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I'm looking for a function $u$ say, multi variable or otherwise, that has the property

$$\frac{1}{u}=\|u\|^2$$

I have a difficult problem I would like to solve by making a substitution with a function that has this property.

Note I mean the standard Archimedean norm or its generalisation to the complex plane or some other $\mathbf{R}^n.$

Does such a function, or one with a similar property, exist?

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    $u(x)=1$ satisfies that. What *other* constraints do you have in mind? – 2017-02-28
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    You are taking $\dfrac1u$, so $u$ is real or complex valued? In that case what is $\|u\|$ supposed to mean, the pointwise absolute value? Which complex numbers $a$ satisfy $\dfrac1a =|a|^2$? This implies $a>0$, so $a=|a|$, and $a^3=1$. – 2017-02-28
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    Yes, this is the intended meaning. – 2017-02-28
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    Ideally, a non trivial example would be nice as a start.A periodic, doubly periodic, quasi periodic etc. function would be even better. – 2017-02-28
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    A nontrivial example doesn't exist based on the stated conditions. – 2017-02-28

2 Answers 2

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The constant $1$ has this property. It is the only one.

No non-constant analytic function can be only real valued in any open region. Just take a point with a non-zero derivative. There must be some direction that will have an imaginary component.

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    Not sure why I didnt just look at the problem instead of asking. Are there any norms for which such a function could exist? – 2017-02-28
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    The OP never says the function must be analytic. – 2017-02-28
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    But btilly didn't say that the function HAD to be. This answer just notes that if you're looking for an analytic one aside from this, you're not gonna find it. – 2017-02-28
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It seems to me that by algebra, taking the absolute value of both sides, you get $$ \frac{1}{\|u\|} = \|u \|^2 $$ so $\|u\| = 1$.

So: the constant function $1$ seems to work pretty well.

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    Not taking absolute values, $u>0$ already, so $u\equiv 1$. – 2017-02-28
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    OK, I should have said "norm"; and even if $u$ is already positive, it's true that $\|u \| = 1$. You may not think that was a good idea, but it was what I wanted to do to get to my answer. :) – 2017-02-28
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    (I prefer "absolute value" to "norm" for a complex number.) Yes it is true. – 2017-02-28
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    And I prefer "modulus. When you write *your* answers, you should use "absolute value", though, and I promise not to complain. Note that nowhere did the OP say that the codomain was the complex plane, so I was just guessing (from the presence of the double-bars) that it was some normed space. And then I picked a particularly simple normed space in which to show an example.:) – 2017-02-28
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    Because of the $\frac1u$ it had to be a codomain where that made sense. The OP seemed to clarify in comments when asked that $u$ is real or complex valued. (I was not complaining about your terminology; you *did* use absolute values, so how could I? But I wouldn't anyway. I put in the parenthetical because the beginning of your response addressed terminology which I had no dispute with. I tried to make clear that I had no dispute and failed.) – 2017-02-28
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    There can be nonconstant functions $u$ satisfying $\|u\|=1$ everywhere. – 2017-02-28
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    @Greg: yes, indeed, there can. But OP asked for a function that had a certain property; I played with the property long enough to discover something that suggested a possible solution to me (namely the constant function), and then named that function. I didn't claim that $\|u \| = 1$ *implied* that $u = 1$, just that $u = 1$ had that property (and the OP's desired property). – 2017-02-28
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    Fair enough. It's the question that's sloppily stated, not your answer. – 2017-02-28
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    @JonasMeyer: Fair enough; I misunderstood your intent, which is awfully easy to do. Apologies. – 2017-02-28