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

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

□