Let $X,Y,Z$ be $\mathbb R$-Banach spaces and $\mathfrak b:X\times Y\to Z$ be bilinear. $\mathfrak b$ is called bounded $:\Leftrightarrow$ $\exists C\ge0$ with $$\left\|\mathfrak b(x,y)\right\|_Z\le C\left\|x\right\|_X\left\|y\right\|_Y\;\;\;\text{for all }(x,y)\in X\times Y\;.\tag1$$

If $X=Y$ is a $\mathbb R$-Hilbert space and $Z=\mathbb R$, such $\mathfrak b$ are considered in the Lax-Milgram theorem. In that case, suppose we have $$\left|\mathfrak b(x,y)\right|\le C\left\|x\right\|_X^{\color{red}2}\left\|y\right\|_X\;\;\;\text{for all }x,y\in X\tag2$$ instead of $(1)$. Moreover, assume that $\mathfrak b$ is coercive and $\ell$ is a bounded linear functional on $X$. Are we still able to show that $$\mathfrak b(x,y)=\ell(y)\;\;\;\text{for all }y\in Y\tag3$$ for some unique $x\in X$?

  • $\begingroup$ My first impression is that your modified condition is equivalent to boundedness for a bilinear form $b(x,y)$. $\endgroup$ – hardmath Mar 3 '17 at 20:51
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    $\begingroup$ A simple homogeneity argument shows that if $\mathfrak b$ satisfies (2) then it vanishes identically. $\endgroup$ – Umberto P. Mar 3 '17 at 20:52
  • $\begingroup$ @hardmath $\mathfrak b$ is bilinear. And I don't think that $(1)$ and $(2)$ are equivalent. $\endgroup$ – 0xbadf00d Mar 3 '17 at 20:56
  • $\begingroup$ As @UmbertoP. shows, (2) implies (1) in an especially strong way. So strongly that (1) does not imply (2), and in that sense they are not equivalent. $\endgroup$ – hardmath Mar 3 '17 at 21:16
  • $\begingroup$ @hardmath What? The especially strong way is that the bilinear form must be identically $0$. $\endgroup$ – 0xbadf00d Mar 3 '17 at 21:19

Suppose that $\newcommand{\b}{\mathfrak b} \b$ is a bilinear form satisfying $$|\b(x,y)| \le C \|x\|^2 \|y\|.$$ Let $x,y \in X$ be arbitrary and let $t > 0$. Then $$|\b(tx,y)| \le C \|tx\|^2 \|y\|$$ from which it follows that $$t|\b(x,y)| \le Ct^2 \|x\|^2 \|y\|$$ and $$|\b(x,y)| \le Ct\|x\|^2 \|y\|.$$ Now let $t \to 0^+$.

  • $\begingroup$ Let $$\mathfrak b(u,v,w):=\mathfrak b(u,v,w):=\langle((u\cdot\nabla)v,w\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v,w\in H_0^1(\Lambda,\mathbb R^d)$$ for a suitable $\Lambda\subseteq\mathbb R^d$ such that $\mathfrak b$ is well-defined. It's well-known that $\mathfrak b$ is a bounded trilinear form on $H_0^1(\Lambda,\mathbb R^d)$. So, $$\left|\mathfrak b(u,u,v)\right|\le C\left\|u\right\|_{H^1(\Lambda,\:\mathbb R^d)}^2\left\|v\right\|_{H^1(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }u,v\in H_0^1(\Lambda,\mathbb R^d)$$ for some $C\ge0$. $\endgroup$ – 0xbadf00d Mar 3 '17 at 21:01
  • $\begingroup$ $(u,v)\mapsto\mathfrak b(u,u,v)$ (call it $\tilde{\mathfrak b}$) is the inertia term occurring in the weak form of the Navier-Stokes equation and this bilinear form motivated my question. It's not identically zero. So, what am I missing? $\endgroup$ – 0xbadf00d Mar 3 '17 at 21:02
  • $\begingroup$ If $b$ is trilinear you get $t^2$ on the left side, obviously. $\endgroup$ – Umberto P. Mar 3 '17 at 21:03
  • $\begingroup$ $\tilde{\mathfrak b}$ is bilinear with $(2)$. With your argumentation, $\tilde{\mathfrak b}=0$. $\endgroup$ – 0xbadf00d Mar 3 '17 at 21:04
  • $\begingroup$ I disagree that $\tilde{\mathfrak b}$ is bilinear. $\endgroup$ – Umberto P. Mar 3 '17 at 21:08

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