Sobolev embedding

The properties of Sobolev spaces are briefly discussed above. If $m$ is a positive integer then $u \in H^m (\mathbb{R}^n)$ `means' that $u$ has up to $m$ derivatives in $L^2 (\mathbb{R}^n).$ The question naturally arises as to the sense in which these `weak' derivatives correspond to old-fashioned `strong' derivatives. Of course when $m$ is not an integer it is a little harder to imagine what these `fractional derivatives' are. However the main result is:

Proof. By definition, $u \in H^m (\mathbb{R}^n)$ means $v \in \mathcal{S}' (\mathbb{R}^n)$ and $\langle \xi \rangle^m \hat{u} (\xi) \in L^2 (\mathbb{R}^n)$. Suppose first that $u \in \mathcal{S} (\mathbb{R}^n)$. The Fourier inversion formula shows that

$$(2 \pi )^n \left\vert u(x) \right\vert = \left\vert \int e^{ix \cdot \xi} \hat{u} (\xi) d \xi \right\vert$$

$$\leq \left( \int_{\mathbb{R}^n} \langle \xi \rangle^{2m} \left\ve... ...left( \sum_{\mathbb{R}^n} \langle \xi \rangle^{-2m} d \xi \right)^{1/2} .$$

Now, if $m> n/2$ then the second integral is finite. Since the first integral is the norm on $H^m (\mathbb{R}^n)$ we see that

$$\sup_{\mathbb{R}^n} \left\vert u(x) \right\vert = \Vert u \Vert _{L^{\infty}} \leq (2 \pi )^{-n} \Vert u \Vert _{H^m} , m> n/2 .$$ (128)

This is all for $u \in \mathcal{S} (\mathbb{R}^n)$, but $\mathcal{S} (\mathbb{R}^n) \hookrightarrow H^m (\mathbb{R}^n)$ is dense. The estimate (9.2) shows that if $u_j \to u$ in $H^m (\mathbb{R}^n),$ with $u_j \in \mathcal{S} (\mathbb{R}^n)$, then $u_j \to u'$ in $\mathcal{C}^0_0 (\mathbb{R}^n).$ In fact $u'=u$ in $\mathcal{S}'(\mathbb{R}^n)$ since $u_j \to u$ in $L^2 (\mathbb{R}^n)$ and $u_j \to u'$ in $\mathcal{C}^0_0 (\mathbb{R}^n)$ both imply that $\int u_j \varphi$ converges, so

$$\int_{\mathbb{R}^n} u_j \varphi \to \int_{\mathbb{R}^n} u \varphi = \int_{\mathbb{R}^n} u'\varphi\ \forall\ \varphi \in \mathcal{S}(\mathbb{R}^n).$$

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Notice here the precise meaning of $u=u',$ $u \in H^m (\mathbb{R}^n) \subset L^2 (\mathbb{R}^n),$ $u' \in\mathcal{C}^0_0 (\mathbb{R}^n).$ When identifying $u \in L^2 (\mathbb{R}^n)$ with the corresponding tempered distribution, the values on any set of measure zero `are lost'. Thus as functions (9.1) means that each $u \in H^m (\mathbb{R}^n)$ has a representative $u' \in\mathcal{C}^0_0 (\mathbb{R}^n).$

We can extend this to higher derivatives by noting that

Proof. First it is enough to show that each $D_j$ defines a continuous linear map

$$D_j : H^m (\mathbb{R}^n) \to H^{m-1} (\mathbb{R}^n) \forall j$$ (129)

since then (9.3) follows by composition.

If $m \in \mathbb{R}$ then $u \in H^m (\mathbb{R}^n)$ means $\hat{u} \in \langle \xi \rangle^{-m} L^2 (\mathbb{R}^n)$. Since $\widehat{D_j u} = \xi_j \cdot \hat{u}$, and

$$\left\vert \xi_j \right\vert \langle \xi \rangle^{-m} \leq C_m \langle \xi \rangle^{-m+1} \forall m$$

we conclude that $D_j u \in H^{m-1} (\mathbb{R}^n)$ and

$$\Vert D_j u \Vert _{H^{m-1}} \leq C_m \Vert u \Vert _{H^m} .$$

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Applying this result we see

Proof. If $\left\vert \alpha \right\vert \leq k$, then ${D^{\alpha}} u \in H^{m-k} (\mathbb{R}^n) \subset \mathcal{C}^0_0 (\mathbb{R}^n)$. Thus the `weak derivatives' ${D^{\alpha}} u$ are continuous. Still we have to check that this means that $u$ is itself $k$ times continuously differentiable. In fact this again follows from the density of $\mathcal{S}(\mathbb{R}^n)$ in $H^m (\mathbb{R}^n)$. The continuity in (9.3) implies that if $u_j \to u$ in $H^m (\mathbb{R}^n)$, $m > \frac{n}{2} +k$, then $u_j \to u'$ in $\mathcal{C}^k_0 (\mathbb{R}^n)$ (using its completeness). However $u=u'$ as before, so $u \in \mathcal{C}^k_0 (\mathbb{R}^n)$.

