# Representation of this function using a single formula without conditions

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 $F=u\cdot 1_{(-\infty,0]}+v\cdot 1_{(0,\infty)}$. – Rasmus Oct 12 '11 at 8:39

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. – Tapu Oct 12 '11 at 9:32
@user7530 you are clever! – NoChance Oct 12 '11 at 9:43
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! – Rasmus Oct 12 '11 at 11:00
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. – user7530 Oct 12 '11 at 11:07
@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.) – Did Oct 12 '11 at 14:40

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. – Rasmus Oct 12 '11 at 12:52
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. – Did Oct 12 '11 at 14:32
Sorry, my mistake, will edit it now. $G = G(x)$ – maximus Oct 13 '11 at 1:14
@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.) – Did Oct 13 '11 at 5:18