Question: Can Lax-Milgram theorem be restated as:
If an inner product $[\cdot,\cdot]$ induces a norm equivalent* to the norm of the Hilbert space $H=(X,\langle\cdot,\cdot\rangle)$, then $H'=(X,[\cdot,\cdot])$ is a Hilbert space too (and thus Riesz representation theorem holds for $H'$).
(*Two norms $||\cdot||_1, ||\cdot||_2$ are equivalent if there exists $a,b>0$ such that $a||x||_1\le||x||_2\le b||x||_1$.)
If so, Lax-Milgram seems like a trivial corollary of Riesz representation, since it boils down to proving: inner products inducing equivalent norms can make $X$ a Hilbert space.
Did I miss anything about Lax-Milgram that makes it significant enough to be a theorem?
More background:
(Only real Hilbert spaces are considered.)
The Lax-Milgram theorem states that:
For a Hilbert space $H=(X,\langle\cdot,\cdot\rangle)$ and a bilinear form $[\cdot,\cdot]:H\times H\to\mathbb R$, if
(1) $|[f,g]|\le C||f||\cdot||g||$ where $||f||=\sqrt{\langle f,f\rangle}$
(2) $[f,f]\ge c||f||^2$
then for each $f\in H$ there exists $f'$ such that $[f',u]=\langle f,u\rangle$ for all $u\in H$.
I noticed that (1) is indeed equivalent to another condition (1’)
(1’): $[f,f]\le C||f||^2$
(1) implies (1’) is trivial, and (1’) implies (1) via Cauchy-Schwarz.
Then, (1) + (2) is no different to saying “$[\cdot,\cdot]$ and $\langle \cdot,\cdot\rangle$ induce equivalent norms”. Thus, it can be easily seen that $H'=(X,[\cdot,\cdot])$ is also a Hilbert space, and the desired conclusion (i.e. the existence of $f'$ above) is an immediate consequence of Riesz representation theorem applied to $H'$, with $\langle f,\cdot\rangle$ being the linear functional.