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In particular we see that

$$H^{\infty} (\mathbb{R}^n) = \bigcap_m H^m (\mathbb{R}^n) \subset \mathcal{C}^\infty (\mathbb{R}^n) .$$ (130)

These functions are not in general Schwartz test functions.

Proof. This follows directly from (9.5) since the left side is contained in

$$\bigcap_k \langle x \rangle^{-k} \mathcal{C}^{k-n}_0 (\mathbb{R}^n) \subset \mathcal{S} (\mathbb{R}^n).$$

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Thus every tempered distribution is a finite sum of derivatives of continuous functions of poynomial growth.

Proof. Essentially by definition any $u \in \mathcal{S}' (\mathbb{R}^n)$ is continuous with respect to one of the norms $\Vert\langle x\rangle^k \varphi\Vert _{\mathcal{C}^k}.$ From the Sobolev embedding theorem we deduce that, with $m>k+ n/2,$

$$\left\vert u(\varphi)\right\vert\leq C\Vert\langle x\rangle^k\varphi\Vert _{H^m}\ \forall\ \varphi\in\mathcal{S}(\mathbb{R}^n).$$

This is the same as

$$\left\vert\langle x \rangle^{-k} u ( \varphi ) \right\vert \leq C\Vert \varphi \Vert _{H^m} \ \forall\ \varphi \in \mathcal{S} (\mathbb{R}^n).$$

which shows that $\langle x \rangle^{-k} u \in H^{-m} (\mathbb{R}^n)$, i.e., from Proposition 8.8,

$$\langle x \rangle^{-k} u = \sum_{\left\vert \alpha \right\vert \leq m} {D^{\alpha}} u_{\alpha} , u_{\alpha} \in L^2 (\mathbb{R}^n) .$$

In fact, choose $j> n/2$ and consider $v_{\alpha} \in H^j (\mathbb{R}^n)$ defined by $\hat{v}_{\alpha} = \langle \xi \rangle^{-j} \hat{u}_{\alpha}$. As in the proof of Proposition 8.14 we conclude that

$$u_{\alpha} = \sum_{\left\vert \beta \right\vert \leq j} D^{\beta}... ...pha , \beta} \in H^j (\mathbb{R}^n) \subset \mathcal{C}^0_0 (\mathbb{R}^n) \, .$$

Thus,¹⁷

$$u= \langle x \rangle^k \sum_{\left\vert \gamma \right\vert \leq M... ...}_{\alpha} v_{\gamma} \, , \ v_{\gamma} \in \mathcal{C}^0_0 (\mathbb{R}^n) \, .$$ (131)

To get (9.9) we `commute' the factor $\langle x \rangle^k$ to the inside; since I have not done such an argument carefully so far, let me do it as a lemma.

Proof. In fact it is convenient to prove a more general result. Suppose $p$ is a polynomial of a degree at most $j$ then there exist polynomials of degrees at most $j+ \left\vert \gamma - \alpha \right\vert$ such that

$$p \langle x \rangle^k D^{\gamma} v = \sum_{\alpha \leq \gamma} D^... ... \gamma} \langle x \rangle^{k-2 \left\vert \gamma - \alpha \right\vert} v) .$$ (132)

The lemma follows from this by taking $p=1$.

Furthermore, the identity (9.11) is trivial when $\gamma =0$, and proceeding by induction we can suppose it is known whenever $\left\vert \gamma \right\vert \leq L$. Taking $\left\vert \gamma \right\vert =L+1$,

$$D^{\gamma} = D_j D^{\gamma'} \ \left\vert \gamma' \right\vert =L.$$

Writing the identity for $\gamma'$ as

$$p \langle x \rangle^k D^{\gamma'} = \sum_{\alpha' \leq \gamma'} D... ... , \gamma'} \langle x \rangle^{k-2 \left\vert \gamma' - \alpha' \right\vert} v)$$

we may differentiate with respect to $x_j.$ This gives

$$p \langle x \rangle^k D^{\gamma} = - D_j (p \langle x \rangle^k ) \cdot D^{\gamma'} v$$

$$+ \sum_{\left\vert \alpha' \right\vert \leq \gamma} D^{\gamma - \... ...mma'} \langle x \rangle^{k-2 \left\vert \gamma - \alpha \right\vert +2} v) .$$

The first term on the right expands to

$$(- (D_j p) \cdot \langle x \rangle^k D^{\gamma'} v - \frac{1}{i} kpx_j \langle x \rangle^{k-2} D^{\gamma'} v) .$$

We may apply the inductive hypothesis to each of these terms and rewrite the result in the form (9.11); it is only necessary to check the order of the polynomials, and recall that $\langle x \rangle^2$ is a polynomial of degree $2$. $\qedsymbol$

Applying Lemma 9.6 to (9.10) gives (9.9), once negative powers of $\langle x \rangle$ are absorbed into the continuous functions. Then (9.8) follows from (9.9) and Leibniz's formula. $\qedsymbol$