Consider the following problem \begin{align} &\max_u F(x,u)\\ \text{s.t. }& u \in [0,\bar u]. \end{align}
Any idea how to merge the two constraints $u \geq 0$ and $\bar u - u \geq 0$ into one constraint $f(u,\bar u) \geq 0$?
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2$\begingroup$ Sure. Define the function $f$ so that $f(u,\bar{u}) = -1$ if $u < 0$ or $\bar{u} - u < 0$, and otherwise let $f(u,\bar{u}) = 0$. This is a well defined function. Perhaps you would like to include that $f$ is differentiable or some similar condition. $\endgroup$– GiskardCommented Mar 29, 2016 at 12:20
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1 Answer
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$$0\leq u \leq \bar u \implies -\frac {\bar u}{2} \leq u - \frac {\bar u}{2} \leq \frac {\bar u}{2}$$
$$\implies \left | u - \frac {\bar u}{2}\right| \leq \frac {\bar u}{2}$$
$$\implies \frac {\bar u}{2} - \left | u - \frac {\bar u}{2}\right| \geq 0$$
ADDENDUM In a comment it was proposed that we could instead use the squared expression to achieve differentiability everywhere,
$$\frac {\bar{u}^2}{4} - \left ( u - \frac {\bar u}{2}\right)^2 \geq 0$$
Let's see: we then are allowed to decompose the square and write
$$\frac {\bar{u}^2}{4} - u^2 + u\bar u - \frac {\bar{u}^2}{4} \geq 0$$
$$\implies -u^2 + \bar u u \geq 0 \implies u(\bar u -u) \geq 0$$
which is nothing more than the multiplication of the two separate constraints.
ADDENDUM II
If we have $u \in [a,b]$ for $a<b$ arbitrary reals, then the general expression is
$$-u^2+(a+b)u-ab\geq 0$$
answered Mar 29, 2016 at 16:23
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3$\begingroup$ Nice! But $$\frac {\bar{u}^2}{4} - \left ( u - \frac {\bar u}{2}\right)^2 \geq 0$$ is even better because it is differentiable everywhere, even in $u = \frac {\bar u}{2}$. $\endgroup$– GiskardCommented Mar 29, 2016 at 16:44
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$\begingroup$ @denesp By all means - and it gets even better. $\endgroup$ Commented Mar 29, 2016 at 17:30
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$\begingroup$ Thanks guys. Is it conincidence that it's just the multiplication of the two seperate constraints, or some general rule? $\endgroup$– cluelessCommented Mar 29, 2016 at 18:20
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$\begingroup$ @AlecosPapadopoulos Good decomposition but actually you have just shown that all our rules so far are incorrect :) Your new condition also holds if $u < 0$ AND $\bar u < u$ :( If $\bar u < 0$ is impossible then we are fine though. $\endgroup$– GiskardCommented Mar 29, 2016 at 18:29
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$\begingroup$ @denesp These inequalities cannot hold at the same time. $\endgroup$ Commented Mar 29, 2016 at 18:31