Does this a function $x+y \rightarrow x^2+y^2$ exist?

Question

I don't know if this exists, but it would make my algebra easier if it did instead of having to use complicated radicals to solve an equation.

If $x \in \mathbb{R}$ and $y \in \mathbb{R}$ and $h(x,y)=x+y$, is there a real or complex function $f$ such that $f(x+y)=x^{2}+y^{2}$? And if so, what is it?

Answer 1

It doesn't exist because $f(0)=f(-1 + 1) = 2$ and $f(0)=f(2+-2)=8$.

Answer 2

It cant be a single value function because if $w = a+ b = c+d$ then $f(w) = a^2 + b^2$ which may not equal $c^2 + d^2$ so it is not well defined.

But in you comments you seem to ask about multivalued "functions" with branches such as $\arcsin$ and $\sqrt{}$ and $\ln$ of complex numbers where if $f(x) = y$ it can actually be argued that there are a set of $\{y_\alpha\}$ where the each of the $y_\alpha$ can be argued is legitimate value for $f(x)$.

This "branching", of course, violates the definition of a function. And for the must part this functions are restricted to only specific outcomes. For example: although $\sin \theta = \sin (\pi - \theta) = \sin (\theta + 2\pi)=x$ it would seem $\arcsin x$ could be all of $\theta, \pi - \theta,$ or $\theta + 2\pi$. But this is not what happens because we restrict, by definition, that $\arcsin x$ must be a value between $-\frac \pi 2$ (exclusive) and $\frac \pi 2$ (inclusive).

Likewis $\sqrt{x}$ is defined to always be non-negative.

This gets iffy with complex numbers that if $e^x = w$ then $e^{x + 2\pi i} = w$ and we don't bother to restrict $\ln w$ to a specific one of the infinite possible $x + 2\pi i$ but instead often say $\ln w = $ all and any of the $x + 2\pi i$ values.

This utterly bollocks the idea of "function" where $f(w)$ really has to be one thing.

Well, the work around is that $\ln$ is not a function of $\ln: \mathbb C \to \mathbb C$ but a function of $\ln:\mathbb C \to \mathbb C/mod 2\pi i$. That is that $\ln$ doesn't map to a complex number but to an equivalence class.

We can say for two different complex $w_i \ne w_j$ that $w_i$ is equivalent to $w_j$ if $w_i - w_j = 2k \pi i$. This relationship is an equivalence relations ship (its reflexive, symmetric and transitive; two elements related in this way might as well be considered for our purposes to be the same). And for a given $w$ then the set of all $\{w_i| w_i - w = 2k\pi i$ for some integer $k\}$ is call an equivalence class.

So $\ln w = w + 2k\pi$ is not a single number, but a single equivalence class.

And $\mathbb C = \cup $ all equivalence classes. And any two equivalence classes are distinct.

So If we want to define $f(w) = a^2 + b^2$ when $w = a+b$ in this way as a multivalued "function" we must consider $a^2 + b^2$ as an equivalence class.

And if we do that, if we want say $w $ is equivalent to $v$ if there is a number $k$ so that $w = (k -e)^2 + e^2$ for some value $a$ and $v = (k-h)^2 + h^2$ for some $h$

But that is not an equivalence relation. And the classes form by this are not distinct.

So this can't make any sense.