Coercive bilinear form on Hilbert space

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.

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. $\Box$

• And now that I have opened my eyes, I realize that the proof by contradiction is actually shorter than the direct one. – Julien May 16 '13 at 23:50
• Julien, si tu as une preuve plus simple, laisse-moi le savoir stp! – Stephane May 17 '13 at 14:43

Here is the answer to my own question. Please feel free to comment. Any suggestion is welcome:

Let $\widetilde{H}$ be the closed subspace generated by $D$ and the one-dimensional subspace of $H$. Since $B$ is positive definite, there exists a one-to-one bounded symmetric operator $L$ on $\widetilde{H}$ such that $B(u,v)=\langle Lu,\,v\rangle$. Let $y\in \widetilde{H}$ such that $\langle Ld,\,y\rangle=0$ for all $d\in D$. Then, any element $u$ of $\widetilde{H}$ has a unique representation $u=ay+d$ for some $a\in \mathbb{R}$ and $d\in D$. Let $\widetilde{L}$ be the restriction of $L$ on $D$, i.e., if $v\in \widetilde{H}$ and $Lv=a'y+d'$ for some $a'\in \mathbb{R}$ and $d'\in D$, then $\widetilde{L}v\equiv d'$. The operator $\widetilde{L}$ is such that $B(d_1,\,d_2)=\langle \widetilde{L}d_1,\,d_2\rangle$ for all $d_1,d_2 \in D$. Since $B$ is coercive on $D$, $\widetilde{L}$ is an isomorphism on $D$ by the Lax-Milgram theorem. Thus the image of $D$ under $L$ is a closed subspace of codimension 1 in $\widetilde{H}$. Since $L$ is one-to-one, it follows that $L$ is an isomorphism on $\widetilde{H}$ and thus $B$ is coercive on $\widetilde{H}$.

• Looks okay to me, but how do you justify the existence of $y$ such that $\langle Ld,y\rangle=0$ for all $d\in D$? In my version of the proof (just posted) I made this a lemma. – 75064 May 16 '13 at 22:56
• Dear 75064, thank you. Yes, the lemma is really needed. – Stephane May 17 '13 at 14:44