Banach space into hilbert space

Question

If the parallelogram equality holds for a given norm ||·|| in a Banach space. What is the formula for the associated scalar product in terms of this norm that makes this Banach space into a Hilbert space?

Answer 1

The way to go about this is to note that your only given is the following relation between the norm and the inner product: $$\|x\|^2 = \langle x,x\rangle$$ To figure out what a product $\langle x,y\rangle$ needs to be, the easiest way is to look at the above identity for $x+y$ and then expand: $$\|x+y\|^2 = \langle x+y,x+y\rangle = \langle x,x\rangle + \langle x,y\rangle + \langle y,x\rangle + \langle y,y\rangle = 2\operatorname{Re}(\langle x,y\rangle)+\|x\|^2+\|y^2\|$$ From this, you can work out the real part of $\langle x,y\rangle$ to be $$\operatorname{Re}(\langle x,y\rangle)=\frac{1}2\left(\|x+y\|^2-\|x\|^2-\|y\|^2\right).$$ You can also, by similar reasoning, rewrite this in a more symmetric form (which, as noted in the comments, is known as the polarization identity): $$\operatorname{Re}(\langle x,y\rangle)=\frac{1}4\left(\|x+y\|^2-\|x-y\|^2\right)$$ if you desire - noting that, either way, we can imagine expanding then cancelling and grouping terms. To figure out the imaginary part, if we are working with complex numbers, one can note that $\operatorname{Im}(z) = \operatorname{Re}(-iz)$ and then note $$\operatorname{Im}(\langle x,y\rangle)=\operatorname{Re}(\langle x,iy\rangle)=\frac{1}2\left(\|x+iy\|^2-\|x\|^2-\|y\|^2\right)=\frac{1}4\left(\|x+iy\|^2-\|x-iy\|^2\right).$$ which gives a complete answer for what $\langle x,y\rangle$ must be by combining the two formulas.