProof 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

$\int _{|\xi |\geq R} |\widehat{Lu}(\xi )|^2 (1 + |\xi |^2)^s\; \mathrm{d}\xi \leq \| Lu\| _s.$

Proof

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.

Proof

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 $y = (1 + |\xi |^2) \varepsilon ^{1/(t - s)}$.

Proof

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

$\| u\| _{s + k} \leq \sum \| \mu _i u\| _{s + k} \leq \sum C(\| \mu _i u\| _s + \| L \mu _i u\| _s) \leq \sum C(\| \mu _i u\| _s + \| \mu _i Lu\| _s + \| [L, \mu _i] u\| _s).$

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

$\| [L, \mu _i] u\| _s \leq C \| u\| _{s + k - 1} \leq \varepsilon \| u\| _{s + k} + C(\varepsilon ) \| u\| _s.$

Proof