We consider a functional $L+J$ on a Hilbert space $\cal H$, i.e., $\forall f \in {\cal H},L(f)+J(f): {\cal H} \mapsto \mathbb{R}$. Both $L$ and $J$ are continuous convex functionals. $J(f) \ge 0$ and $J(0)=0$. We have
$${\cal H}_0 = \left\{ f \mid J(f)=0, f \in {\cal H} \right\}$$
is a closed subspace. If $L$ is coercive on ${\cal H}_0$, is $L+J$ coercive on $\cal H$? I.e., can we prove or disaprove that $\forall f \in {\cal H}_0$ if we have $\|f\| \to \infty$, we have $L(f) \to \infty$, then we have $\forall f \in {\cal H}$ when $\|f\| \to \infty$, $L(f)+J(f) \to \infty$? Or do we need some conditions on $J$ to make the conclusion holds, e.g., $J(f)$ is coercive for $f \in {\cal H} \ominus {\cal H_0}$
Take ${\cal H}=\Bbb{R}^2$ and $L(x,y)=x^2-y$, $J(x,y)=|y|$. Then ${\cal H}_0$ is the $x$-axis and all your assumptions hold (including that $J$ is coercive on ${\cal H}_0^\bot$), but for $y\ge 0$ we have $L(0,y)+J(0,y)=0\not\to+\infty$ as $y\to+\infty$.