# Why is the solution in FEM bounded by zero?

Consider the following problem: $$-\nabla^2 u = f,$$

Referencing to this post: FEM, we write the problem in variational form

I'm assuming Dirichlet boundary conditions here): Find $$u\in H^1_0(\Omega)$$ such that $$a(u,v):= \int_\Omega \nabla u\cdot \nabla v \,dx = \int_\Omega fv\,dx \qquad\text{for all }v\in H^1_0(\Omega).$$ The important property here is that $$a(v,v) = \|\nabla v\|_{L^2}^2 \geq c \|v\|_{H^1}^2 \qquad\text{for all }v\in H^1_0(\Omega). \tag{1}$$ (This follows from Poincaré's inequality.)

Now the classical finite element approach is to replace the infinite-dimensional space $$H^1_0(\Omega)$$ by a finite-dimensional subspace $$V_h\subset H^1_0(\Omega)$$ and find $$u_h\in V_h$$ such that $$a(u_h,v_h):= \int_\Omega \nabla u_h\cdot \nabla v_h \,dx = \int_\Omega fv_h\,dx \qquad\text{for all }v_h\in V_h.\tag{2}$$ The important property here is that you are using the same $$a$$ and a subspace $$V_h\subset H^1_0(\Omega)$$ (a conforming discretization); that means that you still have $$a(v_h,v_h) \geq c \|v_h\|_{H^1}^2 >0 \qquad\text{for all }v_h\in V_h. \tag{3}$$

How can I be sure that $$\|v_h\|_{H^1}^2 >0$$ ?

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