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?

  • $\begingroup$ 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 $f$. $\endgroup$ Oct 18 '15 at 13:57
  • 2
    $\begingroup$ You 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. $\endgroup$ Oct 18 '15 at 13:59
  • $\begingroup$ Use polarization. $\endgroup$
    – DiegoMath
    Oct 18 '15 at 14:05

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.


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.


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