Does the Lax-Milgram theorem hold for a coercive bilinear form $b$ with $|b(x,y)|\le C\left\|x\right\|^{\color{red}2}\left\|y\right\|$?

Question

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 $x\in X$?

Answer 1

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 $t \to 0^+$.