I have a question which seems trivial, but I don't know how to prove or disprove it.
Let $f:[0,1]\to (0,\infty)$ be an increasing function and let $g:[0,1] \to \mathbb{R}$ be a function satisfying $fg$ is decreasing in $[0,1].$ Then $g$ is decreasing in $[0,1].$
Is it true or false? Or what assumption can be imposed in order to make the statement true?
We may give the assumption on $f$ and $g$ which are differentiable functions.
Actually, I'd like to prove the following assertion
Let $f(t)$ be a positive increasing function and $g$ be a sign-changing function such that $g>0$ for $(0,a)$ and $g<0$ for $(a,1)$. If $fg$ is decreasing in $(0,1),$ then $g$ is decreasing in $(0,1).$
I would be grateful if you give any comment for my question. Thanks in advance.
Best Answer
By definition of increasing and decreasing,
$$ f(x) \text{ is increasing } \longleftrightarrow\frac{df}{dx}>0\\ f(x) \text{ is decreasing } \longleftrightarrow\frac{df}{dx}<0 $$
Therefore, $\frac{df}{dx}>0$ and $\frac{d}{dx}(fg)<0$. By the product rule, $f\frac{dg}{dx}+g\frac{df}{dx}<0\to\frac{dg}{dx}<-\frac gf\frac{df}{dx}$. By what we have been given, $f(x)$ and $\frac{df}{dx}$ are both positive, so as long as $g(x)$ is positive for every $x$ in your interval, $\frac{dg}{df} < 0$, which means $g(x)$ is decreasing. If $g(x)<0$ for any $x\in[0,1]$, then your conclusion is false.
Easiest counter example, let $f(x)=x$ and $g(x)=x-2$. $f(x)g(x)=x^2-2x$ and $\frac{d(fg)}{dx}=2x-2$, which is negative for all $x<1$, so it's decreasing in the interval.
Added to the Question:
For $g(x)$ to be a function such that $g(x)>0$ for $x\in(0,a)$ and $g(x)<0$ for $x\in(a,1)$, $g(x)$ must necessarily go from a positive number to a negative number, which means that it has to either be decreasing for some interval containing $a$ or it must have a discontinuity.
A discontinuous counterexample was given in the comments, namely
$$g(x)=\begin{cases}x, &x<a\\x - 2, &x\geq a\end{cases}$$
This function is always increasing, so $g(x)$ must also be specified to be continuous or the conclusion that $g(x)$ is decreasing is false.
For a continuous function, $g(x)$ must be decreasing around $x=a$, but it doesn't have to be decreasing anywhere else. For example, $g(x)=(x-b)(x-a)$, where $b=1 + \frac a2$ which decreases for $x<\frac {a + b}2$ and increases for $x>\frac {a + b}2$, and $f(x)=x^3$. $\frac{d(fg)}{dx}=5x^4-4(a+b)x^3-3abx^2=5x^4-4x^3-6ax^3-3(ab)x^2$. So now, we need to find all the values of $a$ for which the derivative is less than zero.
$$\begin{align} 5x^4-4x^3-6ax^3-3abx^2 &< 0 &(\text{substitution})\\ 5x^4-4x^3-6ax^3-3abx^2 &< 5x^4-4x^3-6ax^2 &(a, b, x^2 > 0)\\ x^3(5x-4-6a) &< 0 \\ 5x-4 &< 6a &((x^3>0)(\forall x \in(0,1))\\ a &> \frac16 &(\max x = 1) \end{align}$$
In other words, as long as $a>\frac16$, $fg$ will be decreasing on $(0,1)$, so we'll pick an $a$ greater than $\frac16$. We also need to find out for what values of $a$ is $\frac{a + b}2<1$, which is simply
$$\begin{align} a + \frac a2 + 1 &< 2\\ \frac32a &< 1\\ a &< \frac23 \end{align}$$
So $a$ also has to be less than $\frac23$.
Since $f(x)$ is positive and increasing for $x\in(0,1)$, $fg$ is decreasing for $x\in(0,1)$, and $g(x)$ is a smooth function that is positive for $x<a$ and negative for $x>a$, but is increasing for $x\in\left(\frac{a + b}2,1\right)$ as long as $a\in\left(\frac16,\frac23\right)$, we have a known counterexample to the statement that $g(x)$ must be decreasing on $x\in(0,1)$.