Find $ f ( 2021 ) $ if the function $ f : \mathbb R \to \mathbb R $ satisfies the following Conditions:
- $ f ( 1 ) = 1 $;
- $ f ( x + y ) = f ( x ) + f ( y ) $, $ \forall x , y \in \mathbb R $;
- $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $, $ \forall x \in \mathbb R \setminus \{ 0 \} $.
I tried it to solve it using the first two conditions:
$ f(1)=1 $ is given. $ \quad $ (using the first condition)
So, if we take $ x = y = 1 $, we will have $ f ( 2 ) = f ( 1 + 1 ) = f ( 1 ) + f ( 1 ) = 1 + 1 = 2 $. $ \quad $ (using the second condition)
Now, if $ x = 2 $ and $ y = 1 $, we have $ f ( 3 ) = f ( 2 + 1 ) = f ( 2 ) + f ( 1 ) = 2 + 1 = 3 $. $ \quad $ (using the second condition and $ f ( 2 ) = 1 $ from above)
The same pattern will give us $ f ( 4 ) = 4 $ and if we apply it generally it will give us $ f ( x ) = x $ for $ x \in \mathbb Z _ + $.
As $ 2021 \in \mathbb Z _ + $, we get $ f ( 2021 ) = 2021 $.
The question seems to be interesting. But I found the above reasoning to be easy, and I'm not sure my solution is correct or not.
Your argument is correct, as is already noted in several comments above. But I want to add some remarks that I think would be helpful to you. First of all, the functional equation $$ f ( x + y ) = f ( x ) + f ( y ) $$ is a well-known one. It's named Cauchy's functional equation after the French mathematician Augustin-Louis Cauchy, and functions that satisfy it are called additive functions. You can take a look at the post with the title "Overview of basic facts about Cauchy functional equation" to find out about the properties of the functions that satisfy it (I won't go into some details here, as you can find them there). The most relevant one for us here is that for an additive function $ f : \mathbb R \to \mathbb R $, a rational number $ q $ and a real number $ x $, we have $$ f ( q x ) = q f ( x ) \text . $$ What you've proven is part of this fact. So, if the question had asked for the value of $ f \left( \frac { 2021 } 2 \right) $ or $ f ( - 2021 ) $, we could find them using the above fact, and we would have $ f \left( \frac { 2021 } 2 \right) = \frac { 2021 } 2 f ( 1 ) $ and $ f ( - 2021 ) = - 2021 f ( 1 ) $.
But the important thing to know is that there are additive functions that have very wild behaviors, and we can't always get $ f ( x ) = x f ( 1 ) $ for arbitrary (not necessarily rational) $ x $. For example, if the question had asked for $ f \left( \sqrt { 2021 } \right) $, you couldn't find it only having $ f ( 1 ) $ and the fact that $ f $ is additive. That's where the third condition comes into play. For an additive function $ f $ that also satisfies $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $, we always have $ f ( x ) = x f ( 1 ) $, whatever $ x $ is. To see why, you can for example take a look at "Cauchy equation $ f ( x + y ) = f ( x ) + f ( y ) $ with an additional condition $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $" or "Confusing functional equation problem (application of Cauchy equation when $ f \left( \frac 1 x \right) = \frac { f ( x ) } { x ^ 2 } $)".