Test functions, distributions, and Sobolev’s lemma

Jordan Bell
May 22, 2014

1 Introduction

If X is a topological vector space, we denote by X* the set of continuous linear functionals on X. With the weak-* topology, X* is a locally convex space, whether or not X is a locally convex space. (But in this note, we only talk about locally convex spaces.)

The purpose of this note is to collect the material given in Walter Rudin, Functional Analysis, second ed., chapters 6 and 7, involved in stating and proving Sobolev’s lemma.

2 Test functions

Suppose that Ω is an open subset of n. We denote by 𝒟(Ω) the set of all ϕC(Ω) such that suppϕ is a compact subset of Ω. Elements of 𝒟(Ω) are called test functions. For N=0,1, and ϕ𝒟(Ω), write

ϕN=sup{|(Dαϕ)(x)|:xΩ,|α|N},

where

Dα=D1α1Dnαn,|α|=α1++αn.

For each compact subset K of Ω, we define

𝒟K={ϕ𝒟(Ω):suppϕK},

and define τK to be the locally convex topology on 𝒟K determined by the family of seminorms {N:N0}. One proves that 𝒟K with the topology τK is a Fréchet space. As sets,

𝒟(Ω)=K𝒟K.

Define β to be the collection of all convex balanced subsets W of 𝒟(Ω) such that for every compact subset K of Ω we have W𝒟KτK; to say that W is balanced means that if c is a complex number with |c|1 then cWW. One proves that {ϕ+W:ϕ𝒟(Ω),Wβ} is a basis for a topology τ on 𝒟(Ω), that β is a local basis at 0 for this topology, and that with the topology τ, 𝒟(Ω) is a locally convex space.11 1 Walter Rudin, Functional Analysis, second ed., p. 152, Theorem 6.4; cf. Helmut H. Schaefer, Topological Vector Spaces, p. 57. For each compact subset K of Ω, one proves that the topology τK is equal to the subspace topology on 𝒟K inherited from 𝒟(Ω).22 2 Walter Rudin, Functional Analysis, second ed., p. 153, Theorem 6.5.

We write 𝒟(Ω)=(𝒟(Ω))*, and elements of 𝒟(Ω) are called distributions. With the weak-* topology, 𝒟(Ω) is a locally convex space.

It is a fact that a linear functional Λ on 𝒟(Ω) is continuous if and only if for every compact subset K of Ω there is a nonnegative integer N and a constant C such that |Λϕ|CϕN for all ϕ𝒟K.33 3 Walter Rudin, Functional Analysis, second ed., p. 156, Theorem 6.8.

For Λ𝒟(Ω) and α a multi-index, we define

(DαΛ)(ϕ)=(-1)|α|Λ(Dαϕ),ϕ𝒟(Ω).

Let K be a compact subset of Ω. As Λ is continuous, there is a nonnegative integer N and a constant C such that |Λϕ|CϕN for all ϕ𝒟K. Then

|(DαΛ)(ϕ)|=|Λ(Dαϕ)|CDαϕNCϕN+|α|,

which shows that DαΛ𝒟(Ω).

The Leibniz formula is the statement that for all f,gC(n),

Dα(fg)=βα(αβ)(Dα-βf)(Dβg),

where (αβ) are multinomial coefficients.

For Λ𝒟(Ω) and fC(Ω), we define

(fΛ)(ϕ)=Λ(fϕ),ϕ𝒟(Ω);

this makes sense because fϕ𝒟(Ω) when ϕ𝒟(Ω). It is apparent that fΛ is linear, and in the following lemma we prove that fΛ is continuous.44 4 Walter Rudin, Functional Analysis, second ed., p. 159, §6.15.

Lemma 1.

If ΛD(Ω) and fC(Ω), then fΛD(Ω).

Proof.

Suppose that K is a compact subset of Ω. Because Λ is continuous, there is some nonnegative integer N and some constant C such that

|Λϕ|CϕN,ϕ𝒟K.

For |α|N, by the Leibniz formula, for all ϕ𝒟K,

Dα(fϕ)=βα(αβ)(Dα-βf)(Dβϕ).

