Let us define
$$
\mathbb{H}^{1} = H^{1}(-L,0) \times H^{1}(0,L) \ \ \text{and} \ \ \mathbb{L}^{2} = L^{2}(-L,0)\times L^{2}(0,L),
$$
where $H^{1}(I) = \big\lbrace u \in L^{2}(I) \ \text{and} \ u\_{x} \in L^{2}(I); I = (a,b) \big\rbrace$.


Besides these,
$$
\mathbb{M} = \big\lbrace (u,v) \in \mathbb{H}^{1}; u(-L) = v(L) = 0 \ \text{and} \ u(0) = v(0) \big\rbrace .
$$
Under the above conditions, we have that the phase space is given by
$$
\mathcal{H} = \mathbb{M} \times \mathbb{L}^{2}.
$$
Note that this space equipped with the inner product
$$
\langle (u\_{1},v\_{1},w\_{1},z\_{1}), (u\_{2},v\_{2},w\_{2},z\_{2}) \rangle = \int\_{-L}^{0}u\_{1\_{x}}\overline{u}\_{2\_{x}} + \int\_{0}^{L}v\_{1\_{x}}\overline{v}\_{2\_{x}} + \int\_{-L}^{0}w\_{1}\overline{w}\_{2} + \int\_{0}^{L}z\_{1}\overline{z}\_{2}
$$
is a Hilbert space.



Inequality involving Sobolev spaces
Just take a derivative, and you get


$$ i \lambda u\_x = f\_x - w\_x $$


So taking the $L^2(-L,0)$ norms on both sides, you get


$$ \lambda^2 \int |u\_x|^2 \leq \int |f\_x - w\_x|^2 $$


The RHS can be expanded and estimated using AM-GM to be


$$ \lambda^2 \int |u\_x|^2 \leq 2 \int |f\_x|^2 + |w\_x|^2 $$


The first term is bounded by $\|F\|^2$, and the second, by your assumption on the $L^2$ bound of $w\_x$, is bounded by $\|U\| \|F\|$.


