Sobolev Norm is Norm
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Theorem
Let $k \in \Z_+$ and $1 \le p \le \infty$.
Let $U \subset \R^n$ be an open set.
Then the Sobolev norm is a norm on the Sobolev space $\map {W^{k, p} } U$.
Proof
Norm Axiom $\text N 1$: Positive Definiteness
Let $u, v \in \map {W^{k, p} } U$.
From P-Seminorm of Function Zero iff A.E. Zero:
- $\norm u_{\map {W^{k, p} } U} = 0$
for $u \in \map {W^{k, p} } U$ if and only if $u = 0$ almost everywhere.
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So Norm Axiom $\text N 1$: Positive Definiteness is satisfied.
$\Box$
Norm Axiom $\text N 2$: Positive Homogeneity
Let $u \in \map {W^{k,p}} U$ and $\lambda \in \R$.
Then:
\(\ds \size \lambda \norm u_{\map {W^{k,p} } U}\) | \(=\) | \(\ds \size \lambda \paren {\sum_{\size \alpha \le k} \norm {D^\alpha u}_{\map {L^p} U}^p}^{1/p}\) | Definition of Sobolev Norm | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\sum_{\size \alpha \le k} \norm {\lambda \cdot D^\alpha u}_{\map {L^p} U}^p}^{1/p}\) | Lp Norm is Norm, Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\lambda \cdot u}_{\map {W^{k,p} } U}\) | Definition of Sobolev Norm |
So Norm Axiom $\text N 2$: Positive Homogeneity is satisfied.
$\Box$
Norm Axiom $\text N 3$: Triangle Inequality
Let $u, v \in \map {W^{k,p}} U$.
Let $1\le p < \infty$.
Then:
\(\ds \norm {u + v}_{\map {W^{k, p} } U}\) | \(=\) | \(\ds \paren {\sum_{\size \alpha \le k} \norm {D^\alpha u + D^\alpha v}_{\map {L^p} U}^p}^{1/p}\) | Definition of Sobolev Norm | |||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {\sum_{\size \alpha \le k} \paren {\norm {D^\alpha u}_{\map {L^p} U} + \norm {D^\alpha v}_{\map {L^p} U} }^p}^{1/p}\) | Lp Norm is Norm, Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {\sum_{\size \alpha \le k} \norm {D^\alpha u}_{\map {L^p} U}^p}^{1/p} + \paren {\sum_{\size \alpha \le k} \norm {D^\alpha v}_{\map {L^p} U}^p}^{1/p}\) | Minkowski's Inequality on Lebesgue Space for $L^p$ norm | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm u_{\map {W^{k,p} } U} + \norm v_{\map {W^{k,p} } U}.\) | Definition of Sobolev Norm |
So Norm Axiom $\text N 3$: Triangle Inequality is satisfied.
$\blacksquare$
Sources
- 2010: Lawrence C. Evans: Partial Differential Equations (2nd ed.) $\S5.2$ Sobolev Spaces, $3.$ Elementary properties, Theorem 2 (Sobolev spaces as function spaces).