Functional Equation again

We claim that the only solution is $f(x)=x+1$ . Let $P(x,y)$ be the assertion, using $P(x+y,0)$ and $P(x,y)$ , we can easily get $f(x+y)+f(0)=f(x)+f(y)$ and then let $g(x)=f(x)-f(0)$ , $g(x+y)=g(x)+g(y)$ since $f$ is continuous, $g$ is also continuous and thus $g$ is linear. Thus, $f(x)=kx+c$ where $k,c \in \mathbb{R}$ . Then we can find that when $k=1$ , since $f(1010)=1011$ , we must have $c=1$ . So, the only solution is $f(x)=x+1 \ \forall x \in \mathbb{R}.$ Hence, $f(2020)=2021.$

Let $P(x,y)$ be the statement $f\left( {f\left( {x + y} \right)} \right) = f\left( x \right) + f\left( y \right){\text{ }}\left( 1 \right)$ .
Moreover, let $f\left( 0 \right) = c$ . $P\left( {x + y,\;0} \right) \Rightarrow f\left( {f\left( {x + y} \right)} \right) = f\left( {x + y} \right) + c{\text{ }}\left( 2 \right)$

$\left( 1 \right),\left( 2 \right) \Rightarrow f\left( {x + y} \right) + c = f\left( x \right) + f\left( y \right){\text{ }}\left( 3 \right)$

Let $g\left( x \right) = f\left( x \right) - c$ , $x \in \mathbb{R}$ .
Then, $\left( 3 \right) \Rightarrow f\left( {x + y} \right) - c{\text{ = }}\left( {f\left( x \right) - c} \right) + \left( {f\left( y \right) - c} \right)$ $\Rightarrow g\left( {x + y} \right){\text{ = }}g\left( x \right) + g\left( y \right).$ Thus, $g$ satisfies Cauchy’s functional equation . Since $f$ is continuous, the same holds for $g$ . It is well known that any continuous solution to Cauchy’s equation is of the form $g\left( x \right) = kx$ , for some $k \in \mathbb{R}$ .

This implies that $f\left( x \right) = kx + c,{\text{ }}x \in \mathbb{R}$ .

Now,
$\begin{gathered} P\left( {x,\;0} \right) \Rightarrow f\left( {f\left( x \right)} \right) = f\left( x \right) + c \\ \Rightarrow k \cdot f\left( x \right) + c = kx + 2c \\ \Rightarrow k \cdot \left( {kx + c} \right) + c = kx + 2c \\ \Rightarrow {k^2}x + kc = kx + c \\ \end{gathered}$

$\Rightarrow \left\{ \begin{matrix} {{k}^{2}}=k \\ kc=c \\ \end{matrix} \right.\Rightarrow \left( \left\{ \begin{matrix} k=0 \\ c=0 \\ \end{matrix}\text{ }or \right.\text{ }\left\{ \begin{matrix} k=1 \\ c\in \mathbb{R} \\ \end{matrix} \right. \right)$

Since $f$ is not the zero function, the first option is rejected. Hence, $f\left( x \right) = x + c,{\text{ }}x \in \mathbb{R}$ .
Plugging $x = 1010$ , we get $f\left( {1010} \right) = 1010 + c \Rightarrow 1011 = 1010 + c \Rightarrow c = 1$ .

We can easily verify that $f\left( x \right) = x + 1$ satisfies all given conditions.
Thus, this is the unique solution to the problem.

For the answer, $f\left( {2020} \right) = \boxed{2021}$ .