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 LL a differential operator of order k1k \geq 1 on Rn\mathbb {R}^n, URnU \subseteq \mathbb {R}^n precompact and LL elliptic over Uˉ\bar{U}.

Lemma

If LL has constant coefficients, i.e.

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

then there is an AA such that for all uCc(U)u \in C_c^\infty (U),

us+kA(us+Lus). \| u\| _{s + k} \leq A(\| u\| _s + \| Lu\| _{s}).

Proof
Observe that we have

Luundefined(ξ)=p(ξ)u^(ξ). \widehat{Lu}(\xi ) = p(\xi ) \hat{u}(\xi ).

for some polynomial pp of degree at most kk. By ellipticity, for some R0R \gg 0 and constant A>0A > 0, we have

Ap(ξ)(1+ξ2)k/2. A |p(\xi )| \geq (1 + |\xi |^2)^{k/2}.

So in the decomposition

u^(ξ)2(1+ξ)s+k  dξ=(ξR+ξR)u^(ξ)2(1+ξ)s+k  dξ, \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+R2)kus(1 + R^2)^k \| u\| _s, and we can bound the second term by

ξRLuundefined(ξ)2(1+ξ2)s  dξLus. \int _{|\xi |\geq R} |\widehat{Lu}(\xi )|^2 (1 + |\xi |^2)^s\; \mathrm{d}\xi \leq \| Lu\| _s.

Proof

Lemma

For any fixed LL and x0Ux_0 \in U, there is some neighbourhood VUV \subseteq U of xx and A>0A > 0 such that for all uCc(V)u \in C_c^\infty (V), we have

us+kA(us+Lus). \| u\| _{s + k} \leq A(\| u\| _s + \| Lu\| _{s}).

Proof
Let L0L_0 be the differential operator with constant coefficients that agree with LL at x0x_0. Then L0L_0 is also an elliptic operator, and the above applies. So for any uu, we have

us+kA1(us+L0us)A(us+(LL0)us+Lus). \| 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 (LL0)us\| (L - L_0)u\| _s. For a tiny δ1\delta \ll 1, pick a neighbourhood VV of x0x_0 such that the coefficients of LL0L - L_0 are bounded by δ\delta . Then

(LL0)usδA2us+k+A3us+k1, \| (L - L_0)u\| _s \leq \delta A_2 \| u\| _{s + k} + A_3 \| u\| _{s + k- 1},

where A2,A3A_2, A_3 are fixed, independent of uu and VV. By the next lemma, for any ε>0\varepsilon > 0, we can bound

A1A3us+k1εus+k+A4(ε)us. A_1 A_3 \| u\| _{s + k - 1} \leq \varepsilon \| u\| _{s + k} + A_4(\varepsilon ) \| u\| _s.

We then deduce that

us+kA1Lus+(δA1A2+ε)us+k+(A1+A4(ε))us. \| 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 δ\delta and ε\varepsilon to be small enough, we are done.

Proof

Lemma

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

(1+ξ2)s(1+ξ2)tε+(1+ξ2)rC(ε) (1 + |\xi |^2)^s \leq (1 + |\xi |^2)^t \varepsilon + (1 + |\xi |^2)^r C(\varepsilon )

for all ξ\xi . Hence

usεut+C(ε)ur. \| u\| _s \leq \varepsilon \| u\| _t + C(\varepsilon ) \| u\| _r.

Proof
The claim is the same as

1(1+ξ2)tsε+(1+ξ2)rsC(ε). 1 \leq (1 + |\xi |^2)^{t - s} \varepsilon + (1 + |\xi |^2)^{r - s} C(\varepsilon ).

Observe that for any yy, we always have

1yts+(1/y)sr. 1 \leq y^{t - s} + (1/y)^{s - r}.

Then take y=(1+ξ2)ε1/(ts)y = (1 + |\xi |^2) \varepsilon ^{1/(t - s)}.

Proof

Theorem

For any LL, there exists AA such that

us+kA(us+Lus). \| u\| _{s + k} \leq A (\| u\| _s + \| Lu\| _s).

Proof
Pick WUˉW \supseteq \bar{U} such that LL is elliptic on WW, and cover WW (and hence Uˉ\bar{U}) with finitely ViV_i where we have a bound as above, say

us+kA(us+Lusk) \| u\| _{s + k} \leq A (\| u\| _s + \| Lu\| _{s - k})

for any uu supported in the ViV_i's. Now pick a partition of unity {μi}\{ \mu _i\} subordinate to {Vi}\{ V_i\} . Then

us+kμius+kC(μius+Lμius)C(μius+μiLus+[L,μi]us). \| 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 us\| u\| _s and Lus\| Lu\| _s. To bound the last term, we use that [L,μi][L, \mu _i] is a differential operator of order k1k - 1, and hence

[L,μi]usCus+k1εus+k+C(ε)us. \| [L, \mu _i] u\| _s \leq C \| u\| _{s + k - 1} \leq \varepsilon \| u\| _{s + k} + C(\varepsilon ) \| u\| _s.

Proof