Equivalency of two optimization problem

Question

On what conditions (constraints) on $x,y$ the following two problems are equivalent?

P1: $\underset{x,y}\max ~\log(1+\frac{x}{y})$

P2: $\underset{x,y}\max ~ x-y$

the domain of the two problems is $x,y>0 , \frac{x}{y}\geq 1$

Answer 1

If $f(x,y)=\log(1+x/y),\ g(x,y)=x-y,$ then the gradient of $f$ is proportional to $(y,-x)$ and the gradient of $g$ is the constant vector field $(1,-1).$ Suppose somehow a restricted domain $D$ is set up for the two problems. Since in the entire plane the only places where the two gradients are proportional are the points of the line $x=y,$ it follows that any max on $D$ which is the same for each of $f,g$ cannot occur at an interior point of $D.$ That is, $D$ must have empty interior, which means it would have to be a peculiar domain in some sense.

There are of course trivial $D$ consisting of single points $D=\{(a,b)\}$ on which at least the point where the maxes of $f,g$ must be the same, and maybe one can rig up such pairs $a,b$ where the function values agree. These trivial cases didn't seem worth looking at.

But to me it looks like setting up reasonably defined $D$ for which the two problems either give the same points where maxes occur, and or the same values of the maxes, would be difficult, at least it would be hard to provably show one had found all such domains $D.$

On second thought the above only says something about where a possible common max might occur in $D$ (i.e. not at an interior point), but does not rule out a region having nonempty interior for which both functions restricted to $D$ have the same max. [I had deleted, but below found what I think is a reasonable answer, so will undelete it...]

Update: Interchange $x,y$ (just because it doesn't matter really) so your initial domain is $x,y>0$ and $y/x\ge1,$ the latter being $y \ge x.$ Now assume $P=(a,b)$ satisfies $0<a<b$ so lies in the initial domain. To make a region $D(a,b)$ for which each of $y-x$ and $y/x$ take on their maxima at the same point $(a,b),$ construct the region bounded by $y=x,$ by $y/x=b/a,$ and by $y-x=b-a.$ This region has one boundary line segment connecting $O=(0,0)$ to the point $P,$ another boundary line which is a ray starting at $P$ and pointing to the right from there with slope $1,$ and finally the boundary line which is the ray starting at $O$ and extending up and to the right at 45 degrees from there.

One cannot include any points other than $P$ on either the segment $OP$ or on the ray from $P$ going up at a 45 degree angle, otherwise one or both of the max problems will have more than one point at which the max is taken on. Other than that restriction (and that $O$ not be included), any subregion of this $D(a,b)$ will have the point $P$ as the location of both the max of $y-x,$ and the max of $y/x,$ [hence the max of $\log(1+y/x).$]

One can make closed subregions for such $D(a,b)$ with linear edges, if desired... these will be finite in extent. Another thing to note is that, given a fixed positive $a,$ there is in fact a particular $b>a$ for which the values $\log(1+b/a)$ and $b-a$ are actually equal as well. For these special points, the two max problems are then *completely$ equivalent [same max, at same point].