Looking for a function that satisfies some kind of mean value property

Question

Given $a<b\in (0,1)$ and $\delta<1/2$, I need to find an integrable function $\gamma :(a-\delta,b+\delta)\to [0,1]$ such that $$\frac{1}{2\delta}\int_{x-\delta}^{x+\delta}\gamma(y)\; dy=\frac{1}{2}\left(\gamma(x+\delta)+\gamma(x-\delta)\right)\quad \forall x\in (a,b)$$ I know that linear polinomials $(\gamma(y)=a_1y+a_0)$ satisfy this equation. However, I need to know if there is any other function (possibly discontinuous) satisfying the equation.

Answer 1

Let $$\gamma(y)=a_2y^2+a_1y+a_0$$ then $$\frac{1}{2\delta}(\frac{a_2}{3}(x^3+3\delta x^2+3\delta^2 x +\delta^3-x^3+3\delta x^2-3\delta^2x-\delta^3)+\frac{a_1}{2}(x^2+2\delta x+\delta^2-x^2+2\delta x-\delta^2)+a_0(x+\delta-x+\delta)\\=1/2(a_2(2x^2+2\delta^2)+a_1(2x)+2a_0)$$

Simplifying :

$$a_2x^2+a_1x+a_0\delta=a_2x^2+a_1x+a_2\delta^2+a_0$$

Hence it is possible with degree 2. For the general case :

Attempt of idea, but impractical : let $\gamma(y)=\sum a_ny^n$.

The equation becomes : $$ \sum a_n(\frac{(x+\delta)^{n+1}-(x-\delta)^{n+1}}{\delta(n+1)}-(\delta+x)^n-(x-\delta)^n)=0$$

The derivative wrt to x gives another series that shall vanish. The latter gives x via Gram-Schmidt orthogonalization, starting with any summable series, seeing the sum as a scalar product. Knowing x and substituting in the original equation, then again orthogonalization and the a's are found.