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$\mathcal{H}$ : real Hilbert space with inner product $(\,,\,)$ and norm $||\,||:=(\,,\,)^{1/2}$

Let $D$ be a linear subspace of $\,\mathcal{H}$ and $\mathcal{E}$ : $D\times D\to \mathbb{R}$ a bilinear map. For $\alpha\geq0$ we set

$\mathcal{E}_{\alpha}(u,v):=\mathcal{E}(u,v)+\alpha(u,v)\quad(\,u,v \in D \times D \,)$

Assume $\mathcal{E}(u,u)\geq 0$ for all $ u \in D $ and $\mathcal{E}(u,u)=0 \Leftrightarrow u=0 $

I want to prove that the following assertions are all equivalent:

$(a)$ $\exists K>0$ s.t. $|\mathcal{E}_{1}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}\quad(\,\forall u, \forall v \in D\,)$

$(b)$ $\forall \alpha>0,\,\exists K_{\alpha}>0$ s.t. $|\mathcal{E}_{\alpha}(u,v)|\leq K_{\alpha} \mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2}\quad(\,\forall u,\forall v \in D\,)$

$(c)$ $\forall \alpha>0,\,\exists K_{\alpha}'>0$ s.t. $|\mathcal{E}(u,v)|\leq K_{\alpha}' \mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2}\quad(\,\forall u,\forall v \in D\,)$

$(b) \Rightarrow (a)$ is clear. Suppose $(a) \Rightarrow (c)$ is proved, I think assertion $(c) \Rightarrow (b)$ is similarly proved.

Therefore, I tried to prove assertion $(a) \Rightarrow (c) $　as follows:

By definition $\mathcal{E}_{1}(u,v)=\mathcal{E}(u,v)+(u,v)$, assertion (a) can be transformed as follows:

$|\mathcal{E}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+|(u,v)|$

Also by definition $\mathcal{E}_{\alpha}(u,v)=\mathcal{E}(u,v)+\alpha(u,v)$ and then

$|\mathcal{E}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+ \left| \frac{\mathcal{E}_{\alpha}(u,v)-\mathcal{E}(u,v)}{\alpha} \right|$

$\left(1-\frac{1}{\alpha}\right) |\mathcal{E}(u,v)|\leq K \mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+\frac{1}{\alpha}| \mathcal{E}_{\alpha}(u,v)| $

Since $\mathcal{E}_{\alpha}(u,v)$ is inner product, from Cauchy-Schwarz inequality

$\left(1-\frac{1}{\alpha}\right) |\mathcal{E}(u,v)|\leq K\mathcal{E}_{1}(u,u)^{1/2}\mathcal{E}_{1}(v,v)^{1/2}+\frac{1}{\alpha}\mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2} $

If $\alpha>1$ then $\mathcal{E}_{1}(u,u)\leq \mathcal{E}_{\alpha}(u,u)$ and $\left(1-\frac{1}{\alpha}\right) |\mathcal{E}(u,v)|\leq \left(K+\frac{1}{\alpha}\right)\mathcal{E}_{\alpha}(u,u)^{1/2}\mathcal{E}_{\alpha}(v,v)^{1/2} $

I can't deal with the case of $\,0<\alpha<1$. Please teach me.

ko4
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  • Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need. –  Jan 31 '14 at 00:48
  • It is not true that ${\cal E}_\alpha$ is an inner product. The Cauchy-Schwarz inequality does not apply. If it did, we wouldn't need the weak sector condition! –  Jan 31 '14 at 16:27
  • I made a mistake. – ko4 Feb 01 '14 at 17:05

1 Answers1

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The form ${\cal E}_\alpha(u,v)$ is the sum of two forms: ${\cal E}(u,v)$ which is not symmetric, and $\alpha(u,v)$ which is symmetric. Now, let's exploit Cauchy-Schwarz on the symmetric part to obtain
$$|\alpha(u,v)|\leq \alpha (u,u)^{1/2}\,(v,v)^{1/2}\leq {\cal E}_\alpha^{1/2}(u,u)\,{\cal E}_\alpha^{1/2}(v,v). $$ You should be able to take it from here.