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I'm reading through Sogge's Lectures on Nonlinear Wave Equations and am confused by the proof of existence for the linear inhomogeneous problem. Sorry for the long setup.

(Theorem) Let $s \in \mathbb{Z}$. Then for every $f \in H^{s+1}(\mathbb{R}^n)$, $g \in H^s(\mathbb{R}^n)$ and $F \in L^1([0,T];H^s(\mathbb{R}^n))$ there is a unique $$ u \in C([0,T];H^{s+1}) \cap C^1([0,T];H^s) $$ solving $$ \begin{cases} Lu=F, &0 < t <T \\ u_{t=0}=f, \partial_tu_{t=0}=g \end{cases} $$ where $$ L=\sum_{j,k=0}^n g^{jk}(t,x)\partial_j\partial_k u + \sum_{j=0}^n b^j(t,x)\partial_j u + a(t,x)u $$ has $C^\infty$ coefficients with uniform bounds on each derivative. Here $g^{jk}$ is symmetric and close to the d'Alembertian in the sense: $$ \sum |g^{jk}(t,x)-g_0^{jk}| < \frac12 $$ where $g_0^{jk}=\text{diag}(1,-1,\dots,-1)$ are the coefficients of $\square$.

The proof makes use of the following energy-type inequality: $$ \sum_{|\alpha|\leq1} \|\partial^\alpha u(t,\cdot)\|_{H^s} \leq C_{s,T}\left( \sum_{|\alpha|\leq1} \|\partial^\alpha u(0,\cdot)\|_{H^s} + \int_0^t \|Lu(\tau,\cdot)\|_{H^s} \; d\tau\right). \tag{1} $$

The relevant text: Proceed assuming $f=g=0$. If $\psi \in C_0^\infty((-\infty,T)\times\mathbb{R}^n)$ then applying the above energy inequality to $L^*$, with $t$ replaced by $T-t$, yields $$ \|\psi(t,\cdot)\|_{H^{-s}} \leq C\int_0^T\|L^*\psi(\tau,\cdot)\|_{H^{-s-1}} \; d\tau. \tag{2} $$ Hence, since $H^s$ and $H^{-s}$ are dual spaces, for fixed $F \in L^1([0,T];H^s)$ we have $$ |\langle F,\psi \rangle| = \left|\int_0^T \langle F(t,\cdot),\psi(t,\cdot)\rangle \; dt\right| \leq C' \int_0^T \|L^*\psi(t,\cdot)\|_{H^{-s-1}} \; dt. \tag{3} $$ So, by the Hahn-Banach Theorem, there is a $u \in L^{\infty}([0,T];H^{s+1})$ satisfying $u=0$ when $t<0$ and, moreover, $$ \langle F,\psi \rangle = \langle u,L^*\psi \rangle, \qquad \forall \psi \in C_0^{\infty}((-\infty,T)\times\mathbb{R}^n). $$ Consequently, $Lu=F$ in $(0,T)\times\mathbb{R}^n$ in the sense of distributions.


  1. When applying the energy inequality (1), how does $L^*\psi$ end up in the $H^{-s-1}$ norm in (2)?
  2. How exactly is the Hahn-Banach Theorem applied? The linear functional $\langle F, \cdot \rangle$ is bounded by the sublinear function on the right hand side of (3) for all $\psi \in C_0^\infty((-\infty,T)\times\mathbb{R}^n)$. I don't understand the details which provide the existence of $u$. A similar question is asked here.
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up vote 2 down vote accepted
  1. Note that in the left hand side of (1) you have basically the $H^{s+1}$ norm of $u$. Hence the $H^{-s-1}$ norm in the right hand side of (2).

  2. I think you are confused because you think that the Hahn-Banach theorem is applied to the functional $\langle F,\cdot\rangle$. What you are actually doing here is you are considering the linear functional that sends $L^*\psi$ to $\langle F,\psi\rangle$. This is well-defined because the estimate (3) shows that if $\psi_1$ and $\psi_2$ are such that $L^*\psi_1=L^*\psi_2$ then $\langle F,\psi_1\rangle=\langle F,\psi_2\rangle$. How do we get existence from Hahn-Banach? Ok. So we have the linear functional $T:L^*\psi\mapsto\langle F,\psi\rangle$, defined and bounded on a subspace of $L^1(0,T;H^{-s-1})$. By you-know-what theorem, this can be extended to a bounded linear functional on $L^1(0,T;H^{-s-1})$. Let us call this extension $u$. By construction, this is an element of the dual of $L^1(0,T;H^{-s-1})$, which can be identified with $L^\infty(0,T;H^{s+1})$. Let us write down what we have: We have an element $u\in L^\infty(0,T;H^{s+1})$ such that $$ \langle u,L^*\psi\rangle = \langle F,\psi\rangle \qquad\textrm{for all nice test functions }\psi. $$ What this means is $Lu=F\,$!

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1. Got it. 2. You're right, I was thinking that. Everything you wrote makes sense. But how do we get the existence of $u$ from Hahn-Banach? I feel like there's some small point I'm still missing. Thanks! – dls Sep 19 '13 at 16:48
@dls: Please have a look at the edit. – timur Sep 19 '13 at 17:01
For some reason I didn't understand that the domain of $T$ was a subspace of $L^1(0,T;H^{-s-1})$. That clears everything up! Thank you for your time! – dls Sep 19 '13 at 17:25

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