# Partial Differential Equations/Sobolev spaces

In this chapter, it will be defined, what Sobolev spaces are. Also, some properties of Sobolev spaces will be shown.

## Weak derivatives and Sobolev spaces

### Weak derivatives

Let $\Omega \subseteq \R^d$ be a domain, $\alpha \in \N_0^d$ a multiindex, and $f \in L^p(\Omega)$, where $p \in [1, \infty]$. We call $g \in L^p(\Omega)$ the weak derivative of order $\alpha$ of $f \in L^p(\Omega)$, if and only if:

$\frac{\partial^\alpha}{\partial x^\alpha} T_f = T_g$

We denote the weak derivative of $f$ with the notation $\frac{\partial^\alpha}{\partial x^\alpha} f$, if it exists.

#### Properties

It can be easily observed that the weak derivatives are linear, i. e. $\frac{\partial^\alpha}{\partial x^\alpha} (\lambda f + g) = \lambda \frac{\partial^\alpha}{\partial x^\alpha} f + \frac{\partial^\alpha}{\partial x^\alpha} g$. Furthermore, they are equal to the actual derivatives if these exist. This follows due to integration by parts.

The weak gradient of $f \in L^p$ is defined as follows:

$\nabla f := \left( \begin{smallmatrix} \frac{\partial^{(1, 0, \ldots, 0)}}{\partial x^{(1, 0, \ldots, 0)}} f \\ \vdots \\ \frac{\partial^{(0, \ldots, 0, 1)}}{\partial x^{(0, \ldots, 0, 1)}}f \\ \end{smallmatrix} \right)$

, i. e. it is the sorted d-dimensional vector of all weak derivatives with $|\alpha| = 1$.

### Sobolev spaces

Let $m \in \N_0, p \in [1, \infty]$, and $\Omega$ be a domain. Then the Sobolev space $\mathcal W^{m, p}(\Omega)$ is defined as follows:

$\mathcal W^{m, p}(\Omega) := \{f \in L^p : \frac{\partial^\alpha}{\partial x^\alpha} f \text{ exists and } \frac{\partial^\alpha}{\partial x^\alpha} f \in L^p\}$

The Sobolev spaces also have a norm: $\|f\|_{W^{m, p}(\Omega)} := \sum_{|\alpha| \le m} \left\| \frac{\partial^\alpha}{\partial x^\alpha} f \right\|_{L^p}$

### Theorem 7.1

This expression is really a norm.

Proof: To show that the expression is a norm, we have to check the three defining properties for a norm:

• Definiteness, i. e. $\|x\| = 0 \Leftrightarrow x = 0$
• Homogeneity, i. e. $\|\alpha x\| = |\alpha|\|x\|$ for every scalar $\alpha$
• and the triangle inequality, i. e. $\|x + y\| \le \|x\| + \|y\|$.

We start with definiteness: If $f = 0$ (i. e. f is the function constantly zero), then obviously $\|f\|_{W^{m, p}} = 0$, since the derivatives of zero (which is a constant function) are zero. Furthermore, if $\|f\|_{W^{m, p}} = 0$, then in particular it follows that $\|f\|_{L^p} = 0$, which implies that $f=0$ since $\|f\|_{L^p}$ is a norm.

Homogeneity follows directly from the linearity of the derivative and the homogeneity of the $\| \cdot \|_{L^p}$-norm.

The triangle inequality follows from the linearity of the derivative, the triangle inequality for the $\| \cdot \|_{L^p}$-norm and the commutation law for addition:

$\|f + g\|_{W^{m, p}} := \sum_{|\alpha| \le k} \left\| \frac{\partial^\alpha}{\partial x^\alpha} f + \frac{\partial^\alpha}{\partial x^\alpha} g \right\|_{L^p} \le \sum_{|\alpha| \le k} \left\| \frac{\partial^\alpha}{\partial x^\alpha} f \right\|_{L^p} + \left\| \frac{\partial^\alpha}{\partial x^\alpha} g \right\|_{L^p} =: \|f\|_{W^{m, p}} + \|g\|_{W^{m, p}}$

### Theorem 7.2

All Sobolev spaces are Banach spaces.

