Estimate for a specific differential operator Denote $\mathbb{T}$ the 1-torus and let  $G : H^1(\mathbb{T}) \to H^1(\mathbb{T})$  be Frechet differentiable such that $G(0) = 0$ and consider the operator $$L = \sqrt{1-\Delta}.$$ 
Can we prove that there exists $C>0$ such that 
$$\|LG(u)\|_{L^2} \leq C\|DG(u) Lu\|_{L^2}, \quad \forall u \in H^1(\mathbb{T}).$$
Example: For the case where  $G$ is linear, we see clearly that $C =1.$ 
Thank you for any hint. 
 A: No.  Pick any $v \in H^1(\mathbb{T})$ such that $Lv \neq 0$ (it's not hard to find some such $v$ using the Fourier transform and the form of $L$ as a Fourier multiplier).  Define the map $G: H^1(\mathbb{T}) \to H^1(\mathbb{T})$ via $G(u) = v$.  In other words, $G$ is the constant map, and as such is smooth.  One readily computes the $DG(u) = 0$.  So, if the estimate you're asking for were to hold then we would have that 
$$
0 < \Vert L v \Vert_{L^2} = \Vert L G(u) \Vert_{L^2} \le C \Vert DG(u) Lu \Vert_{L^2} =C\Vert 0v \Vert_{L^2}=0,
$$
which is a contradiction.
EDIT (after post was edited):
It's not true for all linear maps $G$.  Consider the linear map $G : H^1(\mathbb{T}) \to H^1(\mathbb{T})$ given by 
$$
Gu(x) = \hat{u}(0) + \hat{u}(1),
$$
in other words $Gu$ is the constant function on $\mathbb{T}$ with value given by the sum of these two Fourier coefficients.  If you don't want to work over $\mathbb{C}$, then just take the real part of the right side to get a real map.
Then 
$$
\Vert Gu \Vert_{H^{1/2}} \le  | \hat{u}(0)| + |\hat{u}(1)| \le C \Vert u \Vert_{L^2} \le C \Vert u \Vert_{H^{1/2}},
$$
which shows that $G$ is a bounded linear operator.  Then $G(0)=0$ and $DG = G$.  
Now note that 
$$
DG(u) Lu = G Lu = \widehat{Lu}(0) + \widehat{Lu}(1) = \hat{u}(0) + \sqrt{1+ 1^2} \hat{u}(1) = \hat{u}(0) + \sqrt{2} \hat{u}(0),
$$
so if we pick any $u$ such that 
$$
\hat{u}(0) = \frac{\sqrt{2}}{\sqrt{2}-1} \text{ and } \hat{u}(1) = -\frac{1}{\sqrt{2}-1}
$$
then 
$$
DG(u) Lu = \hat{u}(0) + \sqrt{2}\hat{u}(1) = 0
$$
but 
$$
Gu =  \hat{u}(0) + \hat{u}(1) = 1 \text{ and so } LGu = 1.
$$
We then get the same contradiction to the proposed inequality as above.
