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.



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.

$\blacksquare$


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