Let $M$ be a compact Riemann manifold. Consider the trace map $T:H^1(M) \to H^{\frac 12}(\partial M)$. Is it always the case that $$|Tu|_{H^{\frac 12}(\partial M)} \leq C\lVert \nabla u \rVert_{L^2(M)}$$ i.e., is the fractional seminorm of the trace always bounded by the $H^1$ seminorm of the function?
Edit: I use the Gagliardo norm for the fractional space.
This should follow from the nonhomogeneous trace inequalty $$ \vert T u \vert_{H^{1/2} (\partial M)} \le C \lVert \nabla u \rVert_{L^2} + \lVert u \rVert_{L^2}, $$ and and from the classical Poincaré inequality on (M) $$ \inf_{c \in \mathbb{R}} \lVert u - c \rVert_{L^2} \le \lVert \nabla u \rVert_{L^2}. $$
If you forget the $L^2$ norm on both sides, then yes, that's true. To see this in the simple setting of the half space, use Fourier transforms.
Let $f$ be a smooth and fast decaying function on $\mathbb{R}^n$. By the Fourier inversion formula, we have
\begin{equation} f(x',0) = \int_{\mathbb{R}^n} \hat{f}(\xi',\xi_n) e^{ix' \cdot \xi'} d\xi \end{equation}
where the superscript $'$ refers to the $n-1$ first coordinates and the index $n$ refers to the last.
Using Fubini, it follows that
\begin{equation} f(x',0) = \int_{\mathbb{R}^{n-1}} g(\xi') e^{ix' \cdot \xi'} d\xi' \end{equation}
where $g(\xi') := \int_{\mathbb{R}} \hat{f}(\xi',\xi_n) d\xi_n$.
The last identity tells us that $g$ is no less than the $n-1$ dimensional FOurier transform of the trace of $f$ on the half space ${x_n = 0}$. Thanks to Cauchy-Schwarz, we can bound $g$ by
\begin{equation} |g(\xi')|^2 \leq \int_{\mathbb{R}} |\xi|^{-2} d\xi_n \int_{\mathbb{R}} |\xi|^2 |\hat{f}(\xi',\xi_n)|^2 d\xi_n . \end{equation}
(Be careful to distinguish between $\xi_n, \xi'$ and $\xi$).
The first integral may be computed explicitly : one has
\begin{equation} \int_{\mathbb{R}} |\xi|^{-2} d\xi_n = \frac{C}{|\xi'|} , \end{equation}
with probably $C = \frac{\pi}{2}$, coming from the evaluation of the arctangent integral.
Thus, one arrives at the "pointwise" inequality
\begin{equation} |g(\xi')|^2 \leq \frac{1}{|\xi'|} \frac{\pi}{2} \int_{\mathbb{R}} |\xi|^2 |\hat{f}(\xi',\xi_n)|^2 d\xi_n . \end{equation}
Multiply both sides by $|\xi'|$, integrate over $\xi'$ and you are done (with, in addition, a neat and useless constant).
The case of bounded and smooth (Lipschitz should be enough) domains is treated as usual with partitions of unity and diffeomorphisms. Details are left to the reader. :)
