Let $\Omega$ be a domain in $\mathbb{R}^n$ with smooth boundary $\partial\Omega.$ Is the trace operator $$T:H^1(\Omega) \to H^{\frac 1 2}(\partial\Omega)$$ bounded from below: $$|Tu|_{H^{\frac 1 2}} \geq C|u|_{H^1}\quad\text{for all $u \in H^1$}$$ How to show this fact?
My attempt: Maybe to use the open mapping theorem, but this requires a surjective bounded linear operator which is invertible, but $T$ is not invertible, it only has a partial inverse. So I cannot use it.
This cannot be true, because, as @Guillermo pointed out in the comments, it would implie that the trace operator $T$ is injective. This is obviously not true, as one can see considering for example, any two functions $C^1(\overline{\Omega})$ such that their $H^1(\Omega)$ norm are distinct, but they have the same value in $\partial\Omega$.