Consider $I = [-1,1]$. Let $C(I)$ be the normed space, equipped with norm \begin{align} ||f||_{2} = \left( \int_{-1}^{1} |f(t)|^2 \, dt \right) ^{1/2} \end{align}
Show, that norm is induced by a scalar product. Any hints on how to proceed?
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You want that $\left<f,f\right> = \Vert f \Vert^2$. Remember that $\vert f \vert^2 = ff^$ with $f^$ being the complex conjugate of $f$. – Lionel Ricci Oct 18 '15 at 13:57
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2You might want to look up "polarization identity." And then you might want to think about how lucky you are that your field doesn't have characteristic 2. – John Hughes Oct 18 '15 at 13:59
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Use polarization. – DiegoMath Oct 18 '15 at 14:05
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Any norm in a Hilbert space is induced by a scalar product, to do this you can use the polarization identity:
$$\langle f,g\rangle=\frac{1}{4}\left(||f+g||^2-||f-g||^2\right),$$ where $||f||$ is the norm that you described before.
DiegoMath
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As many has pointed out, what you need is the Polarization identity. Basically, it tells you when the converse of "every inner product induces a norm" is true. The special case where the field of the vector space is real is spelled out in Diego's post. For complex case the formula is $$ \langle f,g \rangle=\frac 14 \sum_{n=0}^3 i^n||f+i^ng||^2 $$ For more discussion and proofs, consider reading this link Norms Induced by Inner Products and the Parallelogram Law.
BigbearZzz
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