# Can we define an inner product in terms of the norm induced by it?

I know that not all norms are induced by any inner product.

But if we have an inner product, $$\langle\cdot,\cdot\rangle$$ we can define a norm $$||v||=\sqrt{\langle v,v\rangle}$$.

My question is, can we somehow reconstruct the inner product from this norm? I.e. can we define an inner product in terms of a norm?

If not, can we construct it if we assume some additional structure on the normed space induced by this inner product?

Indeed we can. Note that we have $$\langle v, v \rangle = \Vert v \Vert^2$$. This is the quadratic form associated to the bilinear form $$\langle \cdot, \cdot \rangle$$. However, by the polarisation formula (see https://en.wikipedia.org/wiki/Polarization_identity), we can recover the bilinear form from its associated quadratic form.
As AGF remarked, we have a criterion to decide when a norm is induced by an inner product. Namely, it is induced by an inner product iff the following identity holds true for all $$x,y$$ in our vector space $$\Vert x + y \Vert^2 + \Vert x-y \Vert^2 = 2(\Vert x \Vert^2 + \Vert y \Vert^2)$$ This is the so-called parallelogram identity with you can find here https://en.wikipedia.org/wiki/Parallelogram_law
• One can also use the polarisation identity to show that a linear isometry (i.e. a linear map such that $\|Tx\| = \|x\|$) must preserve inner products. So, if a linear map preserves distances, it also preserves angles. – Theo Bendit Feb 24 '19 at 11:09