max of product of two functions

Question

Let $f(x)$ and $g(x,y)$ be non-negative continuous functions when $x\in [a,b]$, $y\in [c,d]$,

so they are bounded and there exists $C>0$ such that $$ 0\leq f(x)\leq C \quad \forall x\in [a,b], \quad\quad 0\leq g(x,y)\leq C \quad \forall (x,y)\in [a,b]\times [c,d]. $$

Let $x^{\ast}\in [a,b]$ be arg max of $f(x)$ on $[a,b]$, that is, $$ f(x^{\ast})=\max\limits_{x\in [a,b]}f(x). $$

Is it true that $$ \max\limits_{(x,y)\in [a,b]\times [c,d]}f(x)g(x,y)=\max\limits_{x\in [a,b]}f(x) \cdot \max\limits_{y\in [c,d]}g(x^{\ast},y) \,? $$

I think it is not true but I cannot construct a counter-example. Or it is really true?

As I see, we have $f(x^{\ast})\geq f(x)$, $g(x^{\ast},y^{\ast})\geq g(x^{\ast},y)$. And if we can prove that $g(x^{\ast},y)\geq g(x,y)$ the problem would be solved positively. But I cannot prove the last inequality.

Answer 1

assuming $x,y \in [0,1]$ choose

$$f(x) = x, g(x,y) = (1-x)y$$

The max of $f$ is achieved in $x = 1$, but the max of $f(x)g(x,y)$ is in $(1/2,1)$

Answer 2

If $g$ is constant wrt $y$, then it is easier to find a counterexample.

Consider $f,h:[0,1]\to[0,1]$ defined by $f(t)=t$ and $h(t)=1-t$. Then $x^*=1$ and $$ \max_{x\in[0,1]}f(x)h(x)=\max_{x\in[0,1]}x-x^2=\frac14 $$ while $$ \max_{x\in[0,1]}f(x)h(x^*)=1\cdot 0=0. $$ Finally, you can choose $g:[0,1]\times[c,d]\to[0,1]$ by $g(x,y)=h(x)$ and you get $$ \max_{(x,y)\in[0,1]\times[c,d]}f(x)g(x,y)=\max_{(x,y)\in[0,1]\times[c,d]}x-x^2=\frac14 $$ while $$ \max_{x\in[0,1]}f(x)\max_{y\in[c,d]}g(x^*,y)=1\cdot 0=0. $$