Considering this constraint, determine the real function $f$

Question

The problem

Let $f: \mathbb{R} \to \mathbb{R}$

Determine $f(x)$ knowing that

$ 3f(x) + 2 = 2f(\left \lfloor{x}\right \rfloor) + 2f(\{x\}) + 5x $, where $ \left \lfloor{x}\right \rfloor $ is the floor function and $\{x\} = x - \left \lfloor{x}\right \rfloor$ (also known as the fractional part)

My thoughts

We can observe that for $x = 0$ we obtain $f(0) = 2$.

Considering $f(\left \lfloor{x}\right \rfloor)$ we get $ 3f(\left \lfloor{x}\right \rfloor) + 2 = 2f(\left \lfloor\left \lfloor{x}\right \rfloor\right \rfloor) + 2f(\{\left \lfloor{x}\right \rfloor\}) + 5\left \lfloor{x}\right \rfloor $

And for $f(\{x\})$ we get $ 3f(\{x\}) + 2 = 2f(\left \lfloor\{x\}\right \rfloor) + 2f(\{\{x\}\}) + 5\{x\} $

I did this in the hope of defining $f(\left \lfloor{x}\right \rfloor)$ and $f(\{x\})$ and thus replacing them in the initial condition.

Answer 1

As you explained, $f(0)=2$. Now assume that $x\in [0,1[$. Then $\{x\}=x$ and $[x]=0$, so we have $3f(x)+2=2f(0)+2f(x)+5x$, which yields $f(x)= 5x+2$ for all $x\in [0,1[$.

If $x\in \mathbb{Z}$, we have $[x]=x$ and $\{x\}=0$. Thus the equation simplifies to $3f(x)+2=2f(x)+2f(0)+5x$, yielding $f(x)=5x+2$ again for all $x\in \mathbb{Z}$.

Finally, for a general element in the domain $x\in \mathbb{R}$, we have $[x]\in \mathbb{Z}$ and $\{x\}\in [0,1[$, so we can combine the above two partial results to obtain

$$3f(x)+2=2f([x])+2f(\{x\})+5x = 2(5[x]+2)+2(5\{x\}+2)+5x=$$ $$10([x]+\{x\})+5x+8= 15x+8$$

thus $f(x)=5x+2$ for all $x\in \mathbb{R}$.

Answer 2

The first of your derived equations allows to calculate $f(\lfloor x\rfloor)$ by observing that $\{\lfloor x\rfloor\}=0$. Similarly the second may serve to calculate $f(\{x\}) $ by taking into account that $\lfloor\{x\}\rfloor=0$.