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Let $\mathcal{H}$ be real Hilbert space, $b(\cdot,\cdot)$ is a bi-linear form on $\mathcal{H}$ and satisfies \begin{equation} \begin{split} \sup_{||v||=1}\sup_{||u||=1}b(u,v)\leq C_0\cdots(1)\\ \inf_{||v||=1}\sup_{||u||=1}b(u,v)\geq C_1\cdots(2)\\ \inf_{||u||=1}\sup_{||v||=1}b(u,v)\geq C_2\cdots(3) \end{split} \end{equation} where $C_0,C_1,C_2$ are positive constants, then the problem asks to show that there exists a unique bounded linear operator $B\in\mathcal{L}(\mathcal{H})$ such that $b(u,v)=(u,Bv)$ and $B$ have bounded inverse.

[Observation] The construction of $B$ is straightforward from (1) and Riesz Representation theorem, the problem is to show that $B$ thus defined is both injective and surjective. The injective part can be shown quite easily while the sujective part was still under consideration. I was wondering if there is some $\delta>0$ such that (2),(3) implies $$|b(u,u)|\geq \delta ||u||^2$$ then we can solve the problem by simply applying Lax-Milgram theorem, but it does not come to me very quickly....

Roy Han
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  • Try a the right shift on $\ell_2$, may be this will give a counterexample to surjectivity – Norbert Aug 18 '13 at 15:32
  • @Norbert Thank you for your comment but I am not sure if I had got your idea...Do you mean $b(x,y):=\sum_{i\geq 2}u_i \bar{v_{i-1}}$? – Roy Han Aug 19 '13 at 12:38

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As you said, (1) gives you the existence of $B$. Now you can look at (2), but written as $$ C_1\leq\inf_{\|v\|=1}\sup_{\|u\|=1}\langle u,Bv\rangle=\inf_{\|v\|=1}\|Bv\|. $$ So $\|Bv\|\geq C_1$ for all $v$ with $\|v\|=1$, from where we deduce that $$ \|Bv\|\geq C_1\,\|v\| $$ for all nonzero $v$. This implies that $B$ is injective, and that its inverse (if it exists) is bounded.

Now (3) gives you the same information that (2) but for $B^*$. So $B^*$ is injective, which means that $$ \overline{\mbox{ran}(B)}=\ker(B^*)^\perp=\{0\}^\perp=\mathcal H. $$ So the range of $B$ is dense, and we can define $B^{-1}$ on it. As $B^{-1}$ is bounded by (2), it can be extended to all of $\mathcal H$, and this shows that the range of $B$ was closed to start with. So $B$ is invertible.

Martin Argerami
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