Definition 1 (Functional derivative): Given a function $f \in \mathcal{F}$, the functional derivative of $F$ at $f$, denoted $\frac{\partial F}{\partial f}$, is defined to be the function for which: $$ \begin{aligned} \int \frac{\partial F}{\partial f}(x) \eta(x) d x & =\lim _{h \rightarrow 0} \frac{F(f+h \eta)-F(f)}{h} \\ & =\left.\frac{d F(f+h \eta)}{d h}\right|_{h=0}. \end{aligned} $$
Which we can define $$(\frac{\partial F}{\partial f}, \eta)_{L^2}=\left.\frac{d F(f+h \eta)}{d h}\right|_{h=0}.$$
My question is about finding a $v$ to represent $\frac{\partial F}{\partial f}$, and my example is $$ I_\lambda(u)=\frac{1}{2} \int_M\left|\nabla u\right|^2 d \mu_g-\frac{\lambda}{2 m} \log \left(\int_M e^{2 m u} d \mu_g\right), $$ on $E:=\left\{u \in H^m(M): \int_M u d \mu_g=0\right\}$, here $(M,g)$ is a closed Riemannian manifold.
Then we get $$ \left(\frac{\partial I_\lambda}{\partial u}, \eta\right)_{L^2}=\int_M \nabla u \cdot \nabla \eta - \lambda \eta \frac{e^{2 m u}}{\int_M e^{2 m u} d \mu_g} $$ which equals to $$\tag{1} \int_M (-\Delta u -\lambda \frac{e^{2 m u}}{\int_M e^{2 m u} d \mu_g}) \eta d \mu_g.$$
My problem is: can we find a $v \in E$ such that $$\tag{2} \left(\frac{\partial I_\lambda}{\partial u}, \eta\right)_{L^2}= (v, \eta)_{L^2}. $$
My attempt is $$ I_\lambda(u)=\frac{1}{2} \int_M|\nabla u|^2 d \mu_g-\frac{\lambda}{2 m} \log \left(\int_M e^{2 m u} d \mu_g\right). $$ If we treat $I_\mu^{\prime}(u)$ as a functional from $E$ to $\mathbb{R}$ by $$ \eta \mapsto \int_M \nabla u \cdot \nabla \eta-\lambda \eta \frac{e^{2 m u}}{\int_M e^{2 m u} d \mu_g}, $$ which is linear, so we can apply Riesz representation theorem to get a $v \in E$ such that $$ (v, \eta)_{L^2}=I_\mu^{\prime}(u)\eta, $$ which we can use this $v$ to represent $I_\mu^{\prime}(u)$. Now, recall that $$ \left\|I_\mu^{\prime}(u)\right\|:=\sup _{\|w\| \leq 1}\left\langle I_\mu^{\prime}(u), w\right\rangle, $$ and $$ \left\langle I_\mu^{\prime}(u), v\right\rangle:=\left.\frac{d}{d t} I_\mu(u+t v)\right|_{t=0}, $$ now if this $v$ represents $I_\mu^{\prime}(u)$, $\left\langle I_\mu^{\prime}(u), v\right\rangle$ should be the same as $(\sup _{\|w\| \leq 1}\left\langle I_\mu^{\prime}(u), w\right\rangle)^2$, let's check:
$$ \left\langle I_\mu^{\prime}(u), v\right\rangle= \|v\|_2^{2} $$ and $$ \sup _{\|w\| \leq 1}\left\langle I_\mu^{\prime}(u), w\right\rangle=\sup _{\|w\| \leq 1}\left\langle v, w\right\rangle=\|v\|_2. $$
which means that we can take some $v \in E$ to represent $I_\mu^{\prime}(u)$ in specific computation.
The gap: here the $2-$ norm is not complete on $E$.