Lax-Milgram theorem states that

If $B(,)$ is a symmetric,strictly positive and bounded bilinear form on Hilbert space $V$, then for any continuous functional $l$, there exists $u\in V$ s.t. $B(u,v)=l(v)$.

I am wondering if this result can be extended to the case of Banach space,i.e. $B$ is defined on a Banach space $V$. By the condition of strictly positive and boundededness, we know that the topology of the Banach space is the same as the topology defined by bilinear form $B$, then we can use Riesz reprensentation theorem to prove it.

  • $\begingroup$ This one? $\endgroup$ – user99914 Aug 18 '17 at 7:40
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    $\begingroup$ Surfing the web, I've bumped into this 2011 talk by Hideo Kozono and Taku Yanagisawa. The short answer seems to be: "More hypothesis are needed". Of course, the natural choice of setting would be the use of two distinct Banach spaces because ideally the pairing is a bilinear map $E\times E^*\to \Bbb F$, and Riesz does not yield an isomorphism $E\cong E^*$. $\endgroup$ – user228113 Aug 18 '17 at 7:48

It would not be wrong, but useless.

If a strictly positive bilinear form exists on a Banach space $(V,\|\cdot\|_1)$, then the Banach space is equivalent to a Hilbert space in the sense that they have the same topology.

Proof: As you already said, the topology defined by $B$ is the same as the original topology. If $B$ is symmetric, you can show, that $B$ defines a scalar product on $V$, thus $V$ is a Hilbert space. If $B$ is not symmetric, then $\langle x,y\rangle := B(x,y)+B(y,x)$ defines a scalar product.

So you cannot apply a Banach space version of Lax-Milgram to spaces such as $L^p(\Omega)$ for $p\neq 2$, because you cannot find a $B$ that satisfies the conditions.

  • $\begingroup$ I think technically $B$ is still missing the symmetry required of a scalar product, however using something like $\langle x,y \rangle = B(x,y) + B(y,x)$ should work. $\endgroup$ – mlk Aug 18 '17 at 7:57
  • $\begingroup$ I am thinking if the original norm satisfies Parallelogram law $\endgroup$ – 89085731 Aug 18 '17 at 7:58
  • $\begingroup$ @89085731 If the original norm satisfies a Parallelogram law, then it is a Hilbert space (if its that what you mean). $\endgroup$ – supinf Aug 18 '17 at 7:59
  • $\begingroup$ Maybe a silly question.Why $L^p,p\neq 2$ can not be homeomorphic to a Hilbert space $\endgroup$ – 89085731 Aug 18 '17 at 8:06
  • $\begingroup$ that is a good question and i think it is not so easy to answer. Also for a finite measure space, the $L^p$ spaces are all equivalent, so you need extra conditions for the measure space. Maybe it is worth to ask another MSE question why $L^p$ and $L^q$ are not homeomorphic if $p\neq q$, if there isnt such a question already $\endgroup$ – supinf Aug 18 '17 at 8:12

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