Proof: We use the fact that $L^p$-spaces are Banach spaces. Let $(u_i)_{i \in \N}$ be a Cauchy sequence in $\mathcal W^{m, p}(\Omega)$. Then, since for all multi-indices $\alpha \in \N_0^d$, we find that:

$\left\|\frac{\partial^\alpha}{\partial x^\alpha} u_i - \frac{\partial^\alpha}{\partial x^\alpha} u_j\right\| \le \sum_{|\alpha| \le k} \left\| \frac{\partial^\alpha}{\partial x^\alpha} (u_i - u_j) \right\|_{L^p}$

Therefore, the sequences $\left( \frac{\partial^\alpha}{\partial x^\alpha} u_i \right)_{i \in \N}$ are all Cauchy sequences in $L^p$ and therefore have a limit $u_\alpha$. We choose $u := u_{(0, \ldots, 0)}$ and show that $u$ is the limit of $(u_i)_{i \in \N}$ in $\mathcal W^{m, p}(\Omega)$: Let $\varphi \in \mathcal D(\Omega)$ (and therefore also $\varphi \in \mathcal S$). Then:

$\int_{\R^d} \frac{\partial^\alpha}{\partial x^\alpha} \varphi(x) u(x) dx = \lim_{i \to \infty} \int_{\R^d} \frac{\partial^\alpha}{\partial x^\alpha} \varphi(x) u_i(x) dx = \lim_{i \to \infty} (-1)^{|\alpha|} \int_{\R^d} \varphi(x) \frac{\partial^\alpha}{\partial x^\alpha} u_i(x) dx$

, and due to the dominated convergence theorem,

$\lim_{i \to \infty} (-1)^{|\alpha|} \int_{\R^d} \varphi(x) \frac{\partial^\alpha}{\partial x^\alpha} u_i(x) dx = (-1)^{|\alpha|} \int_{\R^d} \varphi(x) u_\alpha(x) dx$

, which is why $u_\alpha$ is the weak derivative of order $\alpha$ of $u$. This finishes the proof, since it shows that $u_i \to u, i \to \infty$ with respect to the norm of $\mathcal W^{m, p}(\Omega)$.

## Approximation by smooth functions

### Approximation of continuous functions

Recall the definition of the multi-dimensional mollifiers. Let $\Omega \subseteq \R^d$ be a domain and $u \in C_0(\Omega)$, i. e. $u$ is a continuous function with compact support.

Then $u * \eta_\epsilon \in C^\infty(\Omega)$ and $u * \eta_\epsilon \to u$ uniformly as $\epsilon \to 0$.

Proof: First, that $u * \eta_\epsilon \in C^\infty(\Omega)$ follows by the definition of the convolution and the Leibniz integral rule (depending on how you define the convolution, you also need to prove that $u * \eta_\epsilon \in C^\infty(\Omega) = \eta_\epsilon \in C^\infty(\Omega) * u$, so that you apply the derivative to $\eta_\epsilon$). This shows the first part.

We further know that $u$ is a continuous function. Since $u$ has compact support, it is also uniformly continuous. Furthermore:

$(u * \eta_\epsilon) (y) = \int_{\R^d} u(x) \eta_\epsilon(y - x) dx$

Since $u$ is uniformly continuous, we may choose $\epsilon > 0$ such that $\|x - y\| < 2 \epsilon \Rightarrow |u(x) - u(y)| < \delta$. Therefore, we find, since $\int_{\R^d} \eta_\epsilon(x) dx = 1$, the following:

$\left| \int_{\R^d} u(x) \eta_\epsilon(y - x) dx - u(y) \right| \le \int_{\R^d} |u(x) - u(y)| \eta_\epsilon(y - x) dx < \delta$

Since the smallest possible $\delta$ decreases with decreasing $\epsilon$, and is arbitrary, the claim is proven.

### Approximation of Lp-functions

Let $\Omega \subseteq \R^d$ be a domain, $1 \le p < \infty$ and $u \in L^p(\Omega)$, and let furthermore $\text{supp } u + B_\epsilon(0) \subseteq \Omega$.

Then $u * \eta_\epsilon \in C^\infty(\Omega)$ and $u * \eta_\epsilon \to u$ uniformly as $\epsilon \to 0$.

