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Please note, this seems very simple. But I am unable to prove this.

Main Doubt:

$$ \left|\left(\left|x\right|,\left|y\right|\right)\right|=\left(\left|x\right|,\left|y\right|\right) $$

Here, $x$ and $y$ are complex functions and $(x,y)$ is the inner product. $|x| = \left(x\overline{x}\right)^{\frac{1}{2}}$ is the absolute value of $x$. Absolute value is defined as the non-negative square root of a complex number multiplied by its conjugate. (Using the definition of absolute value from Rudin Principles of Mathematical Analysis: Definition 1.32)

From my limited experience with the inner product, I have seen it defined differently depending on the space. But the above should hold in all spaces I presume (just a hunch, could be wrong) ? So without using a specific definition of inner product restricted to a particular space. Are we able to show the above?

Related Question: Perhaps, the real question is whether the inner product of real numbers is always a real number? That is, if the $\left(\left|x\right|,\left|y\right|\right)$ is a real number since $|x|$ and $|y|$ are real numbers. If this is true then, $$\left(\left|x\right|,\left|y\right|\right)=\overline{\left(\left|x\right|,\left|y\right|\right)}$$

This gives,

$$ \left|\left(\left|x\right|,\left|y\right|\right)\right|=\left\{ \left(\left|x\right|,\left|y\right|\right)\overline{\left(\left|x\right|,\left|y\right|\right)}\right\} ^{\frac{1}{2}}=\left(\left|x\right|,\left|y\right|\right) $$

Another confusion or question: Can we say that $x$ and $y$ are vectors which would suggest that absolute value is defined for any vector in a complex vector space?

I am using the notation from Rudin Real and Complex Analysis (RCA).

Thanks in advance. Happy to delete this question if too simple etc.

texmex
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  • What's your definition of absolute value of a complex function? If you mean a norm on complex value function, then your question 1 is true since norm map a vector to non-negative real number – wer Oct 18 '23 at 06:38
  • Please note we can define the absolute value as the non-negative square root of a complex number multiplied by its conjugate. (Using the definition from Rudin Principles of Mathematical Analysis: Definition 1.32) $|x| = (x\overline{x})^{1/2}$ – texmex Oct 18 '23 at 08:28
  • @wer Using the above definition might not need norm or other advanced topics perhaps? – texmex Oct 18 '23 at 08:30
  • you say "x and y are complex functions" so your definition of abs(x(t)) is a real valued function $g(t)=x(t) \bar{x} (t) $ over some set? So you use two inner product here. One is over function, one is over two real numbers. But that's fine. Just take Abs(x(t)) as a non negative number on fixed t. – wer Oct 18 '23 at 09:30
  • Your equality follows from the fact that inner product has to be positive by the positive of $|x|=(x,x)$. Additionally, inner product on $f(x,y) :R \times R \to R $ is in fact of the form $f(x,y)=axy$ in your case , where $a$ is a real number. (you can prove it with 3 properties of inner product) By positive definitness of inner product, it has to be $a>0$ – wer Oct 18 '23 at 09:31
  • @wer My understanding is that the inner product is a complex number. So unless we specify more properties, we cannot consider a complex number as positive or negative right? – texmex Oct 18 '23 at 12:45
  • your equality for question 1 is natrually right, No additional properties are needed. You may take a look at the defination of inner product and norm derived from inner product for more clear understanding. Norm is not very advanced topic, it is a generalization of absolute value. You can try to learn it for a better understanding. – wer Oct 18 '23 at 13:51
  • Linking Related Question: https://math.stackexchange.com/questions/16163/how-are-norms-different-from-absolute-values – texmex Oct 19 '23 at 03:26

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