I have this theorem with a part of the proof:

$\quad$ Let $V$ be a Hilbert space, $U$ an open neighborhood of $u\in V$, and let $\varphi\in C^2(U,\mathbf R)$. Define implicity the linear operator $L:V\to V$ by $$(Lv,w)=\varphi''(u)(v,w).$$ Then $L$ is self-adjoint and we shall identify $L$ with $\varphi''(u)$. If $\varphi''(u)$ is a Fredholm operator, $V$ is the orthogonal sum of $R(\varphi''(u))$ and $\ker(\varphi''(u))$.
$\quad$ Assume now that $u$ is a critical point of $\varphi$. The Morse index of $u$ is defined as the supremum of the dimensions of the vector subspaces of $V$ on which $\varphi''(u)$ is negative definite. The nullity of $u$ is defines as the dimension of $\ker\varphi''(u)$. Finally, the critical point $u$ will be said to be non-degenerate if $\varphi''(u)$ is invertible.

Theorem $\bf 8.3.\,$ Let $U$ be an open neighborhood of $0$ in a Hilbert space $V$ and let $\varphi\in C^2(U,\mathbf R)$. Suppose that $0$ is a critical point of $\varphi$ with positive nullity and that $L=\varphi''(0)$ is a Fredholm operator, so that $V$ is the orthogonal direct sum of $\ker(L)$ and $R(L)$. Let $w+v$ be the corresponding decomposition of $u\in V$. Then there exists an open neighborhood $A$ of $\,0$ in $V$, an open neighborhood $B$ of $\,0$ in $\ker(L)$, a local homeomorphism $h$ from $A$ into $U$, and a function $\hat\varphi\in C^2(B,\mathbf R)$ such that $$h(0)=0,\quad\hat\varphi'(0)=0,\quad\hat\varphi''(0)=0$$ and $$\varphi(h(u))=(1/2)(Lv,v)+\hat\varphi(w)$$ on the domain of $h$.

Proof. $1)$ Let $Q:V\to V$ be the orthogonal projection onto $R(L)$. By the implicit function theorem, we can find $r_1>0$ and a $C^1$-mapping $$g:B(0,r_1)\cap\ker L\to R(L)$$ such that $g(0)=0,$ $g'(0)=0$ and $$Q\nabla\varphi(w+g(w))=0.\tag{13}$$ Let us define $\hat\varphi$ on $B=B(0,r_1)\cap\ker L$ by $$\hat\varphi(w)=\varphi(w+g(w))$$ so that, by direct computation and $(13)$, $$\nabla\hat\varphi(w)=(I-Q)\nabla\varphi(w+g(w))$$ and $$\hat\varphi''(w)=(I-Q)\varphi''(w+g(w))(Id+g'(w)).$$ In particular $$\nabla\hat\varphi(0)=(I-Q)\nabla\varphi(0)=0.$$

And I have this question:

why $\nabla\hat{\varphi}(w)=(I-Q) \varphi(w+g(w))$ ?

Please help me.

Thank you.

  • $\begingroup$ do you mean: why $\nabla\hat{\varphi}(w)=(I-Q) \nabla\varphi(w+g(w))$? $\endgroup$ – robjohn Jan 6 '14 at 13:05
  • $\begingroup$ It would be useful to become acquainted with MathJax. It is preferable to have questions in MathJax versus an image, if for no other reason than it is easier to search MathJax than an image. $\endgroup$ – robjohn Jan 6 '14 at 13:08
  • $\begingroup$ May I ask you whence your image is taken? That book looks familiar to me... $\endgroup$ – Siminore Jan 6 '14 at 13:35
  • $\begingroup$ Critical Point Theory and Hamiltonian Systems by Jean Mawhin Michel Willem $\endgroup$ – Vrouvrou Jan 6 '14 at 13:37

Since $(13)$ says that $Q\nabla\varphi(w+g(w))=0$ and $I\nabla\varphi(w+g(w))=\nabla\varphi(w+g(w))$, we get $$ (I-Q)\nabla\varphi(w+g(w))=\nabla\varphi(w+g(w))-0=\nabla\hat{\varphi}(w) $$

| cite | improve this answer | |
  • $\begingroup$ thank you, but how to find from $\nabla \hat{\varphi}(w)$ ? please $\endgroup$ – Vrouvrou Jan 6 '14 at 13:31
  • $\begingroup$ ohh i understand i tryed to calculate $\nabla \hat{\varphi}$ $\endgroup$ – Vrouvrou Jan 6 '14 at 13:43
  • 1
    $\begingroup$ $\nabla(\varphi(w+g(w)))=(I+\mathcal{J}g)^T(\nabla\varphi)(w+g(w))$ where $\mathcal{J}g$ is the Jacobian of $g$. $\endgroup$ – robjohn Jan 6 '14 at 13:59
  • $\begingroup$ yes but here -Q dont appear, for this i asked the question $\endgroup$ – Vrouvrou Jan 6 '14 at 14:05
  • 1
    $\begingroup$ @Vrouvrou: I think the problem is with ambiguity: I believe $\nabla\varphi(w+g(w))$ in $(13)$ means $\nabla\left(\varphi(w+g(w))\right)=\nabla\hat{\varphi}(w)$ rather than $\left(\nabla\varphi\right)(w+g(w))$. $\endgroup$ – robjohn Jan 6 '14 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.