Coercivity of a functional on $H^1([0,T]^2)$ Let $I: X=H^1(\mathbb{T}^2,\mathbb{C})\rightarrow\mathbb{R}$ be the functional 
$$ I(u)=\int_{[0,T]^2}\vert \nabla u\vert^2 \ dx + \frac{1}{2} \int_{[0,T]^2}(1-\vert u \vert^2)^2 \ dx - \lambda \int_{[0,T]^2} <i\partial_1 u , u >\ dx,$$
where $\lambda \in (0,\sqrt{2}/2)$, $<,>$ stands for the scalar product of $\mathbb{C}$ seen as $\mathbb{R}^2$ and $\partial_1 u := \frac{\partial u}{\partial x_1}$. I wonder if $I$ is coercive, meaning there exists some $C>0$ s.t. $I(u)\geq C \Vert u \Vert _{X}$ and being $\Vert u \Vert_X ^2 = \int_{[0,T]^2} \vert \nabla u\vert^2 \ dx + \int_{[0,T]^2} \vert u \vert^2 \ dx$.
I have proved that the functional $u\mapsto \int_{[0,T]^2}\vert \nabla u\vert^2 \ dx + \frac{1}{2} \int_{[0,T]^2}(1-\vert u \vert^2)^2 \ dx$ is coercive, but I dont know if it is when the term $- \lambda \int_{[0,T]^2} <i\partial_1 u , u >\ dx$ is also considered. Thanks in advance! Any help is appreciated.
 A: As mentioned in the comments, it isn't true that there is $C > 0$ such that $I(u) \geq C \|u\|_X$. This answer shows that $I(u) \to \infty$ as $\|u\|_X \to \infty$ (though it works for a larger range of $\lambda$ so I am suspicious I have made some mistake).
First, notice that $(1-|u|^2)^2 = 1 - 2|u|^2 + |u|^4$ is a quadratic in $|u|^2$ with a positive coefficient on the highest order term. Hence there is a $K_0 > 0$ such that $(1-|u|^2)^2 \geq 2|u|^2 - K_0$ since $x^2 - 4x + 1$ is bounded below. 
As a result,
$$I(u) \geq \int_{[0,T]^2} |\nabla u|^2 + |u|^2 - \frac12K_0 - \lambda \langle i \partial_1 u, u \rangle dx \geq \|u\|_X^2 - K_1 - \int_{[0,T]^2} \lambda \langle i \partial_1 u, u \rangle dx$$
for some other constant $K_1$.
It now suffices to bound above $\lambda \int_{[0,T]^2} \langle i \partial u, u \rangle dx$. For this we have
\begin{align}
\bigg | \lambda \int_{[0,T]^2} \langle i \partial u, u \rangle dx \bigg | \leq& \lambda \|\partial_1 u\|_{L^2(\mathbb{T}^2)} \|u\|_{L^2(\mathbb{T}^2)}
\\ \leq& \lambda \|u\|_X^2
\end{align}
Substituting this bound gives that 
$$I(u) \geq (1-\lambda) \|u\|_X^2 - K_1$$
which tends to $\infty$ as $\|u\|_X \to \infty$ as long as $\lambda < 1$.

In our chat, you indicated that you'd like a solution for $\lambda \geq 1$ also. Here's a way of extending to that case. First fix such a $\lambda$. A minor adaptation of the above shows that for any constant $\alpha > 0$,
$$I(u) \geq \int_{[0,T]^2} |\nabla u|^2 + \alpha |u|^2 dx - K_\alpha - \lambda \|\nabla u\|_{L^2} \|u\|_{L^2}$$
where $K_\alpha$ is a positive constant depending only on $\alpha$. 
Then by Young's inequality with $\varepsilon$, for all $\varepsilon > 0$ we have that $\|\nabla u\|_{L^2} \|u\|_{L^2} \leq \frac{\varepsilon^2 \|\nabla u\|_{L^2}^2}{2} + \frac{\|u\|_{L^2}^2}{2 \varepsilon^2}$. Picking $\varepsilon = \lambda^{-\frac12}$ and substituting this in, we see that
$$I(u) \geq \int_{[0,T]^2} |\nabla u|^2 + \alpha |u|^2 dx - K_\alpha - \frac{1}{2} \int_{[0,T]^2} |\nabla u|^2 dx - \frac{\lambda^2}{2} \int_{[0,T]^2} |u|^2 dx$$
Since $\alpha$ was arbitrary, we can pick $\alpha(\lambda) = \frac{\lambda^2}{2} + \frac12$ so that the above bound gives 
$$I(u) \geq \frac12 \|u\|_X^2 - K_{\alpha(\lambda)} \to \infty$$
as $\|u\|_X \to \infty$ as desired.
