Properties of bilinear forms

Question

Let $V$ be a real vector space with norm $\|\cdot\|_V$ and $W$ a closed, linear subspace of $V$. A bilinear form $a\colon V\times V\rightarrow \mathbb{R}$ is called

1. symmetric, if $a(u, v) = a(v, u)$ for all $u, v \in V$,

2. continuous, if a constant $a_\infty$ exists with $$|a(u, v)| \le a_\infty \|u\|_V \|v\|_V \quad \text{ for all $u, v \in V$}\, ,$$

3. $W$-elliptic, if a constant $a_0 > 0$ exists with: $$a(w, w) \ge a_0 \|w\|_V^2 \quad \text{ for all $w\in W$.}$$

Questions:

• don't we also have to assume that $a_\infty > 0$ ?

• why is the second property called continuous?

  Ok, for $u$ constant, and $\varepsilon > 0$ if we set $\delta = \frac{\varepsilon}{a_\infty \| u \|_V}$ then for any $v, \tilde{v} \in V$ with $\|v - \tilde{v}\|_V < \delta$: $$|a(u, v) - a(u, \tilde{v})| = |a(u, v-\tilde{v})| \le a_\infty \|u\|_V \|v-\tilde{v}\|_V < a_\infty \|u\|_V \delta = \varepsilon \, .$$ Is that all there is to it?

• What is the idea behind the third property? Why don't we have something like $$a(w, \tilde{w}) \ge a_0 \|w\|_V^2 \|\tilde{w}\|_V^2 \quad \text{ for all $w, \tilde{w} \in W$?}$$

Answer 1

1. No, there is no need to assume $a_\infty>0$. If the property holds with some $a_\infty$, it also holds with any greater constant. So, we could assume $a_\infty>0$, but don't need to.

2. Recall that a linear functional $f$ is called continuous if there is $M$ such that $|f(x)|\le M\|x\|$ for all $x$. This property does not look like continuity, but by virtue of linearity is equivalent to it. Same story with bilinear forms. First note that continuity everywhere is equivalent to continuity at $(0,0)$. The property $(2)$ clearly implies continuity at $(0,0)$. Conversely, if $(2)$ fails, there is a sequence of pairs $(u_n,v_n)$ for which $B(u_n,v_n)>n\|u_n\|\,\|v_n\|$. Scaling them carefully, you can achieve $(u_n,v_n)\to 0$ without $B(u_n,v_n)$ approaching $0$.

3. Since $a(u,-v)=-a(u,v)$, asking for a lower bound like $a(u,v)\ge a_0\|u\|\|v\|$ is hopeless. The ellipticity makes sure that moderate values of $B(u,u)$ imply a bound on the norm of $u$. This helps in variational arguments. You have to see this property used to understand why it's there.