# Sobolev trace operator bounded from below??

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.

• I have no idea why this question is off topic!! Dec 31 '13 at 13:11
• I don't think this is true: take any function which vanishes on $\partial \Omega$ but which doesn't have $0$ $H^1$ norm... Dec 31 '13 at 14:29
• @Guillermo That could well be an answer. Dec 31 '13 at 14:39
• But see this accepted answer: math.stackexchange.com/questions/609361/… In the comments this fact is used. Dec 31 '13 at 14:45
• @michael_carbon I think that what the accepted answer in the post you commented tried to say was that you can find a function defined on $H^1(\Omega)$ whose trace is your desired function (provided it is in $H^{1/2}(\partial \Omega)$) and which has the bound you seek. But your bound is not true in general, it is equivalent to asserting the operator is injective, which is false as others have commented. Dec 31 '13 at 19:20

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 $\partial\Omega$.