Proof: The first claim, namely $u * \eta_\epsilon \in C^\infty(\Omega)$, follows exactly the same way as in the subsubchapter about mollifying continuous functions, by the Leibnis rule for integrating under the integral sign.

We will use that $C_0(\Omega)$ is a dense subset of $L^p(\Omega)$. This means by definition that

$\forall f \in L^p(\Omega): \forall \epsilon > 0: \exists g \in C_0(\Omega) : \|f - g\|_{L^p} < \epsilon$

i. e. one can approximate functions in $L^p(\Omega)$ arbitrarily well with functions in $C_0(\Omega)$.

This is proven by approximating the elementary functions, which are used in the definition of the Lebesgue integral, with continuous functions. It is OK that they have only compact support, because all $L^p$-functions $h$ have the property that for arbitrary $\epsilon > 0$ there exists a bounded set $M$ such that

$\left| \int_{\R^d \setminus M} h(x) dx \right| < \epsilon$

.

Let' choose an arbitrary $\epsilon > 0$. We choose now first $g \in C_0(\Omega)$ such that $\|g - f\|_{L^p} < \frac{\epsilon}{4}$.

Then we choose $\delta_1 > 0$ such that $\|g - g * \eta_{\delta_1} \| < \frac{\epsilon}{4}$.

Third, we notice that $f * \eta_\delta - g * \eta_\delta = (f - g) * \eta_\delta$. With the triangle inequality and the reversed triangle inequality, we furthermore obtain:

$|\|(f - g) * \eta_\delta\|_{L^p} - \|f - g\|_{L^p}| \le \|((f - g) * \eta_\delta) - (f - g)\|_{L^p}$

If we choose now $\delta_2 > 0$ such that $\|((f - g) * \eta_{\delta_2}) - (f - g)\| < \frac{\epsilon}{4}$, we obtain, together with the fact that $\|f - g\|_{L^p} < \frac{\epsilon}{4}$, that

$\|f * \eta_{\delta_2} - g * \eta_{\delta_2}\|_{L^p} < \frac{\epsilon}{2}$

We now choose $\delta = \min\{\delta_1, \delta_2\}$ and obtain with the triangle inequality:

$\|f - f * \eta_\delta\|_{L^p} \le \|f - g\|_{L^p} + \|g - g * \eta_\delta\|_{L^p} + \|f * \eta_\delta - g * \eta_\delta\|_{L^p} < \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{2} = \epsilon$

Since $\epsilon > 0$ was arbitrary, this finishes the proof.

### Lemma 7.3

Let $\Omega \subset \R^d$ be a domain, let $r > 0$, and $U \subset \Omega$, such that $U + B_r(0) \subseteq \Omega$. Let furthermore $u \in \mathcal W^{m, p}(U)$. Then $\mu_\epsilon * f$ is in $C^\infty(U)$ for $\epsilon < r$ and $\lim_{\epsilon \to 0} \|\mu_\epsilon * f - f\|_{W^{m, p}(U)} = 0$.

Proof: The first claim, that $\mu_\epsilon * f \in C^\infty(U)$, follows from the fact that if we choose

$\tilde f(x) = \begin{cases} f(x) & x \in U \\ 0 & x \notin U \end{cases}$

Then, due to the above section about mollifying $L^p$-functions, we know that the first claim is true.

The second claim follows from the following calculation, using the one-dimensional chain rule:

$\frac{\partial^\alpha}{\partial x^\alpha} (\mu_\epsilon * f) (y)= \int_{\R^d} \frac{\partial^\alpha}{\partial x^\alpha}\mu_\epsilon(y -x) f(x) dx = (-1)^{|\alpha|} \int_{\R^d} \frac{\partial^\alpha}{\partial y^\alpha}\mu_\epsilon(y -x) f(x) dx$
$=\int_{\R^d} \mu_\epsilon(y -x) \frac{\partial^\alpha}{\partial y^\alpha} f(x) dx = (\mu_\epsilon * \frac{\partial^\alpha}{\partial y^\alpha}f) (y)$

Due to the above secion about mollifying $L^p$-functions, we immediately know that $\lim_{\epsilon \to 0} \|\mu_\epsilon * \frac{\partial^\alpha}{\partial y^\alpha}f - f\| = 0$, and the second statement therefore follows from the definition of the $W^{m, p}(U)$-norm.

