I am trying to prove the following fact:
Suppose $\rho$ is a $C^3$ function of real Hilbert spaces $E$ into $F$. Let $\rho''$ denote its second Frechet derivative at some point $x \in E$. We have $\rho''hk=\sigma\langle h, k\rangle$ for all $h$ and $k\in E$ and constant $\sigma$. Then $\rho=A|x-x_0|^2+B$ for some constants $A, B$.
It is clear that $D_x(\rho'k)=D_x(\sigma\langle id, k\rangle)$ (sorry for the abuse of notation). But I cannot see why this suggests $\rho'k=\sigma\langle x- x_0, k\rangle$ for some $x_0$. I know that if for all elements $x$ in a connected open set $S$, $D_xf=D_xg$, then $f(x)=g(x)+(f(x_0)-g(x_0))$ for some fixed $x_0$ in $S$. But in the case of $\rho'k$, we have $D_x(\rho'k)=D_x(\sigma\langle x, k\rangle)$. Then we have $$\forall k \in E, x \in S, D_x(\rho)(k)=\sigma\langle x, k\rangle +D_{x_0}(\rho)(k)-\sigma\langle x_0, k\rangle$$ How do we get rid of the term $D_{x_0}(\rho)(k)$? I guess I must did something super incorrect.