# Thread: Prove cts function with midpoint property is linear

1. ## Prove cts function with midpoint property is linear

Hi,
I would like to show that a continuous map $fa,b)\to \mathbb{R}" alt="fa,b)\to \mathbb{R}" /> which has the property that $f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$ for $x,y\in (a,b)$ satisfies
$f(x) = Ax + B$ for some constants $A, B$.

I can prove it on the closed interval [0,1]: in this case it suffices to show $f(x)=Ax$ with $f(0)=0$. I prove it holds on the dyadic rationals of the form $\frac{m}{2^k}$ by an induction argument, and extending to the closed interval by continuity. BUT my argument needs the endpoints of the interval: I am progressively subdividing; $f(\frac{1}{2}) = \frac{f(0)+f(1)}{2} = \frac{f(1)}{2}$, etc, and I cannot see how to prove it on an open interval.

Many thanks

2. Originally Posted by jacobi
Hi,
I would like to show that a continuous map $fa,b)\to \mathbb{R}" alt="fa,b)\to \mathbb{R}" /> which has the property that $f\left(\frac{x+y}{2}\right) = \frac{f(x)+f(y)}{2}$ for $x,y\in (a,b)$ satisfies
$f(x) = Ax + B$ for some constants $A, B$.

I can prove it on the closed interval [0,1]: in this case it suffices to show $f(x)=Ax$ with $f(0)=0$. I prove it holds on the dyadic rationals of the form $\frac{m}{2^k}$ by an induction argument, and extending to the closed interval by continuity. BUT my argument needs the endpoints of the interval: I am progressively subdividing; $f(\frac{1}{2}) = \frac{f(0)+f(1)}{2} = \frac{f(1)}{2}$, etc, and I cannot see how to prove it on an open interval.

Many thanks

So, in fact, you can prove it in any interval of the form $[\alpha,\,\beta]\subset (0,1)$ , right? Well, you can choose say $\alpha=\frac{1}{n}\,,\,\beta=1-\frac{1}{n}$ and let $n\to\infty$ ...

Tonio

Pd. Where does continuity come into play in the above?

3. Oh, of course, I see now: given x, y in (0,1) they'll lie in some closed interval. Duh.

Thanks, Tonio!

4. Oh, of course, I see now: given x, y in (0,1) they'll lie in the closed subinterval [x,y]. Duh.

Thanks, Tonio!

Pd. Where does continuity come into play in the above?
If you're referring to my work, I used continuity to extend the dyadic rationals to the interval. If you're referring to your post, you're using that $a_n\to a$ implies $f(a_n)\to f(a)$.

5. Actually I am mistaken. For each $\alpha_n = \frac{1}{n}$, $\beta_n = 1-\frac{1}{n}$ we get different constants $A_n$, $B_n$. How can we guarantee that these converge?

6. Originally Posted by jacobi
Actually I am mistaken. For each $\alpha_n = \frac{1}{n}$, $\beta_n = 1-\frac{1}{n}$ we get different constants $A_n$, $B_n$. How can we guarantee that these converge?
hehe...it took you a while . Never mind, we all passed through these messes. So you have different constants...or do you?

Take n, n+1 and their respective constants...and NOW use continuity!

Tonio