Reason for the integer case and the rational case to be solved differently

Question

Assume $f$ is continuous,$f(0)=1$ , and $f(m+n+1)=f(m)+f(n)$ for all real $m, n$. Show that $f(x) = 1 + x$ for all real numbers $x$.

This is referenced from Terence Tao’s solving mathematical problems and in the exercise he provided a hint;

first prove this for integer $x$, then for rational $x$, then finally for real $x.$

The questions are as follows: Why would there be a separate case to be considered for this question? Wouldn’t a direct method of solving suffice? Is there another way of approaching the question?

Any help would be much appreciated.

Answer 1

Before I proceed with my solution let me tell you an interesting Theorem which is necessary to understand the solution I have given.

Continous Additive functions are linear

Coming back to your Orignal Problem Consider the function $g(x)=f(x)-1$. Note that $g(x)$ is continuous function. Keep it aside for a while.

By the functional equation given $f(y)=f({\color{red}{0}}+{\color{blue}{y-1}}+1)=f({\color{red}{0}})+f({\color{blue}{y-1}})$

Hence $f(y)=1+f(y-1)$

Further consider $f(x+y)$

$$f({\color{red}{x}}+{\color{blue}{y}})=f({\color{red}{x}}+{\color{blue}{y-1}}+1)$$ $$f({\color{red}{x}}+{\color{blue}{y}})=f({\color{red}{x}})+f({\color{blue}{y-1}})$$

Hence we have $f(x+y)=f(x)+f(y)-1$ This is equivalent to $f(x+y)-1=f(x)-1+f(y)-1$

Hence $g(x+y)=g(x)+g(y)$

Hence $g(x)=g(1)x$

Which implies $f(x)-1=g(1)x$

Hence $f(x)=1+(f(x)-1)x$

Because $f(0+0+1)=f(0)+f(0)=2f(0)=2 \times 1=2$

We conclude $f(x)=1+x$

You will understand the hint if you prove why Continous Additive functions are linear. I had added a link above

Answer 2

You need the integer case to prove the rational case. How else do you get a handle on $f(\frac 12)?$ If we take $n=0$ we are told $f(m+1)=f(m)+1$ from which we can derive that $f(x)=x+1$ for the integers.

Once we have that, we can substitute in $m=n=\frac 12$ and get $f(2)=2f(\frac 12)$. Since we know $f(2)=3$ because we solved the integer case already, we get $f(\frac 12)=\frac 32$. We can continue this approach to get the function on all the dyadic rationals. Of course this does not prove that there is not a direct approach

Since the dyadics are dense, we can then extend to all the reals. This is where continuity comes in. Clearly the integers do not suffice for this because they are not dense. As Conrad comments, if you remove the continuity restriction there are other solutions. You need a dense set for continuity to work on. That shows you need to compute it on a dense set before you cover all the reals.