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So I'm reading through some analysis problems and one is discussing $\ell^p$ (the space of $p$-summable sequences $x: \mathbb Z^+ \to \mathbb C$ such that $\sum_{n \in \mathbb Z^+}|x_n|^p < \infty$) and without explanation the author starts manipulations with an inner product $\langle x, y \rangle$. It was my understanding that $L^p$ spaces did not admit inner products except for $p=2$, $$ \langle x , y \rangle = \sum_{n \in \mathbb Z^+} x_n\overline{y_n} $$ (or the conjugate). Is it common convention to use the $\ell^2$ inner product $\langle \cdot , \cdot \rangle$ on elements of $\ell^p$ (which seems in poor taste given the convergence isn't guaranteed a priori), or is there an inner product on $\ell^p$ that isn't generalizable to an arbitrary $L^p$ space?

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    $\begingroup$ You would expect to have $\langle x\vert x\rangle=\lVert x\rVert^2$. This can only be true in a normed space satisfying the parallelogram law, and for $L^p$ it is the case only for $p=2$. $\endgroup$
    – tomasz
    May 1, 2013 at 3:36
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    $\begingroup$ The notation $\langle \cdot, \cdot \rangle$ is also used to denote a bilinear form between two vector spaces that form a dual pair (e.g., 2.3.8 in Pedersen's Analysis Now). So $\langle x, y \rangle$ could be referring to the duality between $\ell_p$ and $\ell_q$, with $x \in \ell_p$, $y \in \ell_q$, and $1/p + 1/q=1$. $\endgroup$
    – Tom Cooney
    May 2, 2013 at 17:44

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You are very correct, however if a sequence is in $l^p\cap l^2$, which many sequences certainly will be, the inner product manipulations will be valid. However I would consider it poor exposition to do such a thing without mentioning this caveat.

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As it was pointed out the parallelogram identity forces $ p = 2 $, take for example sequences $ x = (1,0,0,....),\ y = (0,1,0,0,....) $, Then parallelogram identity gives you in $l_p$ $$ 2^{1+2/p}= \|x+y\|^2_p + \|x-y\|^2_p = 2(\|x\|^2_p + \|y\|^2_p) = 4 $$ Hence $ p = 2 $

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