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.
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 $$ (I-Q)\nabla\varphi(w+g(w))=\nabla\varphi(w+g(w))-0=\nabla\hat{\varphi}(w) $$