Continuous $f : \mathbb R \to \mathbb R$ satisfying $f\left(\sqrt{\frac{x^2 + y^2}{2}}\right) = \frac{f(x) + f(y)}{2}$

Solution 1:

Clearly $f(x) = Ax^2+B$ are solutions. Let us show that those are the only solutions.

Observe for any $\lambda>0$, we see that \begin{align} f\left(\lambda\sqrt{\frac{x^2+y^2}{2}}\right)=f\left(\sqrt{\frac{\lambda^2x^2+\lambda^2y^2}{2}} \right) = \frac{f(\lambda x) + f(\lambda y)}{2}. \end{align}

Set $y=0$ and $\lambda = \sqrt{2}$, then we see that \begin{align} f(|x|) = \frac{f(\sqrt{2} x) + f(0)}{2} = \frac{f(-\sqrt{2}x)+f(0)}{2} \end{align} which means $f(x) = f(-x)$ for all $x$.

Consider $g(x) = f(x)-f(0)$. Clearly, we have that $g(0) = 0$. Moreover, we see that \begin{align} c = f\left(\sqrt{\frac{x^2+y^2}{2}}\right)- f\left(0\right) = \frac{f(x)-f(0)+f(y)-f(0)}{2}= \frac{g(x)+g(y)}{2}. \end{align} At this point, $g(x)$ should be $Ax^2$. Note that we have \begin{align} g\left(\frac{\lambda |x|}{\sqrt{2}} \right) = \frac{1}{2}g(\lambda x) = \frac{1}{2} g(-\lambda x). \end{align}

Let us assume $g(1) \neq 0$. Define $h(x) = g(x)/g(1)$, then $h(1) = 1$ and \begin{align} h\left(\sqrt{\frac{x^2+y^2}{2}}\right) = \frac{1}{g(1)}g\left(\sqrt{\frac{x^2+y^2}{2}}\right) = \frac{1}{g(1)}\frac{g(x)+g(y)}{2}= \frac{h(x)+h(y)}{2}. \end{align}

As of now, $h(x)$ should be $x^2$. Note that $h(x) = 2h(|x|/\sqrt{2})$

Consider \begin{align} F(x)=&\ \frac{1}{2}(h(x+1/2)-h(x-1/2)) = h\left( \frac{|x+1/2|}{\sqrt{2}}\right)-h\left(\frac{|x-1/2|}{\sqrt{2}}\right)\\ =&\ h\left( \sqrt{\frac{|x-1/2|^2+\sqrt{2|x|}^2}{2}}\right)-h\left(\frac{|x-1/2|}{\sqrt{2}}\right)=\frac{1}{2}h(\sqrt{2 |x|}) = h(\sqrt{|x|}). \end{align} It's clear that $F(0) = 0$. Finally, note that \begin{align} F(x+y)-F(x)-F(y) =&\ h(\sqrt{|x+y|})-h(\sqrt{|x|})-h(\sqrt{|y|)}\\ =&\ h(\sqrt{|x+y|}) -\frac{1}{2}[h(\sqrt{2|x|})+h(\sqrt{2|y|)}]\\ =&\ h(\sqrt{|x+y|})-h(\sqrt{\frac{2|x+y|}{2}} ) =0. \end{align}

Note, for the last equality, I used the fact that $h(\sqrt{2}\sqrt{|x|})= h(\operatorname{sgn}(x)\sqrt{2|x|})$.

I will leave the case $g(1) = 0$ to the reader.

Remark: There might be a slicker way.