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Let $H=H_0^1(0,\ell)\times H_*^1(0,\ell)\times H_*^1(0,\ell)$ be equipped with the norms \begin{align*} \|(\varphi,\psi,w)\|_1^2&=A\|\varphi_x+\psi+lw\|_{L^2}^2+B\|w_x-l\varphi\|_{L^2}^2+C\|\psi_x\|_{L^2}^2,\\ \|(\varphi,\psi,w)\|_2^2&=\|\varphi_x\|_{L^2}^2+\|w_x\|_{L^2}^2+\|\psi_x\|_{L^2}^2. \end{align*} where $H_*^1(0,\ell)=\{f\in H^1;\;\int_0^\ell f(x)\;dx=0\}$ and $A$, $B$, $C$, $l$ are positive constants. It's also assumed $l\neq \frac{n\pi}{\ell}$, otherwise the condition $\|U\|_1=0\Rightarrow U=0$ is not satisfied.

Some research papers that deals with systems of PDEs from semigroup point of view (example) states $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent. I've tried to prove it without success, so I'd like help.

It's pretty clear (from Poincaré inequality) that $\|\cdot\|_1\leq K\|\cdot\|_2$ however the converse is not obvious for me. Of course, if we know that $H$ is complete with respect to $\|\cdot\|_1$, (from open mapping theorem) it's not necessary to prove the converse but I'm having troubles to prove the completeness too.

If this question is not appropriate for the site, let me know and I will delete it.

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    $\begingroup$ Consider how you would go about solving the problem $$\phi_x+\psi+lw=f,$$ $$w_x-l\phi=g,$$ $$\psi_x=h,$$ with $f,g,h\in L^2$. The inequality you need will come as a byproduct. $\endgroup$
    – Michael Renardy
    Commented Nov 22, 2014 at 1:37
  • $\begingroup$ If we take the Fourier transform, we get $$\|(l^2-x^2)\hat\phi\|_{L^2}\leq K_1\Big(\|\hat{f}\|_{L^2}+\|\hat{h}\|_{L^2}+\|\hat{g}\|_{L^2}\Big).$$ So, If $l>\ell$ then $\|\hat{\phi}\|_{L^2}\leq K_2\|(l^2-x^2)\hat\phi\|_{L^2}$ with $K_2>0$ and thus (by Plancherel theorem), $\|\phi\|_{L^2}\leq K_3\|(\phi,\psi,w)\|_1$. Working in a similar way with the others functions, we get the desired result. Is this the solution that you suggested? Did you assume $l> \ell$ or am I missing something? $\endgroup$
    – Robert
    Commented Nov 25, 2014 at 18:18

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Lemma: Given $F\in L^2_*$ and $G,H\in L^2$, there exists $(\varphi,\psi,w)\in H$ such that \begin{align} \varphi+\psi+lw&=F,\\ w_x-l\varphi&=G,\\ \psi_x&=H. \end{align}

Proof that $H$ is complete with respect to $\|\cdot\|_1$ (and thus Hilbert as desired):

Let $(\varphi^{n},\psi^n,w^n)$ be Cauchy in $(H,\|\cdot\|_{1})$. Then $(w_x-l\varphi)$, $(\psi^n_x)$ are Cauchy in $(L^2,\|\cdot\|_{L^2})$ and $(\varphi^n_x+\psi^n+lw)$ is Cauchy in $(L^2_*,\|\cdot\|_{L^2})$, where $L^2_*=\{f\in L^2;\int f(x)\;dx=0\}$.

Since $(L^2,\|\cdot\|_{L^2})$, $(L^2_*,\|\cdot\|_{L^2})$ are complete, there exists $f\in L^2_*$ and $g,h\in L^2$ such that $$\|\varphi^n_x+\psi^n+lw-f\|_{L^2}\to0,\qquad \|w_x-l\varphi-g\|_{L^2}\to0,\qquad \|\psi_x^n-h\|_{L^2}\to0.$$

So, $(\varphi^{n},\psi^n,w^n)$ converges to the solution $(\varphi,\psi,w)\in H$ of the the problem \begin{align} \varphi+\psi+lw&=f,\\ w_x-l\varphi&=g,\\ \psi_x&=h. \end{align}

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