I was reading the following example from the book Methods in Nonlinear Analysis (Zhang, Springer) on page 10: First, everything was fine:
Example 2. Let $X = C^1(\overline \Omega, \mathbb R^N)$, $Y = \mathbb R^1$. Suppose that $g \in C^2(\overline \Omega \times \mathbb R^n,\mathbb R^1)$. Define $$ f(u) = \frac 1 2 \int_{\Omega } |\nabla u|^2 + \int_\Omega g(x, u(x)) $$ as $u \in X$. By definition, we have $$ f'(u) \cdot \varphi = \int_\Omega [\nabla u(x)\nabla \varphi (x) + g'_u(x, u(x))\varphi (x)]dx , $$ and $$ f''(u)(\varphi , \psi) = \int_\Omega [\nabla \psi(x)\nabla \varphi (x) + g''_{uu}(x, u(x))\varphi (x)\psi(x)]dx . $$ With some additional growth conditions on $g''_{uu}:$ $$ |g''_{uu}(x, u)| \le a(1 + |u|^{4/(n-2)} ), \ \ \ a>0,\ \forall u \in \mathbb R^N , $$ $f$ is twice differentiable in $H^1_0(\Omega ,\mathbb R^N )$.
Then, suddenly, I got totally lost:
As an operator from $H^1_0(\Omega ,\mathbb R^N )$ into itself, \begin{equation} f''(u) = id + (-\Delta)^{-1} g''_{u}(\cdot , u(\cdot )) . \end{equation} is self-adjoint, or equivalently, the operator $-\Delta+g''_{uu}(x, u(x))\cdot\ $ defined on $L^2$ is self-adjoint with domain $H^2 \cap H^1_0(\Omega ,\mathbb R^N )$.
I just don't understand anything about the second part of the example. For example, where is the $(-\Delta)^{-1}$ in the last equation from? I know it is from integration by parts, but shouldn't $f''(u)$ be an integral?
Please help make it clear. Thanks in advance.