Because fC(Ω), there is some Cα such that |(Dα-βf)(x)|Cα for βα and for xK. Using ϕ(x)=0 for xK, the above statement of the Leibniz formula, and the inequality just obtained, it follows that there is some Cα such that |(Dα(fϕ))(x)|CαϕN for all xΩ. This gives

fϕN=sup|α|NsupxΩ|(Dα(fϕ))(x)|sup|α|NCαϕN=CϕN;

the last equality is how we define C, which is a maximum of finitely many Cα and so finite. Then,

|(fΛ)(ϕ)|=|Λ(fϕ)|CfϕNCCϕN,ϕ𝒟K.

This bound shows that fΛ is continuous. ∎

The above lemma shows that fΛ𝒟(Ω) when fC(Ω) and Λ𝒟(Ω). Therefore Dα(fΛ)𝒟(Ω), and the following lemma, proved in Rudin, states that the Leibniz formula can be used with fΛ.55 5 Walter Rudin, Functional Analysis, second ed., p. 160, §6.15.

Lemma 2.

If fC(Ω) and ΛD(Ω), then

Dα(fΛ)=βα(αβ)(Dα-βf)(DβΛ).

If f:Ω is locally integrable, define

Λϕ=Ωϕ(x)f(x)𝑑x,ϕ𝒟(Ω).

For ϕ𝒟K,

|Λϕ|ϕ0K|f|𝑑x,

from which it follows that Λ is continuous. If μ is a complex Borel measure on n or a positive Borel measure on n that assigns finite measure to compact sets, define

Λϕ=Ωϕ𝑑μ,ϕ𝒟(Ω).

For ϕ𝒟K,

|Λϕ|ϕ0|μ|(K),

from which it follows that Λ is continuous. Thus, we can encode certain functions and measures as distributions. I will dare to say that we can encode most functions and measures that we care about as distributions.

If Λ1,Λ2𝒟(Ω) and ω is an open subset of Ω, we say that Λ1=Λ2 in ω if Λ1ϕ=Λ2ϕ for all ϕ𝒟(ω).

Let Λ𝒟(Ω) and let ω be an open subset of Ω. We say that Λ vanishes on ω if Λϕ=0 for all ϕ𝒟(ω). Taking W to be the union of all open subsets ω of Ω on which Λ vanishes, we define the support of Λ to be the set ΩW.

3 The Fourier transform

Let C0(n) be the set of those continuous functions f:n such that for every ϵ>0, there is some compact set K such that |f(x)|<ϵ for xK. With the supremum norm , C0(n) is a Banach space.

Let mn be normalized Lebesgue measure on Rn:

dmn(x)=(2π)-n/2dx.

Using mn, we define

fLp=(n|f|p𝑑mn)1/p,1p<

and

(f*g)(x)=nf(x-y)g(y)𝑑mn(y).

For tn, define et:n by

et(x)=exp(itx),xn.

The Fourier transform of fL1(n) is the function f^:n defined by

(f)(t)=f^(t)=nfe-t𝑑mn,tn.

Using the dominated convergence theorem, one shows that f^ is continuous.

For fC(n) and N a nonnegative integer, write

pN(f)=sup|α|Nsupxn(1+|x|2)N|(Dαf)(x)|,

and let 𝒮n be the set of those fC(n) such that for every nonnegative integer N, pN(f)<. 𝒮n is a vector space, and with the locally convex topology determined by the family of seminorms {pN:N0} it is a Fréchet space.66 6 Walter Rudin, Functional Analysis, second ed., p. 184, Theorem 7.4. Further, one proves that :𝒮n𝒮n is a continuous linear map.77 7 Walter Rudin, Functional Analysis, second ed., p. 184, Theorem 7.4.

The Riemann-Lebesgue lemma is the statement that if fL1(n), then f^C0(n).88 8 Walter Rudin, Functional Analysis, second ed., p. 185, Theorem 7.5.

