# How to prove that 1 is not an eigenvalue of $T'(x)$?

Given a compact continuous operator $T$ from a Banach space $V_1$ to itself and $T$ maps a convex closed bounded set $\mathcal{B}$ into itself, how can we show that 1 is not an eigenvalue of $T'(x)$ (we do not have a closed form expression of $T'$)? Also how can we show that $x \neq T(x)$ for every $x$?

This problem arises from a study of uniqueness of solutions to a boundary value problem.

More details: Let $V$ be the following Hilbert space: $$V=\lbrace v \in L^2(\Omega)^3, {\rm div} v \in L^2(\Omega), curl v \in L^2(\Omega)^3\ \qquad {\rm and } \qquad v\cdot n=0 on \partial \Omega \rbrace$$ where $\Omega$ is a bounded open set in $\mathbb{R}^3$ $\Omega$ connected. We are given the following BVP: \begin{equation*} \begin{align} curl b =\lambda(x) (b+B_0) + \nabla p & \\\ div b=0 & \\\ b\cdot n =0 & \qquad {\rm on} \qquad \partial \Omega \end{align} \end{equation*} where $n$ is the outgoing normal to $\partial \Omega$ and $B_0 \in L^2(\Omega)^3$ be a potential field. $\lambda$ is a function on $\Omega$. Then the operator $T$ assigns to every $j \in V_1$ ($V_1=\lbrace v \in V \mid div v =0\rbrace$) the unique solution of the above BVP where $\lambda$ is a solution of the problem: $$-\epsilon \Delta \lambda + (j+B_0) \cdot \nabla \lambda =0 \qquad {\rm in }\qquad \Omega$$ $$\lambda = \sigma \qquad {\rm on} \qquad \partial \Omega.$$ PS: Maybe $T$ is not even linear? Note that I assumed in the begining that $T$ is linear, then I figured that this needs not to be the case, my appologies.

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If $F$ is linear, then $F'(x) = F$. Isn't that a closed form expression? –  Dirk Dec 5 '11 at 9:27
I think you should prove that your operator does admit a fixed point, that is $F'(x)\ne x$. You should check the hypothesis of the Banach-Caccioppoli theorem. –  Jon Dec 5 '11 at 9:58
How is that $F'(x)=F$? –  user16974 Dec 5 '11 at 9:59
The derivative of a linear map is a linear map, is it not? –  Yemon Choi Dec 5 '11 at 10:10
Yes the derivative of a linear map is a linear map. But how does this help in showing that 1 is not an eigenvalue of $F'(x)$? –  user16974 Dec 5 '11 at 10:12