functional equation with definite integration

Question

If $g:\mathbb{R}-\{0\} \rightarrow \mathbb{R},g(2020)=1,g(-3)=-1$

and $g(x)\cdot g(y)=2g(xy)-g\bigg(\frac{2020}{x}\bigg)\cdot g\bigg(\frac{2020}{y}\bigg)\forall x,y\in \mathbb{R}-\{0\}$.

Then value of $\displaystyle \int^{2021}_{-1}g(x)dx=$

Here we have given

$\displaystyle g(x)\cdot g(y)=2g(xy)-g\bigg(\frac{2020}{x}\bigg)\cdot g\bigg(\frac{2020}{y}\bigg)\cdots \cdots (1)$

Then using partial fraction

Differentiate with respect to $x,$ we get

$\displaystyle g'(x) \cdot g(y)=2g'(xy)\cdot y+g'\bigg(\frac{2020}{x}\bigg)\cdot \frac{1}{x^2}\cdot g\bigg(\frac{2020}{y}\bigg)\cdots (2)$

Differentiate with respect to $y,$ we get

$\displaystyle g(x) \cdot g'(y)=2g'(xy)\cdot x+g'\bigg(\frac{2020}{y}\bigg)\cdot \frac{1}{y^2}\cdot g\bigg(\frac{2020}{x}\bigg)\cdots (3)$

How do I solve after that, please help me

Answer 1

This answer assumes that $g$ is differentiable.

Substituting $x=1$ and $y=2020$ into $$g(x)g(y)=2g(xy)-g\bigg(\frac{2020}{x}\bigg)g\bigg(\frac{2020}{y}\bigg)\tag1$$ we get $$g(1)g(2020)=2g(2020)-g(2020)g(1)$$ which implies $g(1)=1$.

Substituting $y=1$ into $(1)$, we get $$g(x)=g\bigg(\dfrac{2020}{x}\bigg)\tag2$$

From $(2)$, we also have $g(\frac{2020}{y})=g(y)$, so from $(1)$, we get $$g(xy)=g(x)g(y)\tag3$$

Suppose that there is $a$ such that $a\not=0$ and $g(a)=0$. Then, we have $g(x)=g(a\cdot\frac xa)=g(a)g(\frac xa)=0$ which means that $g(x)$ is identically $0$, which is a contradiction.

Differentiating the both sides of $(3)$ with respect to $y$, we have $$xg'(xy)=g(x)g'(y)$$ Evaluating $y$ at $1$, and letting $ p = g'(1)$ and $z = g(x) $, we have

$$ x\frac{dz}{dx} = pz \implies \int \frac{dz}{z} = p\int \frac{dx}{x}$$ so $$\ln|g(x)|=p\ln|x|+C_1$$ Substituting $x=1$, we get $C_1=0$. So, $g(x)=\pm |x|^{p}$.

From $(2)$, we get $$\pm |x|^{p}=\pm\bigg|\frac{2020}{x}\bigg|^{p}\implies x^{2p}=\pm 2020^{p}$$ which implies $p=0$, so we get $g(x)=\pm 1$.

Substituting $x=y=\sqrt t$ into $(3)$ gives $$g(t)=(g(\sqrt t))^2\gt 0$$ which means that if $x\gt 0$, then $g(x)\gt 0$.

Substituting $x=-3$ and $y=\frac{t}{-3}$ with $t\lt 0$ into $(3)$ gives $$g(t)=g(-3)\underbrace{g\bigg(\frac{t}{-3}\bigg)}_{\text{positive}}\lt 0$$ which means that if $x\lt 0$, then $g(x)\lt 0$.

Therefore, we finally get $$g(x)=\begin{cases}1&\text{if $x\gt 0$}\\ -1&\text{if $x\lt 0$}\end{cases}$$ which is sufficient.

Hence, we obtain $$\int_{-1}^{2021}g(x)dx = -\int_{-1}^0dx + \int_0^{2021}dx = \color{red}{2020}$$

Answer 2

Here is an incomplete solution: by inspection, one can verify that $$ g(x)=\begin{cases} -1,\, \text{if $x<0$} \\ +1,\, \text{if $x>0$} \end{cases} $$ satisfies the functional equation. Therefore, for this solution, $$ \int_{-1}^{2021}g(x)dx = -\int_{-1}^0dx + \int_0^{2021}dx = 2020. $$