Banach Space with coercive bilinear form is a Hilbert Space

Question

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.

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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.

Answer 1

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 $v \in V$, $$ k\|v\| \leq \|v\|_{IP} \leq K\|v\|. $$