The inversion theorem99 9 Walter Rudin, Functional Analysis, second ed., p. 186, Theorem 7.7. is the statement that if g𝒮n then

g(x)=ng^ex𝑑mn,xn,

and that if fL1(n) and f^L1(n), and we define f0C0(n) by

f0(x)=nf^ex𝑑mn,xn,

then f(x)=f0(x) for almost all xn. For g𝒮n, as g^𝒮n, the function f(t)=g^(-t) belongs to 𝒮n. The inversion theorem tells us that for all xn,

g(x)=ng^(t)ex(t)𝑑mn(t)=ng^(-t)ex(-t)𝑑mn(t)=nf(t)e-x(t)𝑑mn(t),

and hence that g=f^. This shows that :𝒮n𝒮n is onto. Using the inversion theorem, one checks that

nfg¯𝑑mn=nf^g^¯𝑑mn,f,g𝒮n,

and so fL2=fL2 for f𝒮n. It is a fact that 𝒮n is a dense subset of the Hilbert space L2(n), and it follows that there is a unique bounded linear operator L2(n)L2(n), that is equal to on 𝒮n, and that is unitary. We denote this :L2(n)L2(n).

It is a fact that 𝒟(n) is a dense subset of 𝒮n and that the identity map i:𝒟(n)𝒮n is continuous.1010 10 Walter Rudin, Functional Analysis, second ed., p. 189, Theorem 7.10. If L1,L2(𝒮n)* are distinct, then there is some f𝒮n such that L1fL2f, and as 𝒟(n) is dense in 𝒮n, there is a sequence fj𝒟(n) with fjf in 𝒮n. As

(L1i)(fj)-(L2i)(fj)=L1fj-L2fjL1fj-L2fj0,

there is some fj with (L1i)(fj)(L2i)(fj), and hence L1iL2i. This shows that LLi is a one-to-one linear map (𝒮n)*𝒟(n). Elements of 𝒟(n) of the form Li for L(𝒮n)* are called tempered distributions, and we denote the set of tempered distributions by 𝒮n. It is a fact that every distribution with compact support is tempered.1111 11 Walter Rudin, Functional Analysis, second ed., p. 190, Example 7.12 (a).

4 Sobolev’s lemma

Suppose that Ω is an open subset of n. We say that a measurable function f:Ω is locally L2 if K|f|2𝑑mn< for every compact subset K of Ω. We say that Λ𝒟(Ω) is locally L2 if there is a function g that is locally L2 in Ω such that Λϕ=Ωϕg𝑑mn for every ϕ𝒟(Ω).

The following proof of Sobolev’s lemma follows Rudin.1212 12 Walter Rudin, Functional Analysis, second ed., p. 202, Theorem 7.25.

Theorem 3 (Sobolev’s lemma).

Suppose that n,p,r are integers, n>0, p0, and

r>p+n2.

Suppose that Ω is an open subset of Rn, that f:ΩC is locally L2, and that the distribution derivatives Djkf are locally L2 for 1jn, 1kr. Then there is some f0Cp(Ω) such that f0(x)=f(x) for almost all xΩ.

Proof.

To say that the distribution derivative Djkf is locally L2 means that there is some gj,k:Ω that is locally L2 such that

DjkΛf=Λgj,k.

Suppose that ω is an open subset of Ω whose closure K is a compact subset of Ω. There is some ψ𝒟(Ω) with ψ(x)=1 for xK, and we define F:n by

F(x)={ψ(x)f(x)xΩ,0xΩ;

in particular, for xK we have F(x)=f(x), and for xsuppψ we have F(x)=0. Because suppψΩ is compact and f is locally L2,

FL2=(suppψ|ψf|2𝑑mn)1/2ψ0(suppψ|f|2𝑑mn)1/2<,

and using the Cauchy-Schwarz inequality, FL1FL2mn(suppψ)1/2<, so

FL2(n)L1(n).

Then,

n|F^|2𝑑mn<. (1)

Because ΛF=ψΛf in Ω, the Leibniz formula tells us that in Ω,

DjrΛF=Djr(ψΛf)=s=0r(rs)(Djr-sψ)(DjsΛf)=s=0r(rs)(Djr-sψ)(Λgj,s),

hence, defining Hj:n by

Hj(x)={s=0r(rs)(Djr-sψ)(x)gj,s(x)xΩ,0xΩ,

we have DjrΛF=ΛHj in Ω. It is apparent that HjL2(n)L1(n).

