I have seen a proposition that
Let $(X, \|.\|)$ is a normed space. If this norm can be induced by an inner product (i.e. it satisfies parallelogram property) then is induced by at most one inner product.
Is it true? How can I prove it? Should I use property below?
$x,y \in X$
$$4\langle x,y\rangle= \|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2$$
Best Answer
The key point is that the polarisation identity expresses the inner product solely in terms of the norm, hence the norm uniquely determines the inner product. To be very verbose, suppose that $\lVert \cdot \rVert$ can be induced by both the inner product $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot , \cdot \rangle_2$. Then by the polarisation identity for all $x, y$ $$ \langle x, y \rangle_1 = \frac{1}{4}\left(\lVert x + y \rVert^2 - \lVert x - y \rVert^2 + i \lVert x + iy \rVert^2 - i \lVert x - i y \rVert^2\right) = \langle x, y \rangle_2 $$ Hence $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot , \cdot \rangle_2$ are the same inner product, thus if a norm is induced by an inner product then the inner product is unique.