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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.

89085731
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1 Answers1

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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.

supinf
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  • 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. – mlk Aug 18 '17 at 07:57
  • I am thinking if the original norm satisfies Parallelogram law – 89085731 Aug 18 '17 at 07:58
  • @89085731 If the original norm satisfies a Parallelogram law, then it is a Hilbert space (if its that what you mean). – supinf Aug 18 '17 at 07:59
  • Maybe a silly question.Why $L^p,p\neq 2$ can not be homeomorphic to a Hilbert space – 89085731 Aug 18 '17 at 08:06
  • 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 – supinf Aug 18 '17 at 08:12
  • The reason I ask this question is because u say we can not define bilinear form on $L^p,p\neq 2$, so I think it is equivalent to the question I raise. – 89085731 Aug 18 '17 at 08:39
  • that was just an example, so i did not add any proof why $L^p$ is not homeomorphic to a hilbert space. If you want easier examples, consider $L^1$ or $L^\infty$ which are not reflexive, therefore not topologically equivalent to a hilbert space. – supinf Aug 18 '17 at 08:46
  • sorry what I mean is that why we couldn't have such bilinear form on $L^p,p\neq 2$ – 89085731 Aug 18 '17 at 08:48
  • Let us continue this discussion in chat. – supinf Aug 18 '17 at 08:50