Let $f:\mathbb{R}\to \mathbb{R}$ be continuous and non affine ($f$ is not $mx+q$ for any $m,q\in\mathbb{R}$), I need to show that there exist $a,b\in \mathbb{R}$ such that $$f\bigl(\frac{a+b}{2}\bigr)\not=\frac{f(a)+f(b)}{2}$$ Could you please tell me how you would prove it? I think I have proven it but I probably made it too complicated.
Also if you are able to point it out in your proof, is it necessary for $f$ to be continuous? I have added it as an hypothesis becouse I needed it in the proof.
My attempt (briefly):
I will suppose that $$f\bigl(\frac{a+b}{2}\bigr)=\frac{f(a)+f(b)}{2}$$ holds for all $a,b$ and show that $f$ is affine ( I'll show the contrappositive).
What I want to do is prove that $f$ is affine on $[-2^k,2^k]$, $\forall k\in \mathbb{N}$; in fact, if that holds, I can say that there exists $m_k,q_k$ such that $f(x)=m_kx+q_k$ on $[-2^k,2^k]$, so that $f(0)=q_k$ $\forall k$ and $f(1)=m_k+f(0)$ $\forall k$. This means that i can define $m:=f(1)-f(0)$ and $q:=f(0)$, such that $$\lim_{k\to +\infty}(m_kx+q_k)\chi_{[-k,k]}(x)=(mx+q)\chi_{\mathbb{R}}(x)=mx+q$$ so that $\forall x\in \mathbb{R}$, $f(x)=mx+q$ is affine.
So we are left with proving that $f$ is affine on $[-2^k,2^k]$, $\forall k\in \mathbb{N}$.
First I prove that $$D:=\biggl{\{} \frac{m}{2^n} \mbox{ s.t. } -k2^n \le m \le k2^n \biggr{\}}$$ is a dense subset in $[-2^k,2^k]$. (I won't write it out here to keep it simple, but if interested a similar proof is here.)
Knowing this I want to construct this affine function $y$ and show it is $f$.
Let $$y(x):=\frac{f(2^k)-f(-2^k)}{2^{k+1}}(x+2^k)+f(-2^k)$$ I will show that $y(m/(2^n))=f(m/(2^n))$, so by continuity of $f$ and denity of these numbers we are done.
Note: starting frome $-2^k$ and $2^k$ and diving by $2$ continuously, we obtain the whole set $D$, in the sense that $D=\lim_{n\to +\infty}D_n$ where $D_n$ is defined inductively such that:
$$D_0=\{ -2^k,2^k \} \mbox{ and } D_n=D_{n-1} \cup \{ x:x=\frac{(a+b)/2}{2} \mbox{, with} a,b\in D_{n-1} \}$$ (This definition means that $D_n$ is obtained from $D_{n-1}$ by adding all of the mean points between two any points of it.)
By induction on $n$:
-if $n=1$ then we have the two elements of $D_0$ plus their middle point which is $0$: the two points of $D_0$ obviously satisfy $f(x)=y(x)$ since $y$ is constructed as the line passing trough $(-2^k,f(-2^k))$ and $(2^k,f(2^k))$, for $0$ we see that $$y(0)=\frac{f(2^k)-f(-2^k)}{2^{k+1}}(2^k)+f(-2^k)=\frac{f(2^k)-f(-2^k)}{2}+f(-2^k)=$$ which by hypothesys is $$=\frac{f(2^k)-f(-2^k)}{2}=\frac{f(2^k)+f(-2^k)}{2}=f\biggl(\frac{2^k-2^k}{2}\biggr)=f(0)$$ as we wanted.
-now, supposing (induction) that $$y\biggl( \frac{m}{2^n} \biggr)=f\biggl( \frac{m}{2^n} \biggr) \mbox{ , } \forall m \mbox{ s.t. } -k2^n\le m\le k2^n$$ we have that for $-k2^{n+1}\le \overline{m}\le k2^{n+1}$ $$y\biggl( \frac{\overline{m}}{2^{n+1}} \biggr)=y\biggl( \frac{\overline{m}/2}{2^n} \biggr)$$ where $-k2^n\le \overline{m}/2\le k2^n$, so by induction hypothesis, $$y\biggl( \frac{\overline{m}/2}{2^n} \biggr)=f\biggl( \frac{\overline{m}/2}{2^n} \biggr)=f\biggl( \frac{\overline{m}}{2^{n+1}} \biggr)$$ which is exactly what we wanted, so we are finished.
