1
$\begingroup$

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

  • 3
    I guess the paring between linear functionals $\ell\in X^*$ with $x\in X$ is a generalization of inner product.2012-11-11
  • 1
    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 $p = 2$.2012-11-11
  • 0
    What properties of new inner product do you expect? If you expect all of them, then it is impossible.2012-11-11

1 Answers 1

0

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, 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.