Why $\left.\frac{d }{d \varepsilon}I(\bar u+\varepsilon\varphi)\right|_{\varepsilon=0}=0$?

Question

Let $\Omega \subset \mathbb R^n$ a bounded open set with Lipschitz boundary. Let $\bar u\in u_0+W_0^{1,2}(\Omega )$ s.t. $$I(\bar u)=\inf\left\{\frac{1}{2}\int_\Omega |\nabla u|^2\ :\ u\in u_0+W_0^{1,2}(\Omega )\right\}.$$ I recall that $u\in u_0+W_{0}^{1,2}(\Omega )$ if $u|_{\partial \Omega }=u_0$. I want to show that $$\int_\Omega \left<\nabla \bar u,\nabla \varphi\right>=0,$$ for all $\varphi\in W_0^{1,2}(\Omega )$. I don't understand the proof :

Let $\varepsilon\in \mathbb R$ and $\varphi\in W_0^{1,2}(\Omega )$. In particular, $\bar u+\varphi \varphi\in u_0+W_0^{1,2}(\Omega )$ and thus $$I(\bar u)\leq I(\bar u+\varepsilon\varphi)=\frac{1}{2}\int_\Omega |\nabla u+\varepsilon\nabla \varphi|^2=I(\bar u)+\varepsilon\int_\Omega \left<\nabla \bar u,\nabla \varphi\right>+\varepsilon^2I(\varphi),$$ and thus $$0=\lim_{\varepsilon\to 0}\frac{I(\bar u+\varphi\varepsilon)-I(\bar u)}{\varepsilon}=\int_\Omega \left<\nabla \bar u,\nabla \varphi\right>.$$

Question : I understand why $\lim_{\varepsilon\to 0}\frac{I(\bar u+\varphi\varepsilon)-I(\bar u)}{\varepsilon}=\int_\Omega \left<\nabla \bar u,\nabla \varphi\right>$ but not why is it $0$.

Answer 1

If $\bar u$ is a minimizer of $I$, then $\bar u$ is a stationary point of $I$.

Answer 2

The limit is just the directional derivative of $I$ at $\bar u$ in direction $\phi$. This is $=0$ since $I$ attains its minimum there.