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It is known to all that for inner product space $H$, Cauchy-Schwarz inequality hold: $| \langle x,y \rangle | \leq \| x \| \cdot \| y \| \ \forall x,y\in H$, and for a normed space $X$ without inner product structure, we have two similar "Cauchy inequality":1.real case(use some algebraic substitution); 2.complex case

My question is that for a normed space but not inner product space $X$, noting that $$f(x,y)=\sum_{k=1}^{n} a_k(\|x+b_k y\|^2-\|x\|^2-\|b_k y\|^2) \ (a_k,b_k\ne 0,k=1,2,···,n )$$ which satisfy that $$\sum_{k=1}^{n}a_k\overline{b_k}=1; \ \sum_{k=1}^{n}a_kb_k=0 $$in complex case,or $$\sum_{k=1}^{n}a_kb_k=\frac{1}{2}$$ in real case,

(In summary, for an inner product space there is $f(x,y)=\langle x,y \rangle$ )

then in addition to the two examples mentioned above, what conditions are satisfied by sequence $\{a_k\},\{b_k\},\{c_k\}$ that enable inequality $|f(x,y)|\le\|x\|\|y\| \ \forall x,y\in X$ to hold?

                                                     ----2023.12.17

Okay, I admit that I asked the question in order to find the condition that makes the normed space $X$ isometric isomorphic to the inner product space, here's the link to the question I asked on MathOverFlow. Thank if you can give me any help.

                                                     ----2024.02.02

I'm now focus on a particular case about this problem:

$$f(x,y)=\frac{1}{2}\sum_{k=1}^{n}a_{k}(\|x+b_{k}y\|^2-\|x-b_{k}y\|^2)$$ where $\sum_{k=1}^{n}a_{k}b_{k}=\frac{1}{2}$, is $|f(x,y)|\le \|x\| \|y\|$ hold?

Anyone who can give a proof of this case might be instructive.

                                                     ----2024.02.28
anyon
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    tl;dr. Is the question not just this: When does a norm arise from an inner product? The answer to this question is well-known. – Kurt G. Dec 17 '23 at 09:23
  • Now I am reading your post a bit closer. Even if you find $a_k,b_k,c_k$ so that $|[x,y]|\le|x||y|$ holds: that construct $[x,y]$ that you defined there does not look bilinear in $x,y$ to me, hence it can never ever be an inner product. – Kurt G. Dec 17 '23 at 09:29
  • @KurtG.:I edited it for express my question more clearly. I think you misunderstood my question, my question wasn't about how to give the inner products, it was just formally close, perhaps $[x,y]$ is formally misleading, I used $f(x,y)$ instead – anyon Dec 17 '23 at 12:15
  • Maybe you're looking for something like semi-inner product? – Jakobian Feb 02 '24 at 03:26
  • @Jakobian : Thank, I learned a little bit about semi-inner product. Given that there are infinite varieties of s.i.p. in any normed linear space, the question can then be transformed to whether one can find a proper s.i.p. $[·,·]$ that satisfies $[x,y_f]=f(x,y_f)$, is that what you are referring to? – anyon Feb 03 '24 at 07:12

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