To clarify: I am working with specific $f$ and $g$. They are algebraically very tedious, which makes solving for FOC and SOC rather difficult. I believe the global extremum of my $f+g$ exists only at the endpoints and I'm asking what conditions do I need to check to prove that. Hypothetical counter-examples aren't that helpful because my $f$ and $g$ are specific.


Original Post

Suppose I have two functions $f$ and $g$, both on [0,1]. Suppose I know $f$ is strictly decreasing and convex. $g$ is strictly increasing and concave. And both functions are negative on [0,1]. Is there anything I'd know about $f+g$?

I'm basically trying to show that the global maximum of $f+g$ lies always on 0 and/or 1(the endpoints). Any clues on how to show that or what would I need to show that would be greatly appreciated!

P.S. I do expressions of these functions but they are algebraically very tedious. Meaning it's very difficult to literally "solve" for anything but I suppose I could check for certain properties.



1 Answer 1


With only the given properties of $f$ and $g$, there need not be a global maximum at $0$ or $1$. To show this, let $f(x)=-2x$ and $g(x)=-x^2+3x-2$. It is easy to see that these functions meet your criteria and that $f+g\,$ has a unique maximum at $x=\frac{1}{2}$. A small modification will even allow $f\,$ to be strictly convex.

Note that the condition that both $f$ and $g$ be negative is irrelevant: Since both $f$ and $g$ are bounded, you can always add a suitable constant to make both $f$ and $g$ negative, because all other mentioned properties are invariant under addition of a constant, in particular the location of the maxima.

  • $\begingroup$ Thanks overglow! The properties I mentioned aren't necessarily the only given properties of $f$ and $g$. As said, I do have explicit expressions of $f$ and $g$ and I'm basically asking what other conditions do I need to verify to prove that the maximum locates at the endpoints. $\endgroup$ Jun 24, 2021 at 9:45

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