Is it possible to bound $\|u\|_{H^{1/2}(\partial K)}$ by $\|u\|_{L^2(\partial K)}$ for polynomial $u$?

Question

Let $u\in P^k(K)$ where $P^k$ is polynomial space with maximum degree $\leq k$ and $K$ is bounded convex polyhedron if $K\subset \mathbb{R}^3$ or bounded convex polygonal if $K\subset \mathbb{R}^2$.

Then, is it possible to bound

\begin{equation*} \|u\|_{H^{1/2}(\partial K)}\lesssim \frac{1}{h^\alpha}\|u\|_{L^2(\partial K)} \end{equation*} where $\alpha$ is positive power and $h=|\partial K|$ (side length).

I guess that it should be like

\begin{equation*} \|u\|_{H^{1/2}(\partial K)}\lesssim \frac{1}{h^{1/2}}\|u\|_{L^2(\partial K)} \end{equation*} but I'm not sure since I'm not familiar with the fractional Sobolev space.

EDIT:

As @zaq mentioned, this certainly cannot be bounded by the whole side length. Then what about

\begin{equation*} \|u\|_{H^{1/2}(\partial K)}\leq \sum_{e\subset \partial K}\frac{C(\kappa,n)}{|e|^{1/2}}\|u\|_{L^2(\partial K)} \end{equation*}

where $e$ denotes an edge (or a face) of the domain and $\kappa$ denotes minimum angle of domain and $n$ denotes number of edges (or faces). This certainly satisfies @zaq's example.

Answer 1

No, one cannot control $H^{1/2}$ based only on the size of the boundary: its shape comes into play too. For example, let $K$ be the rectangle $[0,1]\times [0,\epsilon]$ and let $u$ be $0$ on all sides except the top one, where it is $1$. Clearly, $\|u\|_{L^2}=1$.

The $H^{1/2}$ norm of $u$ is equal to $\|\nabla U\|_{L^2(K)}$ where $U$ is the harmonic extension of $u$. Using the Cauchy-Schwarz yields $$ \int_0^1\int_0^\epsilon U_y^2(x,y)\,dy\,dx \ge \int_0^1 \epsilon^{-1} \left(\int_0^\epsilon U_y(x,y)\,dy\right)^2\,dx \\ = \int_0^1 \epsilon^{-1} \,dx = \epsilon^{-1} $$ Therefore, $\|u\|_{H^{1/2}}\to\infty$ as $\epsilon\to 0$, even though the perimeter of $K$ does not change much: it is $2+2\epsilon$.