Finding $f(x)$ with $\frac{1}{2a}\int_{x-a}^{x+a}f(t)dt=f(x)$

Problem

Let $f(x)$ be a differentiable function with $f(0)=1$ and $f(1)=2$. For any $a,x$ where $a \neq 0$, it holds $$\frac{1}{2a}\int_{x-a}^{x+a}f(t)dt=f(x).$$ Find $f(x)$.

Attempt

Since $$f(x)=\frac{\displaystyle\int_{x-a}^{x+a}f(t)dt}{2a}$$ holds for all $x \in \mathbb{R}$ and $a \neq 0$. One can fix $x$ and take the limit as $a \to 0$. Thus $$\lim_{a \to 0}f(x)=\lim_{a \to 0}\frac{\displaystyle\int_{x-a}^{x+a}f(t)dt}{2a}=\frac{f(x+a)+f(x-a)}{2},$$ where we applied L'Hopital's rule. Thus $$f(x)=\frac{f(x+a)+f(x-a)}{2}.$$ How to go on from here?

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