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I am liking know if exist a any generalization for inner product in the set $L^p(\mathbb{R}^2)$ , for any $p\geq 1$.

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I guess the paring between linear functionals $\ell\in X^*$ with $x\in X$ is a generalization of inner product. – Hui Yu Nov 11 '12 at 3:55
What do you mean by "generalization" of an inner product? There exists an inner product $\langle \cdot, \cdot\rangle$ on $L^p$ satisfying $\sqrt{\langle x, x \rangle} = \Vert x \Vert_p$ if and only if $p = 2$. – Jesse Madnick Nov 11 '12 at 6:08
What properties of new inner product do you expect? If you expect all of them, then it is impossible. – Norbert Nov 11 '12 at 13:20
up vote 1 down vote accepted

On any real Banach space $X$ one can define semi-inner products $$\begin{split} \langle x,y\rangle_+ &= \sup \{f(x):f\in X^*, \|f\|=1, f(y)=\|y\|\} \\ \langle x,y\rangle_- &= \inf\{f(x):f\in X^*, \|f\|=1, f(y)=\|y\|\} \end{split} \tag1$$ There is an equivalent definition in terms of directional derivative of the norm.

The semi-inner products are not linear in $y$ unless we are in a Hilbert space. In general, they are not linear in $x$ either: $\langle x,y\rangle_+$ is convex and $\langle x,y\rangle_-$ is concave. But on smooth Banach spaces, such as $L^p$ for $1<p<\infty$, the set of functionals $f$ in (1) consists of just one element. This implies $\langle x,y\rangle_+ = \langle x,y\rangle_-$ and subsequently, the semi-inner product $\langle x,y\rangle$ becomes linear in $x$.

The Cauchy-Schwarz inequality holds in this form: $$-\|x\|\,\|y\|\le \langle x,y\rangle_- \le \langle x,y\rangle_+ \le \|x\|\,\|y\| \tag2$$

Reference: Nonlinear Functional Analysis by Deimling.

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