Proving value that maximizes function increases when function is multiplied by increasing function

Question

Let $k$ be the value that maximizes $f(x)$. Let $g(x)$ be increasing and let $h(x)=f(x)g(x)$. I am pretty sure that if $j$ is the value that maximizes $h(x)$ then $j\geq k$, but I am not sure how to prove this. Any suggestions?

Answer 1

The function $g(x) = x$ is increasing. The function $f(x) = \begin{cases} -10 & x = -1 \\ 1 & x= 1 \\ 0 & \text{otherwise}\end{cases}$ has a maximum at $1$. But $h(x) = g(x)f(x)$ has a maximum at $-1 <1$. Thus your proposition doesn't hold in general.

But if we suppose all our functions are nonnegative, then the proposition holds. Suppose $y \leq k$. Then $g(y) \leq g(k)$ and $f(y) \leq f(k)$, hence $h(y) = f(y) g(y) \leq f(k)g(k) = h(k)$. Thus any maximum must be at some $j \geq k$.