## 2.1 Local Sobolev spaces

To apply functional analytic techniques to PDE problems, we need an appropriate Hilbert space of functions. The right notion is that of a Sobolev space.

Let $U \subseteq \mathbb {R}^n$ be an open set, $s \in \mathbb {R}$ (possibly negative) and $f, g \in C_c^\infty (U)$. We define

$\begin{aligned} (f, g)_s & = \int _{\mathbb {R}^n} (1 + |\xi |^2)^s \overline{\hat{f}(\xi )}\hat{g}(\xi ) \; \mathrm{d}\xi \\ \| f\| _s & = (f, f)_s \end{aligned}$We define $H^s(U)$ to be the Hilbert space completion of $C_c^\infty (U)$ under $\| \cdot \| _s$.

If no confusion arises, we will omit the $(U)$ in $H^s(U)$.

Basic properties of Fourier transforms imply that for $s \in \mathbb {Z}_{>0}$, we have the following more intuitive definition:

If $s$ is a non-negative integer, and $f \in C_c^\infty (U)$, then $\| \cdot \| _s$ is equivalent to the norm

$\| f\| _s' = \sum _{|\alpha | \leq s} \| \mathrm{D}^\alpha f\| _{L^2}.$For any $s$, $\mathrm{D}_\alpha$ extends to a continuous map $H^s \to H^{s - |\alpha |}$.

The natural map $C_c^\infty (U) \hookrightarrow L^2$ induces an isomorphism $H^0 \cong L^2$.

If $s \geq t$ and $f \in C_c^\infty$, then $\| f\| _t \leq \| f\| _s$. Hence there is a continuous map $H^s \to H^t$.

The map is injective.

This lets us view the $H^s$ as nested subspaces of $L^2$. In fact, we get something better.

Let $k$ an integer such that $s > k + \frac{n}{2}$. Then there exists a constant $A$ such that for any $u \in C^{\infty }(U)$, we have

$\| u\| _{C^k} \leq A \| u\| _s.$Thus, there is a continuous inclusion $H^s \hookrightarrow C^k$.

The second term is exactly $\| u\| _s$, and the first term is a constant, which by elementary calculus, is finite iff $s > \ell + \frac{n}{2}$.

Admittedly, the proof above does not make much sense. The following more direct proof of a special case shows why we should expect a result along these lines to be true:

for some constant $A$. In other words, if $u \in H^1$, we want to be able to make sense of $u$ as a continuous function. The trick is to notice we can make sense of $\frac{\mathrm{d}u}{\mathrm{d}x}$ as an $L^2$ function, and hence we can write

$u(t) = \int _0^t \frac{\mathrm{d}u}{\mathrm{d}x}\; \mathrm{d}x + C$for some integration constant $C$. This can be determined by

$C = \int _0^1 u(x)\; \mathrm{d}x - \int _0^1 \int _0^t \frac{\mathrm{d}u}{\mathrm{d}x}\; \mathrm{d}x\; \mathrm{d}t.$This lets us control $\sup u$ in terms of integrals of $u$ and $\frac{\mathrm{d}u}{\mathrm{d}x}$.

Another crucial theorem is the following:

If $U \subseteq \mathbb {R}^n$ is precompact and $s > t$, then the natural map $H^s \to H^t$ is compact.

$\hat{u}_k$ that equicontinuous and uniformly bounded on compact sets.

For any fixed $R$, the first term vanishes in the limit $k, \ell \to \infty$ by uniform convergence. For the second term, we bound

$(1 + |\xi |^2)^t \leq (1 + |\xi |^2)^s (1 + R^2)^{t - s}.$So we know that

$\int _{|\xi | > R}|\hat{u}_k - \hat{u}_\ell |^2 (1 + |\xi |^2)^t \; \mathrm{d}\xi \leq (1 + R^2)^{t - s}(\| u_k\| _s^2 + \| u_\ell \| _s^2) \leq 2 (1 + R^2)^{t - s}.$This can be made small with large $R$, and we are done.

To prove the claim, we use the following trick — pick a bump function $a \in C_c^\infty (\mathbb {R}^n)$ such that $a|_U \equiv 1$. Then trivially, $u_k = au_k$, and thus we have

$\hat{u}_k = \hat{a} * \hat{u}_k.$Then controlling $\hat{u}_k$ would be the same as controlling $\hat{a}$, which is fixed. For example, to show that $\hat{u}_k$ is bounded, we write

$|\hat{u}_k| = |\hat{a} * \hat{u}_k| \leq \int |\hat{a}(\xi - \eta ) \hat{u}_k(\eta )|\; \mathrm{d}\eta \leq \left(\int |\hat{a} (\xi - \eta )|^2 (1 + |\eta |^2)^{-s}\; \mathrm{d}\eta \right)^{1/2} \| u_k\| _s$by Cauchy–Schwarz. Since $\hat{a}$ is a Schwarz function, the first factor is finite and depends continuously on $\xi$. So $\hat{u}_k$ is uniformly bounded on compact subsets. Equicontinuity follows from similar bounds on $\mathrm{D}_j \hat{u}_k = (\mathrm{D}_j \hat{a}) * \hat{u}_k$.

