Sorry about the edits guys, I forgot to add binomial coefficients, I hope I didn't cause any needless confusion.

Edit(again): I've been thinking about this a bit and perhaps I should clarify the question a bit more. I'm essentially asking the following question: Does generalizing the inner product allow one to use properties of polynomials, such as discriminants, in order to derive analogues of the Cauchy-Schwartz inequality by means of inequality $(II)$? If my talk of "generalizing the generalized Cauchy Schwartz" is utter nonsense please focus on this question instead.

Another edit: I've been pondering this question all morning and it occurs to me that it may be rather trivial, most of the people who use this website are mathematicians right? This means that something like "generalizing the inner product and exploring its consequences" has surely been thought of before. This tells me that my question is probably highly imperfect, and even possibly unnatural. I apologize if this is the case. If this more general framework exists and allows one to come to a better understanding of a variety of questions posed which are similar to mine, could someone perhaps reveal this to me. Stated differently, what branch of mathematics can I study that would allow me to see just how naive this question is? Again, I'm very sorry if I have offended anyone by not knowing more mathematics.

Take the following proof of Cauchy-Schwartz. enter image description here The intuition behind this proof becomes clear when viewed from the perspective of an inner product.

Start with an inner product, that is, a function $f:V^2\rightarrow\mathbb{R}$, $V$ a vector space, that satisfies the following axioms for scalars k.

  • $(i) f(u,v) = f(v,u)$
  • $(ii) f(ku,v) = kf(u,v)$
  • $(iii) f(u+v,w) = f(u,w) + f(v,w)$
  • $(iv) f(u,u) \ge 0$ Where equality in $(iv)$ holds iff $u = 0$.

Leveraging each of these axioms gives us the following inequality. $$f(u,v)^2 \le f(u,u)f(v,v) \tag{A}$$ since for scalars $t$ $$0 \le f(ut+v,ut+v)$$ implies $$0 \le f(u,u)t^2 + 2f(u,v)t + f(v,v)\tag{I}$$ whose discriminant must be less than or equal to zero.

If $f(u,v) = \sum u_kv_k $ then $(A)$ reduces to Cauchy-Schwartz.

Now, generalize the inner product as follows, let $n$ be even and $f:V^n \rightarrow \mathbb{R}$ satisfy the following axioms

  • $(i*) f(u_1,u_2,\cdots,u_n) = f(v_1,v_2,\cdots,v_n)$

whenever $(v_1,v_2,\cdots,v_n)$ is a permutation of $(u_1,u_2,\cdots,u_n)$

  • $(ii*) f(u_1,u_2,\cdots,ku_m, \cdots, u_n) = kf(u_1,u_2,\cdots,u_m, \cdots, u_n)$

  • $(iii*) f(u_1,u_2,\cdots,u_m + u_m', \cdots, u_n) = f(u_1,u_2,\cdots,u_m, \cdots, u_n) + f(u_1,u_2,\cdots,u_m', \cdots, u_n)$

  • $(iv*) f(u,u,\cdots,u) \ge 0$ where equality holds iff $u=0$

Such functions obviously exist, take $f(u,v,w,s) = u_1v_1w_1s_1 + u_2v_2w_2s_2 + \cdots + u_rv_rw_rs_r$ for example.

Now, axiom $(iv*)$ tells us that for scalars $t$.

$$0 \le f(ut+v,ut+v,\cdots,ut+v)$$

which implies (using axioms $(i*),(ii*)$, and $(iii*)$) that $$0 \le f(u,u,\cdots, u)t^n + \binom{n}{1}f(v,u,\cdots, u)t^{n-1} + \cdots + f(v,v,\cdots, v) \tag{II}$$

Notice that $n$ in $V^n$ must be even, otherwise inequality $(II)$ is contradictory. If $n$ were odd then the image of RHS must be $\mathbb{R}$, which is absurd.

Also notice that letting $n = 2$ reduces $(II)$ to $(I)$

To give a specific example suppose that n=4 so that $$0 \le f(u,u,u,u)t^4 + 4f(v,u,u,u)t^3 + 6f(v,v,u,u)t^2 + 4f(v,v,v,u)t + f(v,v,v,v)$$

I'm wondering if one can use, say in the case of n=4, well known properties of quartic polynomials in order to derive more fundamental inequalities than Cauchy Schwartz. Perhaps this could tell us something about the function $u_1v_1w_1s_1 + u_2v_2w_2s_2 + \cdots + u_rv_rw_rs_r$ for example? More generally, can one use properties of polynomials, such as discriminants (I'm not very familiar with discriminants to be honest) in order to derive more fundamental inequalities than Cauchy-Schwartz on the basis of $(II)$?

If not, what kind of useful and important inequalities can be produced by means of $(II)$. Just how far does inequality $(II)$ go? In the case of $n=2$ we obtain Cauchy-Schwartz, which is quite profound. What about $n=4, 6, 8, \ldots$? Just what does $(II)$ reveal to us in these cases?

  • $\begingroup$ It seems to me that many answers to your questions are contained in this Terry Tao blog post, which does not look like an easy read, unfortunately. $\endgroup$ – Giuseppe Negro Jan 9 '18 at 16:17
  • $\begingroup$ Wow, "Higher order Hilbert spaces" sounds cool. This might take a while to understand, I don't even know what a tensor product is, not in any meaningful way that is. $\endgroup$ – Sid Jan 9 '18 at 16:22
  • $\begingroup$ Don't let tensor products scare you, the definition is given in the text. That is simply an algebraic device to turns multilinear mappings into linear ones. For example, a real scalar product $\langle, \rangle\colon V^2\to \mathbb R$, which is bilinear (sesquilinear, in the complex case), can be seen as a linear functional on $V\otimes V$, in the obvious way: $$\langle \sum_{j=1}^n v_j \otimes w_j \rangle := \sum_{j=1}^n \langle v_j, w_j\rangle. $$ The author of the linked blog post only needs this, so that he can deal with linear functionals instead of multilinear ones. $\endgroup$ – Giuseppe Negro Jan 10 '18 at 11:46
  • $\begingroup$ Also, the more I see the linked page, the more it seems to me that it is the development of exactly the same idea you had here. $\endgroup$ – Giuseppe Negro Jan 10 '18 at 11:47
  • $\begingroup$ If it is indeed the development of exactly the same idea as mine, then I will definitely try to understand it. My lack of mathematical maturity is certainly a hindrance though (I haven't even taken linear algebra or real analysis yet). $\endgroup$ – Sid Jan 10 '18 at 15:14

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