Are there non-linear functions that satisfy $f(1)=3$ and $f(1+x)=f(1)+f(x)$?

Question

I solved this question: $$f(1)=3\\f(1+x)=f(1)+f(x)\\f(50)=?$$ my try is in below $$f(2)=f(1)+f(1)=3+3=2\times 3\\f(3)=f(1)+f(2)=3+6=3\times 3\\ \vdots\\f(50)=3\times50$$ and $f(x)=3x$ work in here. Actually, this was a multiple choice and it passed. But my question is about the different types (non-linear functions) that fit in this relation. Can anyone help me?

To say as clear as possible:

Is there another type of function that covers those conditions?

There are non-linear functions which satisfy $f(x+y)=f(x)+f(y)$ for all $x,y\in\Bbb R$. See this post. — nejimban, Feb 17 '23 at 10:30
Your method clearly would work for all positive integers and could be tuned into a proof by induction which is easily extended to $0$ and negative integers. Non-linear solutions would only give different values for non-integers. — Henry, Feb 17 '23 at 10:39
f(x) = 3x is the only possible solution, you can prove this by induction as well — Aadi, Feb 17 '23 at 10:41

Umesh Shankar · Accepted Answer · 2023-02-17T10:50:29.720

1

With the condition $f(1+x)=f(1)+f(x)$, specifying the value of $f(1)$ will specify the function value of every integer.

The best we can do is if we specify $f(x)$ for each $x\in (0,1)$, we will be able to retrieve the value of $f(x)$ for $x\in \mathbb R$.

A function that satisfies the given is $f(x)= 3\lfloor x \rfloor \text { if } x \in \mathbb R $ but it is not exactly continuous or linear.

edited Feb 17 '23 at 10:50

answered Feb 17 '23 at 10:37

Umesh Shankar

2,787

Do You mean something like floor function ? – Khosrotash Feb 17 '23 at 10:48
Yes, I shall edit it to reflect that – Umesh Shankar Feb 17 '23 at 10:49

Are there non-linear functions that satisfy $f(1)=3$ and $f(1+x)=f(1)+f(x)$?

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