I need to show the two following results. If true, it must be a simple proof but I do not seem to be able to make it work. Thank you in advance.
Consider a continuous symmetric bilinear form $B$ on a real Hilbert space $H$. Let $D$ be a closed subspace of that Hilbert space over which the form is coercive, i.e, there exists $\alpha >0 \in \mathbb{R}$ such that $B(d,d)\geq \alpha \|d\|^2$ for all $d\in D$. Take the new closed subspace of $H$ generated by $D$ and a one dimensional subspace. Assume $B$ is definite positive on the new subspace. Then I need to show that $B$ is also coercive on the new subspace.
It seems like the logical thing to do is to use the Lax-Milgram theorem but I am not exactly sure how to do this.
Best Answer
Here is a proof from scratch, in particular, without the Lax-Milgram.
Lemma. Let $\widetilde H$ be a real vector space with positive definite inner product (completeness not assumed). Suppose that $K$ is a subspace of $\widetilde H$ which is complete. Then any element $v\in \widetilde H$ can be written as $v_1+v_2$ where $v_1\in K$ and $v_2\in K^\perp$.
Proof. Let $v_1$ be the nearest element of $K$ to $v$; the standard proof of the existence of $v_1$ in a Hilbert space applies here as well, because it uses only the completeness of $K$ (e.g., Theorem I.2.5 in A Course in Functional Analysis by Conway). Let $v_2=v-v_1$. If there is $u\in K$ such that $\langle u,v_2\rangle>0$, then $$\|v-(v_1+tu)\|^2=\|v_2- tu\|^2 = \|v_2\|^2 -2t\langle u,v_2\rangle +O(t^2)<\|v_2\|^2$$ for small $t>0$, contradicting the fact that $v_1$ is the nearest element of $K$ to $v$. $\Box$
Apply the lemma with $K=D$ and $\widetilde H$ being the space generated by $D$ and another vector; the inner product is given by $B$. (Note that $D$ is complete with respect to $B$-inner product, because of coercitivity). The lemma provides a nonzero vector $v\in \widetilde H$ such that $v$ is $B$-orthogonal to $D$. Every element of $\widetilde H$ can be written as $d+tv$ for some $d\in D$ and $t\in \mathbb R$. We have $$ B(d+tv,d+tv) = B(d,d)+t^2B(v,v) \tag1$$ and $$\|d+tv\|^2 \le 2 \|d\|^2 +2t^2\|v\|^2\tag2$$ Let $\gamma$ be the smallest of the numbers $\alpha/2$ and $\frac12 \|v\|^2/ B(v,v)$. Then $$ B(d+tv,d+tv) \ge 2\gamma\|d\|^2 + 2 \gamma t^2\|v\|^2\ge \gamma \|d+tv\|^2 \tag3$$ as required. $\Box$