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Consider the inhomogeneous boundary value problem on the infinite strip $(x,y)\in \mathbb{R}\times [0,1]$ defined by

$$\begin{cases}\partial_{x}u + \partial_{y}v=f & {(x,y)\in \mathbb{R}\times (0,1)}\\ \partial_{y}u-\partial_{x}v=g & {(x,y)\in\mathbb{R}\times(0,1)} \\ v|_{y=0}=u|_{y=1}=0 & {}\end{cases}\tag{BVP} $$

Define a Banach space of analytic functions $Y_{\sigma,s}(\mathbb{R})$ by the norm

$$\|U\|_{Y_{\sigma,s}}^{2} := \int_{\mathbb{R}}e^{2\sigma|k|}(1+k^{2})^{s}|\widehat{U}(k)|^{2}dk$$

Let $H^{r}$ denote the usual $L^{2}$ based Sobolev space with regularity parameter $r$. Now define the Banach space $$\mathbb{K}_{\sigma,s}^{r} := H^{r}((0,1); Y_{\sigma,s}(\mathbb{R}))$$

In the boundary value problem (BVP), we impose the condition $f,g\in \mathbb{K}_{\sigma,s}^{0}\cap\mathbb{K}_{\sigma,s-1}^{1}$, which we equip with the max norm

$$\|f\|_{\mathbb{K}_{\sigma,s}^{0}\cap \mathbb{K}_{\sigma,s-1}^{1}} := \max\{\|f\|_{\mathbb{K}_{\sigma,s}^{0}}, \|f\|_{\mathbb{K}_{\sigma,s-1}^{1}}\}$$

One can show using the Fourier transform/Fourier series the following result:

Claim. If $f,g\in \mathbb{K}_{\sigma,s}^{0}\cap \mathbb{K}_{\sigma,s-1}^{1}$, then the solutions $u,v$ to (BVP) belong to $\mathbb{K}_{\sigma,s+1}^{0}\cap\mathbb{K}_{\sigma,s}^{1}$.

My interest lies in the image of the trace operator $v\mapsto v|_{y=1}$. In Appendix A of the paper which I'm reading, the authors make reference to "the trace theorem" at several points to assert that $v|_{y=1}\in Y_{\sigma,s}$. Actually, they make the general claim (see p. 48) that the trace operator is a bounded linear operator

$$\mathbb{K}_{\sigma, s}^{0}\cap \mathbb{K}_{\sigma,s-1}^{1} \rightarrow Y_{\sigma,s-\frac{1}{2}}$$

However, they neither include the statement of a specific trace theorem in the paper, nor include a reference to a specific trace theorem.

Reading between the lines, I'm guessing they have the classical Sobolev trace theorem in mind and an argument something like as follows. If $u\in \mathbb{K}_{\sigma,s+1}^{0}\cap\mathbb{K}_{\sigma,s}^{1}$, then by complex interpolation

$$u\in \mathbb{K}_{\sigma,s+\frac{1}{2}}^{\frac{1}{2}} = H^{\frac{1}{2}}((0,1); Y_{\sigma,s+\frac{1}{2}})$$

This is too crude, though, as $H^{\frac{1}{2}}\not\subset C_{b}^{0}$, where $C_{b}^{0}$ is the space of bounded continuous functions. The only argument that I can think of to make their claim is the following real interpolation equality

$$(L^{2}, H^{1})_{\theta,q} = B_{2,q}^{\theta}, \quad 0<\theta<1, 1\leq q\leq \infty$$

Taking $\theta=\frac{1}{2}$ and $q=1$, we have $(L^{2}, H^{1})_{1,\frac{1}{2}} = B_{2,1}^{\frac{1}{2}}$. So "morally",

$$(\mathbb{K}_{\sigma,s+1}^{0}, \mathbb{K}_{\sigma,s}^{1})_{\frac{1}{2},1} = B_{2,1}^{\frac{1}{2}}((0,1); Y_{\sigma,s+\frac{1}{2}})$$

Since $B_{2,1}^{\frac{1}{2}} \subset C_{b}^{0}$, it follows that $v|_{y=1} \in Y_{\sigma,s+\frac{1}{2}}$.

I tried proving $(\mathbb{K}_{\sigma,s+1}^{0}, \mathbb{K}_{\sigma,s}^{1})_{\frac{1}{2},1} \subset B_{2,1}^{\frac{1}{2}}Y_{\sigma,s+\frac{1}{2}}$ using the definition of the real interpolation K-functional, but it's not clear to me how to arrive at the desired result.

Question 1. Is it true that the $(\mathbb{K}_{\sigma,s+1}^{0}, \mathbb{K}_{\sigma,s}^{1})_{\frac{1}{2},1}\subset B_{2,1}^{\frac{1}{2}}Y_{\sigma,s+\frac{1}{2}}$?

I consider real interpolation to be a nontrivial tool, and I would prefer to avoid and thus be very grateful for a simpler argument

Question 2. Is there a non-interpolation argument for arriving at the conclusion that $v|_{y=1}\in Y_{\sigma,s+\frac{1}{2}}$?

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  • $\begingroup$ Just wondering, do you mean $\partial_y u - \partial_x v = g$? $\endgroup$
    – user78249
    Mar 27, 2017 at 1:44
  • $\begingroup$ @james.nixon Yes, thank you for catching that. I have corrected the statement of (BVP). $\endgroup$ Mar 27, 2017 at 2:09

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