# Proving $||u_h||_1^2=(f,u_h)$ for mixed finite elements

I'm trying to figure out of something from Claes Johnson's "Numerical solution of partial differential equations by the finite element method" Chapter 11.

Given the problem: Find $(u_h,p_h)\in V_h\times H_h$ such that

$$(\nabla u_h,\nabla v)-(p_h,\operatorname{div}v)=(f,v)\quad\forall v\in V_h\\ (q,\operatorname{div}u_h)=0\quad\forall q\in H_h$$ where all inner products are $L_2$-inner products. Taking $v=u_h$ and $q=p_h$ and adding them together I get $$(\nabla u_h,\nabla u_h)=(f,u_h)$$ and Johnson claim I should get $$||u_h||_1^2=(f,u_h)$$ where $$||w||_1^2=\sum_i\int_\limits\Omega(|w_i|^2+|\nabla w_i|² )\,\mathrm{d}x$$ However, since the inner products are $L_2$-inner products, I do not see from where I should get the first term in the integral. Any guidance appreciated.

• Do not see, where this $L^2$ term should come from. Your derivation is correct. Maybe in the book it is written $|u_h|_1^2$ ($H^1$-seminorm) ? – daw Feb 5 '18 at 13:57
• On the space $H^1_0$, the norm $\|u\|^2:=\int |\nabla u|^2 \, dx$ is equivalent to the standard norm, due to the Poincare inequality $\|u\|_{L^2} \leq C\|\nabla u\|_{L^2}$. Are you certain on the definition of the norm? – Jeff Feb 5 '18 at 16:24
• The book specifically mentions, in relation to this derivation, that the norm is the $H^1$ norm and that the inner products have the form $(\nabla w,\nabla v)=\sum_{i=1}^2\int\limits_\Omega\nabla w_i\cdot\nabla v_i\,\mathrm{d}x$, but I'm starting to doubt that perhaps the book is wrong. More precisely the book states that $||u_h||_1^2=(f,u_h)\leq||f||_{-1}||u_h||_1$. I realize now that I can get to the end result using the Poincaré inequality, but the equal-sign will be incorrect. I guess this must be a mistake in the book then? – sigvaldm Feb 6 '18 at 3:37