In the last couple of weeks, I have been dealing with a two-variable optimization problem, which I have been unable to solve. In the problem given below, I am attempting to show that the unique local maximum in the interior is a global maximum. Any comments to my solution attempt is much appreciated, as well as answers to the questions specified in the text. The optimization problem is as follows:

$\max_{x,y}F(x,y)=(f(x)+cx)y-g(y)$ where $x\in[0,\overline{x}]$ with $0<\overline{x}$, $y\geq 0$ and $c>0$.

  • Assumptions imposed on $f$: $f$ is continuous, monotonically decreasing and concave in $x$ with $f(0)>f(\overline{x})>0$, $f'(0)=0, f'(x)<0 \forall x\in (0,\overline{x}], \lim_{x\to\overline{x}}f'(x)=-\infty$ and $f''(x)<0\forall x$.
  • Assumptions imposed on $g$: $g$ is continuous, monotonically increasing and convex in $y$ with $g(0)=0, g'(0)=0, g'(y)>0\forall y>0,g''(y)>0\forall y$ and $\lim_{y\to\infty}g'(y)=\infty$.

As the Hessian is not negative semidefinite for all $x$ and $y$, I am attempting an alternative solution:

My solution:

The stationary points are given by

$F_{x}(x,y)=0\iff (f'(x)+c)y=0$

$F_{y}(x,y)=0\iff f(x)+cx-g'(y)=0$

It follows that the single stationary point in the domain is such that $-f'(x)=c$ and $f(x)+cx=g'(y)$. Note that $F_{x}(x,y)=0$ implies that a second stationary point could be located at $y=0$, but then $F_{y}(x,y)=0$ requires that $f(x)=-cx$, which is not possible.

To show that our stationary point is a local maximum, it is sufficient to show that:

  • $F_{xx}(x^{0},y^{0})<0\iff f''(x^{0})y^{0}<0$ which holds since $f''<0$ and $y^{0}>0$

  • $F_{yy}(x^{0},y^{0})<0\iff -g''(y^{0})<0$ which also holds since

  • $F_{xx}(x^{0},y^{0})F_{yy}(x^{0},y^{0})-(F_{xy}(x^{0},y^{0}))^{2}>0\iff -g''(y^{0})f''(x^{0})y^{0}-(f'(x^{0})+c)^{2}$ which holds since the first term is strictly positive and the second is $0$ at $(x^{0},y^{0})$.

I know wish to show that $(x^{0},y^{0})$ is a global maximum. As far as I understand, I need to check the following:

(i) $F(x^{0},y^{0})\geq F(x,0), \forall x$,

(ii) $F(x^{0},y^{0})\geq \lim_{y\to\infty}F(x,y),\forall x$,

(iii) $F(x^{0},y^{0})\geq F(0,y), \forall y$, and

(iv) $F(x^{0},y^{0})\geq F(\overline{x},y), \forall y$.

Solution to (i) and (ii): Note that $F(x,0)=0$, so (i) holds if $F(x^{0},y^{0})\geq 0$. Further, as far as I understand, the conditions on $g$ implies that $\lim_{y\to\infty}g(y)=\infty$, so $\lim_{y\to\infty}F(x,y)\leq 0$. So once again, (ii) holds if $F(x^{o},y^{0})\geq 0$. Substituting $g'(y^{0})$ into $F(x^{0},y^{0})$, it becomes

$g'(y^{0})y^{0}-g(y^{0})\geq 0\iff g'(y^{0})\geq\frac{g(y^{0})}{y^{0}}$, which holds by the assumptions imposed on $g$. Thus, (i) and (ii) holds. We are left to show that (iii) and (iv) holds.

Solution to (iii): Let $x=0$ as in (iii). It follows that the optimal $y(0)$ is such that $g'(y(0))=f(0)$. Thus (iii) holds if the following condition holds:

$g'(y^{0})y^{0}-g(y^{0})\geq g'(y(0))y(0)-g(y(0))$

Which should hold since $g(y)$ is increasing and convex in $y$. For example, let $g(y)=\frac{1}{2}y^2$. Then the condition is equivalent to:

$\frac{1}{2}y^{0^{2}}\geq \frac{1}{2}y(0)^{2}$ which holds since $y^{0}>y(0)\geq 0$.

Solution to (iv): Now, my main problem is the final possible case: Let $x=\overline{x}$ as in (iv). It follows that the optimal $y(\overline{x})$ is such that $g'(y(\overline{x}))=f(\overline{x})+c\overline{x}$. Thus (iv) holds if the following condition holds:

$g'(y^{0})y^{0}-g(y^{0})\geq g'(y(\overline{x}))y(\overline{x})-g(y(\overline{x}))$

However, by my argumentation in case (iii), this inequality should not hold.

EDIT: What I meant with the sentence above is that the inequality above holds if and only if $y^{0}\geq y(\overline{x})$ due to the assumptions imposed on $g$ (i.e., $g',g''>0$). Is this correct? Furthermore, it should be the case that $y^{0}\geq y(\overline{x})$ if and only if $g'(y^{0})\geq g'(y(\overline{x}))$, right? When I wrote the sentence, I was convinced that $y^{0}<y(\overline{x})$ and thus $g'(y^{0})<g'(y(\overline{x}))$. However, now that I think about it, this would imply that

$f(x^{0})-cx^{0}<f(\overline{x})+c\overline{x}\iff f(x^{0})-f(\overline{x})<c(\overline{x}-x^{0})$

which would mean that the (absolute) decrease in $f$ by moving from $x^{0}$ to $\overline{x}$ is smaller than the increase in $c(\overline{x}-x^{0})$. However, the condition $\lim_{x\to\overline{x}}f'(x)=-\infty$ implies that the decrease is going to larger than the increase. Thus, $y^{0}>y(\overline{x})$. Is this correct?


My qusetions:

  1. List item

  2. Are my steps (i)-(iii) correct?

  3. Does (iv) imply that the maximum is located at $(\overline{x},y(\overline{x}))$? Or does $\lim_{x\to\overline{x}}f'(x)=-\infty$ rule such a boundary solution out?

  4. If no, is there any condition that can be imposed to ensure that $(x^{0},y^{0})$ is indeed a global maximum?

Many thanks in advance!

  • $\begingroup$ Can you make your argument in point (iv) more explicit? I mean, can you expand your "by my argumentation in case (iii)"? $\endgroup$ – Taneli Huuskonen Feb 15 '18 at 13:28
  • $\begingroup$ I have edited my thoughts on (iv). Thanks for the question. I think that made me realize a mistake I was making. $\endgroup$ – math_learner Feb 15 '18 at 13:49
  • $\begingroup$ Your conditions (i) through (iv) look OK, and your expanded version of (iv) looks promising. I'm afraid I don't have the time to check your reasoning more carefully now, though. $\endgroup$ – Taneli Huuskonen Feb 15 '18 at 14:01
  • $\begingroup$ Thank you very much. If you at some later point find yourself with more time your hand, it would be greatly appreciated. $\endgroup$ – math_learner Feb 15 '18 at 14:16

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