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Every single proof I have found online of the inequality proves it for sums, where the inner product is assumed to be the Euclidean inner product on $\Bbb{R}^n$. However, I have gathered from comments on this site and from Wikipedia that the statement:

$$\langle u,v\rangle^2\le\langle u,u\rangle\cdot\langle v,v\rangle$$

Holds true for any inner product/Hilbert space. This means it holds for an arbitray inner product, perhaps defined over more abstract vector spaces, such as spaces of functionals or operators. Therefore all the proofs I've seen are incomplete, since the statement above is not the same (unless I'm very mistaken) as the statement below:

$$\left(\sum_i u_iv_i\right)^2\le\left(\sum_i u_i^2\right)\left(\sum_i v_i^2\right)$$

Any proof of that statement is just a proof of a very particular special case of the inequality. Moreover, Wikipedia states that equality is only ever achieved, again for any abstract Hilbert space, if $u,v$ are linearly dependent.

A comment I read on this site suggested that the proof of the inequality for abstract spaces was a huge step in the study of functional analysis. I cannot for the life of me find this proof.

I am looking for a proof (or reference to one) of the general statement and/or a proof/reference of the statement: "equality is only achieved with $u,v$ linearly dependent".

Many thanks!

FShrike
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    Consider the discriminant of the quadratic $\langle u+\lambda v,u+\lambda v\rangle$. – user10354138 Sep 02 '21 at 10:34
  • There must be something about that that makes it unsuitable as a proof in arbitrary Hilbert space @user10354138 - that strikes me as too elementary and far too brief to be a "fundamental step forward" – FShrike Sep 02 '21 at 10:36
  • There is nothing unsuitable for the proof of arbitrary real inner product space. The only thing it doesn't immediately give you is for complex inner product space, but for that you just need to change $y$ to $e^{i\theta}y$ for suitable $\theta\in\mathbb{R}$ to bring $\langle x,e^{i\theta}y\rangle\in[0,\infty)$, or proceed like this answer. – user10354138 Sep 02 '21 at 10:58
  • @user10354138 So this approach proves the inequality for all abstract spaces too? For example, $$\left(\int f(x)g(x),dx\right)^2\le\left(\int f^2(x),dx\right)\left(\int g^2(x),dx\right)$$ Assuming of course that $f,g$ are such that this is an inner product. – FShrike Sep 02 '21 at 11:11

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