I have something like this $$ h(x) = f(x) g(x) $$ where $f$ and $g$ are some nice functions such that $f$ is $[0,1]$-bounded, concave function and $g$ is some negative, decreasing, concave function.
Is $h$ concave? If not, then can we impose more conditions on $f$ and $g$ such that $h$ is concave?
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$\begingroup$Let's take $$ f(x)=g(x)=x, $$which is concave (but not strictly concave).
Then $h(x)=x^2$, which is convex.
NB I take the definition of concavity from Wikipedia. It might be the other way round in your textbook.
You can then easily find strictly concave examples. Take$$ f(x)=g(x)=x^{2/3}, $$which are both concave, but $f\times g$ is clearly convex.
$\endgroup$ $\begingroup$No. Let $f$ be decreasing concave and bounded and $g(x)=-x$. Then $fg$ is actually convex for $x >0$. You can see this by computing the second derivative.
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