Coercive bilinear form

Remarks

Example: the inner product $a(u, v) = \ip{u}{v}$ is coercive with $\alpha = 1$, since $a(u, u) = \norm{u}^2$. Non-example: on $\R^2$, the form $a(u, v) = u_1 v_1$ is not coercive, because $a(u, u) = 0$ for $u = (0, 1)$, while $\norm{u}^2 = 1$. The same definition is used on a real normed space. On a complex Hilbert space, a sesquilinear form $a$ is called coercive if $\operatorname{Re} a(u, u) \ge \alpha \norm{u}^2$ for all $u$. Some authors call a coercive form elliptic. In the Lax–Milgram theorem the form is also assumed bounded: $\abs{a(u, v)} \le M \norm{u} \norm{v}$ for some constant $M$ and all $u, v \in H$.