A question about the derivative of functional.

Question

I'm reading Existence of solutions to a higher dimensional mean-field equation on manifolds, in this paper they defined $$ E:=\left\{u \in H^m(M): \int_M u d \mu_g=0\right\}, $$ and $$\|u\|:=\left(\int_M\left|\Delta_g^{\frac{m}{2}} u\right|^2 d \mu_g\right)^{\frac{1}{2}}.$$ They then found a $u_0$ such that $$ I\left(u_0\right)<0 \quad \text { and } \quad\left\|u_0\right\| \geq 1 . $$ then defined the inner product $$\tag{1} \left\langle I_\mu^{\prime}(u), v\right\rangle:=\left.\frac{d}{d t} I_\mu(u+t v)\right|_{t=0} $$ and the norm $$ \left\|I_\mu^{\prime}(u)\right\|:=\sup _{\|v\| \leq 1}\left\langle I_\mu^{\prime}(u), v\right\rangle . $$

Consider the set of paths $$ P:=\left\{\gamma \in C^0([0,1] ; E): \gamma(0)=0, \gamma(1)=u_0, \gamma(t) \in C^{\infty}(M) \text { for } 0 \leq t \leq 1\right\}. $$ In the last line of page10, they choose $\varphi \in C^{\infty}(\mathbb{R})$ such that $0 \leq \varphi \leq 1$ with $\varphi \equiv 1$ on $[-1, \infty)$ and $\varphi \equiv 0$ on $(-\infty,-2]$. For $n \in \mathbb{N}$ and $u \in E$ set $$ \varphi_n(u):=\varphi\left(\frac{I_{\mu_n}(u)-c_\mu}{\mu_n-\mu}\right), $$

then they defined a sequence $\gamma_n \in P$ and set $$ \tilde{\gamma}_n(t):=\gamma_n(t)-\sqrt{\mu_n-\mu} \varphi_n\left(\gamma_n(t)\right) \frac{I_\mu^{\prime}\left(\gamma_n(t)\right)}{\left\|I_\mu^{\prime}\left(\gamma_n(t)\right)\right\|} \in E \cap C^{\infty}(M), $$ I'm confused about this step, why $I_\mu^{\prime}\left(\gamma_n(t)\right)$ can be viewed as a function, isn't it an operator? In my understanding, it must be accompanied by a test function $v$ such as (1) to be meaningful.

Answer 1

From the definition (1), you have $$ I_\mu'(u) \in E^*$$ for all $u \in E$, i.e., the derivative $I_\mu'$ maps $E$ into $E'$.

Now, there are two possibilies for the final formula.

1. Since $E$ is a Hilbert space, you can identify $E^*$ with $E$. Thus, you have $R^{-1} I_\mu'(u)$ appearing in the final formula, where $R : E \to E^*$ is the Riesz isomorphism. In your case $R$ should be something like $\Delta_g^m$.

2. We use the embedding $E \hookrightarrow L^2(M) \hookrightarrow E^*$, i.e., every $v \in E$ can be viewed as a functional $$ E \ni w \mapsto (w, v)_{L^2(M)}.$$ Now, you could ask whether there exist $v \in E$ with $$ (w,v)_{L^2(M)} = I_\mu'(u) w \qquad\forall w \in E.$$ In this case, we can say $I_\mu'(u) \in E$.

I do not know which interpretation is used in the paper (I did not look into it), but the difference is crucial, since in "1.", you have an additional $\Delta_g^m$ sitting around.