# Existence for linear wave equation using energy inequality.

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.

Questions:

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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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 $Lu=F\,$!
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
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