Let $\Lambda\subseteq\mathbb R^2$ be bounded and open, $$V:=\left\{u\in H_0^1(\Lambda,\mathbb R^d):\nabla\cdot u=0\right\}$$ and $H$ denote the completion of $V$ with respect $\left\|\;\cdot\;\right\|_{L^2(\Lambda,\:\mathbb R^2)}$. Note that $V=H\cap H_0^1(\Lambda,\mathbb R^2$. Let $\operatorname P_H$ denote the orthogonal projection of $L^2(\Lambda,\mathbb R^2)$ onto $H$, $$B(u,v):=-\operatorname P_H(u\cdot\nabla)v\;\;\;\text{for }u,v\in V$$ and $$B(u):=B(u,u)\;\;\;\text{for }u\in V.$$
Can we show that $B$ is Fréchet differentiable considered as a mapping (a) from $V$ to $H$ and (b) from $V$ equipped with the seminorm induced from $L^2(\Lambda,\mathbb R^2)$ to $H$?
$B:V\rightarrow H$ is a quadratic form between normed spaces.
In case (a), when $V$ is equipped with the $H^1$-norm, then $B$ is continuous. Further just using that $B$ is quadratic yields $$B(u+h)-B(u) - 2 B(u,h) = B(h,h) = o(\Vert h\Vert_{H^1})$$ and thus $B$ is Fréchet-differentiable with differential $dB_u(h)=2B(u,h).$
In case (b), when $V$ carries the $L^2$-norm, then $B$ even fails to be continuous, so it cannot be Fréchet-differentiable.