If we are lazy, we will often restrict to the case where $s \in \mathbb {Z}$. If further $s \geq 0$, then the definition of the Sobolev norm in terms of derivatives can be very useful. For $s < 0$, we will exploit the following duality:

The pairing

$\begin{aligned} C_c^\infty \times C_c^\infty & \to \mathbb {C}\\ (f, g) & \mapsto \int \overline{f(x)} g(x) \; \mathrm{d}x. \end{aligned}$satisfies

$|(f, g)| \leq \| f\| _s \| g\| _{-s}.$Thus, it extends to a map $H^s \times H^{-s} \to \mathbb {C}$. Moreover,

$\| f\| _s = \sup _{g \not= 0} \frac{|(f, g)|}{\| g\| _{-s}}.\tag {$\dagger $}$Thus, this pairing exhibits $H^s$ and $H^{-s}$ as duals of each other.

To show $(\dagger )$, one direction of the inequality is clear. For the other, if $f \in C^\infty$, take $\hat{g} = \hat{f} (1 + |\xi |^2)^s$. For general $f$, approximate.

A sample application of this is to show that differential operators induce maps between Sobolev spaces. We already said that $\mathrm{D}_\alpha$ tautologically induces a map $H^s \to H^{s - |\alpha |}$. In a general differential operator, we differentiate, then multiply by a smooth function. Thus, we want to show that multiplication by a smooth function is a bounded linear operator. This is easy to show for $s \in \mathbb {Z}_{\geq 0}$, and duality implies the result for negative $s \in \mathbb {Z}$.

Let $s \in \mathbb {Z}$, $a \in C_c^\infty (\mathbb {R})$ and $u \in H^s(U)$. Then

$\| au\| _s \leq C\| a\| _{C^{|s|}} \| u\| _s.$for some constant $C$ independent of $a$. Thus, if $L$ is a differential operator of compact support of order $k$, then it extends to a map $L: H^{s + k} \to H^s$.

For negative $s$, if $s > 0$, then

$\| au\| _{-s} = \sup _{v \not= 0} \frac{|(au, v)|}{\| v\| _s} = \sup \frac{|(u, \bar{a}v)|}{\| v\| _s} \leq \sup \frac{\| u\| _{-s} \| \bar{a}v\| _s}{\| v\| _s} \leq \| u\| _{-s} \| a\| _{C^s}.$Let $s \in \mathbb {Z}$, $a \in C_c^\infty (\mathbb {R})$ and $u \in H^s(U)$. Then

$\| au\| _s \leq \| a\| _{C^0} \| u\| _s + C \| a\| _{C^{|s| + 1}} \| u\| _{s - 1}$for some constant $C$ independent of $a$.

For $s < 0$, we proceed by downward induction on $s$, using the isometry $H^{s + 2} \to H^s$ given by extending $1 + \Delta$. This is an isomorphism, because in the Fourier world, this is just multiplication by $(1 + |\xi |^2)$. Thus, we check the claim for $u$ of the form $Lv$. We have

$\| a Lv\| _s \leq \| [a, L]v\| _s + \| L(av)\| _s \leq \| [a, L]v\| _s + \| av\| _{s + 2}.$The second term is bounded, by induction, by

$\| a\| _{C^0} \| v\| _{s + 2} + C \| a\| _{C^{|s + 2| + 1}} \| v\| _{s + 1} = \| a\| _{C^0} \| Lv\| _s + C\| a\| _{C^{|s + 2| + 1}} \| Lv\| _{s - 1}$Observing that $|s + 2| \leq |s|$ since $s < 0$, we are happy with this term. To bound the first term, we have

$[a, L]v = [a, \Delta ]v = -\nabla a \cdot \nabla v.$So we get a bound (omitting constant multiples)

$\| [a, L]v\| _s = \| \nabla a \cdot \nabla v\| _s \leq \| \nabla a\| _{C^{|s|}} \| \nabla v\| _{s} \leq \| a\| _{C^{|s| + 1}} \| v\| _{s + 1} = \| a\| _{C^{|s| + 1}} \| Lv\| _{s - 1}.$