### Partition of unity

Let $\Omega \subseteq \R^d$ be an open set, and $\bigcup_{\alpha \in A} M_\alpha$ be an open cover of $\Omega$. Then there exists a sequence of functions $\eta_i$ such that the following conditions are satisfied:

1. $\forall i \in \N : \forall x \in \Omega : 0 \le \eta_i(x) \le 1$
2. $\forall x \in \Omega : \exists \text{ only finitely many } i \in \N : \eta_i(x) \neq 0$
3. $\forall i \in \N : \exists \alpha \in A : \text{supp } \eta_i \subseteq M_\alpha$
4. $\forall x \in \Omega : \sum_{i=0}^\infty \eta_i(x) = 1$

Proof: We will proof this by explicitly constructing such a sequence of functions. First, we construct a sequence of closed balls $B_i$ with the properties

1. $\forall i \in \N : \exists \alpha \in A : B_i \subseteq M_\alpha$
2. $\forall x \in \Omega : \exists \text{ only finitely many } i \in \N : x \in B_i$
3. $\bigcup_{i \in \N} \text{interior}( B_i) = \Omega$.

In order to do this, we first start with the definition of a sequence compact sets:

$K_i := \{x \in \Omega : \text{dist}(\partial \Omega, x) \ge \frac{1}{i}, \|x\| \le i\}$

This sequence obviously has the properties

• $\bigcup_{j \in \N} K_j = \Omega$
• $K_j \subset \text{interior}(K_{j+1})$

We now construct our sequence of closed balls such that

$K_1 \subset \bigcup_{1 \le i \le k_1} \text{interior}(B_i) \subseteq \text{interior}(K_2)$
and $K_{j+1} \setminus \text{interior}(K_j) \subset \bigcup_{k_j < i \le k_{j+1}} B_i \subseteq \text{interior}(K_{j+2}) \setminus K_{j-1}$

We do this in the following way: To meet the first condition, we first cover $K_1$ with balls by choosing for every $x \in K_1$ a ball $B_x$ such that $B_x \subseteq ( M_\alpha \cap \text{interior}(K_2))$. Since these balls cover $K_1$, and $K_1$ is compact, we may choose a finite subcover $B_1, \ldots B_{k_1}$.

To meet the second condition, we do just the same thing, noticing that $K_{j+1} \setminus \text{interior}(K_j)$ is compact and $\text{interior}(K_{j+2}) \setminus K_{j-1}$ is open.

This sequence of open balls fulfills the conditions which we wanted.

We recall now the Definition of the multi-dimensional mollifier and note that we can recenter it to an arbitrary $x_0 \in \R^d$ by using the formula

$\eta_{\epsilon, x_0} (x) := \eta_\epsilon (x - x_0)$

What we do now, is that for each $B_i$, we choose $\tilde \eta_i$ as the mollifier which is adjusted to the ball, i. e. the ball is it's support. Then we choose

$\eta(x) := \sum_{i=0}^\infty \tilde \eta_i(x)$

and

$\eta_i := \frac{\tilde \eta_i}{\eta}$,

and we have found a sequence of functions which satisfies the conditions we wanted.

### Approximation of Wm, p-functions with smooth functions on open sets

Let $\Omega \subseteq \R^d$ be an open set. Then for all functions $v \in W^{m, p}(\Omega)$, there exists a sequence of functions in $C^\infty(\Omega) \cap W^{m, p}(\Omega)$ approximating it.

Proof:

Let's choose

$U_i := \{x \in \Omega : \text{dist}(\partial \Omega, x) > \frac{1}{i} \wedge \|x\| < i\}$

and

$V_i =\begin{cases} U_3 & i = 0 \\ U_{i+3} \setminus \overline{U_{i+1}} & i > 0 \end{cases}$

One sees that the $V_i$ are an open cover of $\Omega$. Therefore, we can choose a sequence of functions $(\tilde \eta_i)_{i \in \N}$ (partition of the unity) such that

