# Inequality involving Sobolev spaces

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.

• Why is the inequality deleted? Nov 13 at 16:50

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 $$\|U\| \|F\|$$.

• why the condition of $|\lambda |> 1$? of your answer, it is valid for all $\lambda \in \mathbb{R}$. Sep 6 at 13:57
• Nothing in your question requires $|\lambda | > 1$; there may be other consideration in whatever book/paper you are reading. But since you didn't provide that info, I cannot answer. Sep 6 at 15:46
• In the thesis of the question he affirms the inequality for $|\lambda |> 1$. I wrote this information. But thanks for the reply. Sep 6 at 16:10