Problem on continuity based on functional relation

Question

Given that $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies

$2f^3(x)-3=2x-3f(x)$ , $x\in \mathbb{R}$, show that $f$ is continuous on $\mathbb{R}$.

How can we handle this problem?

Answer 1

Let us define $$F(x):=2f^3(x)+3f(x)$$ then $$F(x)=3+2x$$ hence $F$ is continuous. Now let us define $$ G(y):= 2y^3 + 3y $$ then, since $G'(y)=6y^2+3>0\ $, (plus computing the limits in $\pm\infty$, $G:\mathbb{R}\rightarrow \mathbb{R}$ is a diffeomorphism (i.e. $t\mapsto G^{-1}$ exists and is continuous).

We have $$ F=G\circ f$$ so $$ f=G^{-1}\circ F$$ hence $f$ is continuous.

Answer 2

The solution of @Netchaiev is short and smart. I managed to get to a lengthy alternative below:

For $x_0 \in \mathbb{R}$ we have $2f^3(x_0)=2x_0-3f(x_0)+3$

Subtract the latter from the given $2f^3(x)=2x-3f(x)+3$ to take

$2(f^3(x)-f^3(x_0)) +3(f(x)-f(x_0))= 2(x-x_0)$

or

$2(f(x)-f(x_0))(f^2(x)+f(x)f(x_0)+f^2(x_0))+3(f(x)-f(x_0))=2(x-x_0)$

or

$(f(x)-f(x_0))(2f^2(x)+2f(x)f(x_0)+2f^2(x_0)+2+1)=2(x-x_0)$

Let $A(x) = 2f^2(x)+2f(x)f(x_0)+2f^2(x_0)+2$

the above is quadratic with respect to $f(x)$ with discriminant

$\Delta = (2f(x))^2-4\cdot2\cdot(2f^2(x)+2)=4f^2(x)-8\cdot2\cdot f^2(x)-16= -12f^2(x)-16<0$

thus $A(x)>0, \forall x$ and

$(f(x)-f(x_0))(A(x)+1)=2(x-x_0)$. Then

$\left|f(x)-f(x_0)\right|=\frac{\left|2(x-x_0)\right|}{A(x)+1}\leq \left|2(x-x_0)\right|$, since $A(x)+1>1$

or

$-\left|2(x-x_0)\right| \leq f(x)-f(x_0) \leq \left|2(x-x_0)\right|$

We have that $\lim_{x \to x_0} 2(x-x_0) = 0$

then from the squeeze theorem it follows that $\lim_{x \to x_0} (f(x)-f(x_0)) = 0$, thus $f(x)$ is continuous