Given $\Omega\subset\mathbb R^N$ is open, we say $\omega$: $\Omega\to [0,+\infty)$ belongs to Muckenhoupt class $A_1$ if there exists some $C>0$ such that $$ \frac{1}{|{B}|}\int_{B(x,r)} \omega(y)\,dy\leq C\omega(x) $$ for all balls $B\subset\Omega$.
For simplification we take $N=1$, the real line, and we take $\Omega=(0,1)$.
My question is, given $p(x)$ is a continuous piece-wise affine function and $p(x)\geq 1$, do we always have $p(x)\in A_1$? It looks to me the answer should be yes. Since $p(x)\geq 1$ and is piece-wise affine continuous, we should have $p(x)$ is bounded. Define $M=\max p(x)$, we have
$$ \frac{1}{|{B}|}\int_{B(x,r)} p(y)\,dy\leq \frac{1}{2r}\int_{x-r}^{x+r}Mdy=M\leq Cp(x) $$ where we take $C=M$ and bring in the fact that $p(x)\geq 1$.
It should be an OK proof, but something else went wrong with this result. So I am hoping somebody can help me to verify this result. Thank you!
Here is a counterexample on $(0, \infty)$ (it can be symmetrically extended to all of $\mathbb{R}$. Let $\omega(x)$ be a function s.t. for all integers $n \geq 0$ $$\omega(x) = 1 \,\,\,\,\,\, \forall x \in [n, n + 1/2)$$ and $$\omega(x) = 1 + 2n\left(x - (n + 1/2)\right) \,\,\,\,\,\, \forall x \in [n + 1/2, n + 1)$$
Then on each segment $[n, n+1)$, the function is a constant one on the first half of the segment, and rises up (linearly) to $n + 1$ on the second half.
Now consider the set of all balls $B(n + 1/2, 1/2)$. We have: $$\frac{1}{|{B}|}\int_{B(x,r)} \omega(y)\,dy = \frac{1}{2} + \frac{n}{4}$$ But $\omega(n + 1/2) = 1$ for all $n$. We cannot find a mutual constant s.t. $$\frac{1}{2} + \frac{n}{4} \leq C\cdot 1$$ for all $n$.