I have this theorem and in the first part of the proof , I don't understand why $d_z f(\theta_1+\theta_2)=\theta_1$ ?
Theorem $5.1$. Suppose that $U$ is a neighborhood of $\theta$ in a Hilbert space $H$ and that $f\in C^2(U,\mathbb{R}^1)$. Assume that $\theta$ is the only critical point of $f$ and that $A=d^2f(\theta)$ with kernel $N$. If $0$ is either an isolated point of the spectrum $\sigma(A)$ or not in $\sigma(A)$, then there exist a ball $B_\delta$, and a $C^1$ mapping $h:B_\delta\cap N\to N^\perp$ such that $$ \tag{5.1}f\circ\phi(z+y)=\frac12(Az,z)+f(h(y)+y),\quad\forall x\in B_\delta, $$ where $y=P_Nx$, $z=P_{N^\perp}x$, and $P_N$ is the orthogonal projection onto the subspace $N$.
Proof. $1$. Decomposing the space $H$ into $N\oplus N^\perp$, we have $$ d_zf(\theta_1+\theta_2)=\theta_1\quad(\theta_1=P_{N^\perp}\theta, \theta_2=P_N\theta), $$ and $$ d_z^2f(\theta_1+\theta_2)=A|_{N^\perp}. $$ Because of the implicit function theorem, there is a function $h:B_\delta\cap N\to N^\perp$, $\delta\gt0$, such that $$ d_zf(y+h(y))=\theta_1 $$ Let $u=z-h(y)$, and let $$ \begin{align} &F(u,y)=f(z+y)-f(h(y)+y)\tag{5.2}\\ &\quad\qquad F_2(u)=\frac12(Au,u).\tag{5.3} \end{align} $$ Then we obtain $$ F(\theta_1,y)=0\\ d_uF(\theta_1,y)=d_zf(h(y)+y)=\theta_1,\\ d_u^2F(\theta_1,\theta_2)=d_z^2f(\theta)=A|_{N^\perp}. $$
Thank you.