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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. PI
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1 Answers1

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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.

Glitch
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  • This is the trivial case... There is another assumption on $G$ that I forgot to mention. I've just edited the question. Thank you – A. PI Oct 16 '18 at 18:21
  • Okay, you've excluded constants, but the result is false even for linear maps. See my edit. – Glitch Oct 16 '18 at 19:09
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    @A.PI I have fully answered your posted question twice and you have neither upvoted nor accepted my answers. Instead you have changed your question twice. You seem to be new on math.SE, so allow me to offer some advice: we are not mind readers. If you have pertinent information to a problem, then post it from the beginning. Constantly changing the problem results in nothing but frustration and an unwillingness from the community to help you. – Glitch Oct 17 '18 at 01:22
  • But you answered the question that I wanted to ask with the assumption $G(0) =0.$ I might put the original Problem in a new question. Thank you again – A. PI Oct 17 '18 at 09:19
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    @A.PI The current version of the post says you want $G(0)=0$. Any linear map will obviously satisfy this. If you want $G(0) \neq 0$, then you can combine my two examples and consider an affine map. – Glitch Oct 17 '18 at 13:10
  • I mean you answered the version that I really wanted. – A. PI Oct 17 '18 at 13:19