Let ϕ𝒟(n). There are ϕ1,ϕ2𝒟(n) with ϕ=ϕ1+ϕ2 and suppϕ1Ω, suppϕ2nsuppψ.1313 13 ϕ1 and ϕ2 are constructed using a partition of unity. See Walter Rudin, Functional Analysis, second ed., p. 162, Theorem 6.20. We have just established that (DjrΛF)ϕ1=ΛHjϕ1. For ϕ2, it is apparent that

(DjrΛF)ϕ2=ΛF(Djrϕ2)=n(Djrϕ2)(x)F(x)𝑑mn(x)=0

and

ΛHjϕ2=nϕ2(x)Hj(x)𝑑mn(x)=0.

Hence (DjrΛF)(ϕ)=ΛHjϕ. It is apparent that ΛHj has compact support, so DjrΛF=ΛHj are tempered distributions. Let ξ𝒮n, and take ϕ𝒮n with ξ=ϕ^. Then,

(DjrΛF)ϕ = ΛFDjrϕ
= n(Djrϕ)(x)F(x)𝑑mn(x)
= n(Djrϕ)(y)F^(y)𝑑mn(y)
= n(iyj)rξ(y)F^(y)𝑑mn(y),

and

ΛHjϕ=nϕ(x)Hj(x)𝑑mn(x)=nξ(y)Hj^(y)𝑑mn(y).

It follows that (iyj)rF^(y)=Hj^(y) for all yn. But Hj^L2(n), so

nyi2r|F^(y)|2𝑑mn(y)<,1in. (2)

Using (1), (2), and the inequality

(1+|y|)2r<(2n+2)r(1+y12r++yn2r),yn,

we get

J=n(1+|y|)2r|F^(y)|2𝑑mn(y)<.

Let σn-1 be surface measure on Sn-1, with σn-1(Sn-1)=2πn/2Γ(n/2). Using the Cauchy-Schwarz inequality and the change of variable y=tu, uSn-1, t0,

(n(1+|y|)p|F^(y)|𝑑mn(y))2 = (n(1+|y|)r|F^(y)|(1+|y|)p-r𝑑mn(y))2
Jn(1+|y|)2p-2r𝑑mn(y)
= J(2π)-n/20Sn-1(1+t)2p-2rtn-1𝑑σn-1(u)𝑑t
= 2JΓ(n/2)0(1+t)2p-2rtn-1𝑑t.

This integral is finite if and only if 2p-2r+n-1<-1, and we have assumed that r>p+n2. Therefore,

n(1+|y|)p|F^(y)|𝑑mn(y)<,

from which we get that yαF^(y) is in L1(n) for |α|p.

Define

Fω(x)=nF^ex𝑑mn,xn.

(Note that F depends on ω.) F,F^L1(n) so by the inversion theorem we have F(x)=Fω(x) for almost all xn. FωC0(n). If p1, then we shall show that FωCp(Ω). Take ek to be the standard basis for n. For 1k1n and ϵ0,

Fω(x+ϵek1)-Fω(x)ϵ = 1ϵnF^(y)(exp(iϵek1y)-1)exp(ixy)𝑑mn(y)
= niyk1F^(y)eiϵyk1-1iϵykex(y)𝑑mn(y).

But |iyk1F^(y)eiϵyk1-1iϵyk1ex(y)||yk1F^(y)| and yk1F^(y) belongs to L1(n) (supposing p1) so we can apply the dominated convergence theorem, which gives us

(Dk1Fω)(x)=limϵ0Fω(x+ϵek1)-Fω(x)ϵ=niyk1F^(y)ex(y)𝑑mn(y).

From the above expression, it is apparent that Dk1Fω is continuous. This is true for all 1k1n, so FωC1(n). If p2, then yk1yk2F^(y) is in L1(n) for any 1k2n, and repeating the above argument we get FωC2(n). In this way, FωCp(n).

For all xω, f(x)=F(x), so f(x)=Fω(x) for almost all xω. If ω is an open subset of Ω whose closure is a compact subset of Ω and ωω, then Fω,FωCp(n) satisfy f(x)=Fω(x) for almost all xω and f(x)=Fω(x) for almost all xω, so Fω(x)=Fω(x) for almost all xωω. Since Fω,Fω are continuous, this implies that Fω(x)=Fω(x) for all xωω. Thus, it makes sense to define f0(x)=Fω(x) for xω. Because every point in Ω has an open neighborhood of the kind ω and the restriction of f0 to each ω belongs to Cp(ω), it follows that f0Cp(Ω). ∎