1. $\forall i \in \N : \forall x \in \Omega : 0 \le \tilde \eta_i(x) \le 1$
2. $\forall x \in \Omega : \exists \text{ only finitely many } i \in \N : \tilde \eta_i(x) \neq 0$
3. $\forall i \in \N : \exists j \in \N : \text{supp } \tilde \eta_i \subseteq V_j$
4. $\forall x \in \Omega : \sum_{i=0}^\infty \tilde \eta_i(x) = 1$

By defining $\Eta_i := \{\tilde \eta_j \in \{\tilde \eta_m\}_{m \in \N} : \text{supp } \tilde \eta_j \subseteq V_i\}$ and

$\eta_i(x) := \sum_{\eta \in \Eta_i} \eta(x)$, we even obtain the properties
1. $\forall i \in \N : \forall x \in \Omega : 0 \le \eta_i(x) \le 1$
2. $\forall x \in \Omega : \exists \text{ only finitely many } i \in \N : \eta_i(x) \neq 0$
3. $\forall i \in \N: \text{supp } \eta_i \subseteq V_i$
4. $\forall x \in \Omega : \sum_{i=0}^\infty \tilde \eta_i(x) = 1$

where the properties are the same as before except the third property, which changed. Let $|\alpha| = 1$, $\varphi$ be a bump function and $(v_j)_{j \in \N}$ be a sequence which approximates $v$ in the $L^p(\Omega)$-norm. The calculation

$\int_\Omega \eta_i(x) v_j(x) \frac{\partial^\alpha}{\partial x^\alpha} \varphi(x) dx = - \int_\Omega \left(\frac{\partial^\alpha}{\partial x^\alpha}\eta_i(x) v_j(x) + \eta_i(x) \frac{\partial^\alpha}{\partial x^\alpha} v_j(x)\right) \varphi(x) dx$

reveals that, by taking the limit $j \to \infty$ on both sides, $v \in W^{m, p}(\Omega)$ implies $\eta_i v \in W^{m, p}(\Omega)$, since the limit of $\eta_i(x) \frac{\partial^\alpha}{\partial x^\alpha} v_j(x)$ must be in $L^p(\Omega)$ since we may choose a sequence of bump functions $\varphi_k$ converging to 1.

Let's choose now

$W_i = \begin{cases} U_{i+4} \setminus \overline{U_i} & i \ge 1 \\ U_4 & i = 0 \end{cases}$

We may choose now an arbitrary $\delta > 0$ and $\epsilon_i$ so small, that

1. $\|\eta_{\epsilon_i} * (\eta_i v) - \eta_i v\|_{W^{m, p}(\Omega)} < \delta \cdot 2^{-(j+1)}$
2. $\text{supp } (\eta_{\epsilon_i} * (\eta_i v)) \subset W_i$

Let's now define

$w(x) := \sum_{i=0}^\infty \eta_{\epsilon_i} * (\eta_i v)(x)$

This function is infinitely often differentiable, since by construction there are only finitely many elements of the sum which do not vanish on each $W_i$, and also since the elements of the sum are infinitely differentiable due to the Leibniz rule of differentiation under the integral sign. But we also have:

$\|w - v\|_{W^{m, p}(\Omega)} = \left\|\sum_{i=0}^\infty \eta_{\epsilon_i} * (\eta_i v) -\sum_{i=0}^\infty (\eta_i v)\right\|_{W^{m, p}(\Omega)} \le \sum_{i=0}^\infty \|\eta_{\epsilon_i} * (\eta_i v) - \eta_i v\|_{W^{m, p}(\Omega)} < \delta \sum_{i=0}^\infty 2^{-(j+1)} = \delta$

Since $\delta$ was arbitrary, this finishes the proof.

### Approximation of Wm, p-functions with smooth functions on compact sets

Let $\Omega$ be a bounded domain, and let $\partial \Omega$ have the property, that for every point $x \in \partial \Omega$, there is a neighbourhood $\mathcal U_x$ such that

$\Omega \cap \mathcal U_x = \{(x_1, \ldots, x_d) \in \R^d : x_i < f(x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{d-1}) \}$

for a continuous function $f$. Then every function in $W^{m, p}(\Omega)$ can be approximated by $C^\infty(\overline{\Omega})$-functions in the $W^{m, p}(\Omega)$-norm.

Proof:

to follow