# 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 $x > 0$.
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 $x>0$.
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