Define $\nabla_h u=\frac{\tau_h u-u}{h}$, where $\tau_h u=u(x_1+h,x_2,...,x_n)$. We use $\|\cdot\|$ to denote the norm in $L^2(R^n)$.
We have the following lemma:
Lemma 1. If $u$ belongs to $H^1(R_+^n)$, then $\|\nabla_hu\|\leq\|\frac{\partial u}{\partial x_1}\|$.
Then we will have the following two statements:
- From $\|\nabla_h u\|\leq\|\frac{\partial u}{\partial x_1}\|$, we can deduce that $\tau_hu$ is a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$.
My question is:
Why is $\tau_hu$ a continuous linear operator from $H^1(R_+^n)$ to $L_2(R_+^n)$? I know that continuity of the operator can be deduced from the boundedness of the operator, but I don't know why the operator $\tau_h u$ is bounded.
I think that boundedness of $\tau_hu$ means that there exists a constant $C$ such that $$\sup\frac{\|\tau_hu\|_{H^1(R_+^n)}}{\|u\|_{L_2(R_+^n)}}\leq C$$ But I can't figure out how to prove this.
Regarding the first question:
$$\sup_{u\in H^1} \frac{\|\tau_h u\|_{L_2}}{\|u\|_{H^1}}=\sup_{u\in H^1} \frac{\|\tau_h u-u+u\|_{L_2}}{\|u\|_{H^1}}\leq\sup_{u\in H^1} \frac{\|\tau_h u-u\|_{L_2}+\|u\|_{L_2}}{\|u\|_{H^1}}\leq\sup_{u\in H^1} \frac{h\cdot\|\tfrac{\partial u}{\partial x_1}\|_{L_2}+\|u\|_{L_2}}{\|u\|_{H^1}}\leq\max(1,h)\sup_{u\in H^1} \frac{\|\tfrac{\partial u}{\partial x_1}\|_{L_2}+\|u\|_{L_2}}{\|u\|_{H^1}}\leq\\\max(1,h)\sup_{u\in H^1} \frac{\|u\|_{H^1}}{\|u\|_{H^1}}=\max(1,h)$$
Here I have used the triangle inequality and Lemma 1. Note that in your definition of the boundedness the norms are swapped.
Regarding your second question, it seems to be the limit for $h\to0$.