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Let $(x_n)$, $(y_n) \in \mathbb{R}^\infty$ be sequences.

Is it possible to define an inner product $\langle \cdot , \cdot \rangle$ whereby $\langle (x_n) , (y_n) \rangle = c, c \in \mathbb{R}$?

I am asking this question because while this operation is natural for vectors of arbitrarily finite dimensions, I have never seen it being done for sequences.

Can we possibly take as definition:

$\langle (x_n) , (y_n) \rangle = x_1y_1 + x_2y_2 + \ldots + x_ny_n + \ldots$?

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  • $\begingroup$ Yes we can definitely define one on sequences. Your suggestion won't work though since nothing says that your sum converges. $\endgroup$
    – user370967
    Oct 16 '17 at 5:52
  • $\begingroup$ See this en.m.wikipedia.org/wiki/Sequence_space $\endgroup$
    – Guy Fsone
    Oct 16 '17 at 5:54
  • $\begingroup$ @GuyFsone Where does your link address the issue? At first glance, it seems just a small survey of the most famous Banach space structures on subspaces of $\Bbb R^{\Bbb N}$, rather than the whole $\Bbb R^{\Bbb N}$. $\endgroup$
    – user228113
    Oct 16 '17 at 5:57
  • $\begingroup$ did really went through the link? it is the same issue . juste he notation of space of sequences is difference. note that if he define the scalar product the way he did it automatically become a banach space $\endgroup$
    – Guy Fsone
    Oct 16 '17 at 6:07
  • $\begingroup$ @GuyFsone Nope. If $u = (1, 1, 1, 1, \ldots)$, then $\left< u, u \right>$ is undefined. $\endgroup$
    – Adayah
    Oct 16 '17 at 6:53
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Well, a reason why you have never seen it for sequences is that, extending the existing formula for the regular inner product, we would have the following: $$\langle(x_n),(y_n)\rangle=\sum_{k=1}^\infty x_ky_k$$ which is an infinite sum, so there is mouch doubt about its convergence - we do not want any vector of our space $v$ with $\langle v,v\rangle=\pm\infty$ or, even worse, $\langle v,u\rangle$ not defined for some $u,v$.

To avoid such problems, we introduce the following spaces: $$l^p:=\left\{x\in\mathbb{R}^\infty\left|\sum_{k=1}^\infty|x_k|^p<+\infty\right.\right\}$$ One can show that $l^p$ is a normed space for every $p\geq1$ (and, actually, for $p=+\infty$), but that only for $p=2$ there is an inner product that generates the norm of that space. So, in the case of $l^2$, the inner product of two sequences $x,y\in l^2$ is: $$\langle x,y\rangle=\sum_{k=1}^\infty|x_ky_k|$$ from which we can take also the formula of the norm: $$\lVert x\rVert^2=\langle x,x\rangle=\sum_{k=1}^\infty|x_k|^2$$ Note that this infinite sum is convergent, since $x\in l^2$. For further reading you can read also this link.

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  • $\begingroup$ There is one superfluous $| \cdot |$ in the definition of $\left< \cdot, \cdot \right>$ in $\ell^2$. Also, while you discuss the reasons why $\mathbb{R}^{\mathbb{N}}$ is not usually equipped with an inner product, you actually haven't answered wheter it's possible to do so. $\endgroup$
    – Adayah
    Oct 16 '17 at 7:26
  • $\begingroup$ The $|\cdot|$ is a matter of habbit, since the same norms and inner products can be defined over complex sequences, where $z^2\neq|z|^2$. In general, you are right about the "spirit" of the question, so I will try to come up with a complete answer later on today, due to lack of time right now. $\endgroup$ Oct 16 '17 at 8:04
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If you are only interested in some inner product and don't care what properties are satisfied by the norm/metric/topology then the question involves only the cardinality. Take a vector space isomorphism from your space onto $l^2$ for example and define the inner product by taking the inner product of the images. Of course, this doesn't give a formula for the inner product.

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