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I have a Hilbert space that has an inner product and is complete. But it has no norm. Is this acceptable? The axioms of the Hilbert space says complete inner product vector space, so I assume it is one. But it threw me off having no norm.

If I define the norm as $||x||=\sqrt{\langle x,x\rangle}$, then $||x||$ does not follow the triangle inequality. Everything else seems to work. Is it still a Hilbert space? Can I $\sqrt{\langle x,x\rangle}$ to define Cauchy sequences and prove completeness, instead of a norm?

My specific example is as follows. I define a vector of $n\times n$ matrices, then the inner product as follows:

$$ \langle u,v\rangle=\frac{1}{4}(||u+v||^n-||u-v||^n) $$

where

$$ ||x||^n=\sum_{M \in x}\det M $$

A vector space with said inner product follows all axioms of the Hilbert space, except that the "norm" $||x||^n$ does not respect the triangle inequality.

Anon21
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    If the triangle inequality fails, then your "inner product" is not an inner product. Your "inner product" does not seem to be linear in the two components. – N. S. Jun 06 '21 at 01:53
  • @N.S. "If the triangle inequality fails, then your "inner product" is not an inner product." Can you explain why that is, because looking at the 5 axioms of the definition here https://en.wikipedia.org/wiki/Inner_product_space#Definition, none of them mention the triangle inequality. – Anon21 Jun 06 '21 at 02:00
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    An inner product immediately gives a norm. This follows directly from the definition of an inner product, so if you don't get a norm, you can't have given an inner product – Calvin Khor Jun 06 '21 at 02:01
  • The Norm section of the Wikipedia page you linked mentions that an inner product gives a norm. This is proven (in the real case) in various places including Garrett Thomas's notes on Norms and Inner Products. – Mark S. Jun 06 '21 at 02:06
  • @Anon21 The axioms of definition imply the Cauchy--Schwarz inequality (this is a very standard exercise) and the Cauchy--Schwarz inequality implies the triangle inequality. – N. S. Jun 06 '21 at 02:10
  • Again, based on your definition I highly doubt that your "inner product" satisfies $$\langle u, v+w \rangle = \langle u, v \rangle+\langle u, w \rangle$$ and vely likely it also fails $$\langle u, u\rangle \geq 0 ,.$$ – N. S. Jun 06 '21 at 02:18
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    In addition, the zero vector is not the only vector with zero norm because there are other matrices with zero determinant (any non-invertible matrix) – Calvin Khor Jun 06 '21 at 02:24

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Your inner product is not an inner product. For one, it fails to be bilinear. For instance $ \langle 2u,u\rangle $ should be $2\langle u,u\rangle$ but according to your formula \begin{align}\langle u,u\rangle &= \frac14(\|2u\|^n-\|0\|^n) = \frac{2^{n}}4 \|u\|^n, \\ \langle 2u,u\rangle&= \frac14(\|3u\|^n -\|u\|^n)= \frac{3^n-1}4\|u\|^n ,\end{align} and $2^{n+1}\neq 3^n-1$ unless $n=2$.

Alternatively, one just notes that it is a Theorem that every inner product induces a norm. So if you don't get a norm, you could not have started with an inner product.

Calvin Khor
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  • In his definition $| x |^n$ does not seem to mean $| x|$ to the power $n$. – N. S. Jun 06 '21 at 02:12
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    @N.S. I did see that, but it seems the scaling still works out this way; determinants of $n\times n$ matrices are e.g. $n$-linear in the rows, and I can't fully grasp the meaning of the sum over $M\in x$, but sums should be linear – Calvin Khor Jun 06 '21 at 02:13
  • @N.S. Oh, I have understood the notation $\sum_{M\in x}$. OP is treating the vector of $n\times n$ matrices $x\in (\mathbb K^{n\times n})^k$ (apparently $k$ is not explicitly determined) as a subset of size $k$ of $\mathbb K^{n\times n}$. So the scaling is indeed right – Calvin Khor Jun 06 '21 at 02:22