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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?

$x,y \in X$ $$4\langle x,y\rangle= \|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2$$

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  • $\begingroup$ Well... yes. In fact the result follows immediately from the property you just listed. Can you see why? $\endgroup$ – bitesizebo Oct 6 '18 at 14:15
  • $\begingroup$ @bitesizebo I cannot see clearly. Should I take two different inner products that satisfy polarisation property? Could you please write openly then I can check myself? Thanks in advance :) $\endgroup$ – user519955 Oct 7 '18 at 17:32
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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.

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  • $\begingroup$ Thank you so much :)) $\endgroup$ – user519955 Oct 7 '18 at 19:57

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