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Question:

Suppose we have a Banach space $V$ with a coercive, bounded, bilinear form $a:V \times V \rightarrow \Bbb R$.

Prove that $V$ is a Hilbert space.


Attempt:

I have no idea how to use the fact that the bilinear form is coercive and bounded. I was trying to make up some sort of inner product for the Hilbert space, such as

$$\langle u,v \rangle = \frac{a(u,v)+a(v,u)}{2}$$

but nothing seems to work.

Any hints would be much appreciated.

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  • $\begingroup$ Specifically what does seem not to work? $\endgroup$
    – Berci
    Feb 3 '20 at 23:05
  • $\begingroup$ Do you want to say that $V$ is a Hilbert space under an equivalent norm or that the original norm is itself given by an inner product? The former is very easy to verify. $\endgroup$ Feb 3 '20 at 23:20
  • $\begingroup$ That there exists an inner product such that $V$ is a Hilbert space $\endgroup$ Feb 4 '20 at 0:08
  • $\begingroup$ The "sort of inner product for the Hilbert space" is indeed the inner product that you're after. $\endgroup$ Feb 4 '20 at 7:09
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Your function $\langle u,v \rangle$ is indeed an inner product that makes your Banach space into a Hilbert space (there was no need to divide by $2$ though). In order to prove that $V$ is a Hilbert space under $\langle \cdot , \cdot \rangle$, it suffices to show the following:

  • $\langle u, v \rangle$ defines an inner product over $V$ (note: the positive definite property is implied by coerciveness)
  • The norm $\|\cdot\|$ of the Banach space is equivalent to the norm $\|\cdot\|_{IP}$ induced by your inner product. That is, there exist $K \geq k > 0$ such that for all $v \in V$, $$ k\|v\| \leq \|v\|_{IP} \leq K\|v\|. $$
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  • $\begingroup$ Why is it sufficient to show that $|| \cdot ||_{IP}$ satisfies this inequality? $\endgroup$ Feb 4 '20 at 10:24
  • $\begingroup$ Does this somehow imply that $||v||_{IP} = ||v||$ for all $v \in V$? $\endgroup$ Feb 4 '20 at 10:25
  • $\begingroup$ @glow It does not imply that $\|v\|_{IP} = \|v\|$. It does imply, however, that $\|\cdot\|$ and $\|\cdot \|_{IP}$ induce the same topology. In other words, A sequence $x_n$ will converge to $x$ relative to $\|\cdot\|$ if and only if it converges relative to $\|v\|_{IP}$. So, a map into or out of $V$ will be continuous relative to one norm if and only if it is continuous relative to the other. In this case, the resulting Hilbert space is "the same" as the original Banach space. $\endgroup$ Feb 4 '20 at 11:25

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