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Is it possible to represent the following function with a single formula, without using conditions? If not, how to prove it?

$F(x) = \begin{cases}u(x), & x \le 0, \ v(x) & x > 0 \end{cases}$

So that it will become something like that: $F(x) = G(x)$ With no conditions?

I need it for further operations like derivative etc.

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    The only thing you can do is writing something like $F=u\cdot 1_{(-\infty,0]}+v\cdot 1_{(0,\infty)}$.2011-10-12

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What operations are allowed in the formula?

$$G(x) = \frac{x + |x|}{2x} v(x) + \frac{x - |x|}{2x} u(x)$$ will work (away from 0), but any "trick" along these lines is not going to help make taking derivatives any easier.

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    Btw, $|x|$ is itself conditional, according to the question. So, I think in strict sense, the answer should be no.2011-10-12
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    @user7530 you are clever!2011-10-12
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    Why do people upvote this? The function does not have $x$ has an input variable, so its definition using $x$ instead of just $u$ and $v$ does not make sense!2011-10-12
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    I've rewritten G as a functional of all three $u, v,$ and $x$ to make it clearer, though I don't see why a minor detail of notation requires downvotes.2011-10-12
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    Thanks @user7530. Sorry for my snarky previous comment. I didn't ask for downvotes, though. Actually, I think it would be best to write just $G(x)=\ldots$. After all, you want $G$ to be a usual function in $x$, right? Not a functional in $u$ and $v$.2011-10-12
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    I would also like to remark that $\lvert x\rvert$ is just an abbreviation for $$\begin{cases}x & x\geq 0,\\ -x & x<0.\end{cases}$$2011-10-12
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    @user7530, A function of $u$, $v$ and $x$ such as the one you provided is a beast altogether different from a function of $(u(x),v(x))$, which the OP asked for. To describe the difference as *rewriting G as a functional of all three $u$, $v$, and $x$ to make it clearer* is misleading and, technically speaking, your post does not answer the question. (But this is no big deal and I did not downvote it.)2011-10-12
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    @Didier Well, in my mind it answers what the OP *intended*, according to his question in English. He also posted a speculative attempt at formalizing his question, and as you say this formula (now edited) interpreted strictly did not match his English description of his problem. I don't think there a need for us to argue at too great length about which was the "true question" and I assure you I did not intend to mislead anyone :)2011-10-13
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    @user7530, I have no way to know what is in your mind nor what the OP *intended*, only what the OP *wrote*. And it happens that neither the English description of the problem nor the formulas attached to it fitted (at any moment before the last revision) the description in your last comment (and I do not know either where you got the bizarre idea that they did). Since the true question (without scare quotes) is [there](http://math.stackexchange.com/revisions/b7f4b263-d377-44a2-ab85-5e3d6f42174c/view-source) for all to see, indeed there is no need to *argue*. Enough with this stuff for me.2011-10-13
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    I read the OP's question and thought, "he wants a 'single formula' (non-piecewise) reformulation of his $f$." It didn't even cross my mind at the time that there might be other interpretations; I acknowledge that the question I answered is not the same as finding a $G$ that is *only* a function of $u$ and $v$. I never would have imagined this question would be so controversial, and apologize if my posts here have been bizarre or aggressive. I hope there are no hard feelings moving forward.2011-10-13
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Note: This answers the original question, asking whether a formula like $F(x)=G(u(x),v(x))$ might represent the function $F$ defined as $F(x) = u(x)$ if $x \leqslant 0$ and $F(x)=v(x)$ if $x > 0$.
The OP finally reacted to remarks made by several readers that another answer did not address this, by modifying the question, which made the other answer fit (post hoc) the question.


Just to make sure @Rasmus's message got through: for any set $E$ with at least two elements, there can exist no function $G:\mathbb R^2\to E$ such that for every functions $u:\mathbb R\to E$ and $v:\mathbb R\to E$ and every $x$ in $\mathbb R$, one has $G(u(x),v(x))=u(x)$ if $x\leqslant0$ and $G(u(x),v(x))=v(x)$ if $x>0$.

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    Thanks @Didier, that's what bothered me as well.2011-10-12
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    Thanks to you @Rasmus, since this simply reproduces your remark. I do not understand either how a function of $x$ can answer the query for a function of $(u(x),v(x))$. Well... nevermind.2011-10-12
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    Sorry, my mistake, will edit it now. $G = G(x)$2011-10-13
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    @maximus, This is shocking, and akin to *rewriting history*. You should have posted the second question as another post. (And we still do not know whether you realized the two versions of the question are completely different, or not.)2011-10-13