Let a bounded open set $\Omega\subset \mathbb R^N$ be given. Let $f$: $\Omega\to (0,+\infty]$ be given and we assume $f$ is locally integrable and there exists a constant $C>0$ so that $$ M(f)(x)\leq Cf(x)\tag 1 $$ for all $x\in\Omega$, where $M(f)$ denotes the maximal function of $f$ defined by $$ M(f)(x):=\sup_{x\in B}\frac{1}{|B|}\int_Bf(y)dy. $$
My question: can we prove that $f$ is bounded below by a positive constant? i.e., $f\geq c$ for some $c>0$.
Update: Please see below for a nice counter example by @zhw. Also, I think if instead of $f$: $\Omega\to (0,+\infty]$, I define $f$: $\mathbb R\to(0,+\infty]$, then the function $f$ should be bounded below by a positive constant in $\Omega$.
Here is my prove: since $\Omega$ is bounded, so $\bar \Omega$ is closed an bounded and hence compact. Thus we could cover $\bar\Omega$ by finitely many balls $(B_i)_{0\leq i\leq M}$. $(B_i)$ also covers $\Omega$. Thus, for any $x\in\Omega$, we have there exists $0\leq i\leq M$ so that $x\in B_i$ and by $(1)$ we have $$ f(x)\geq CM(f)(x)\geq C\frac{1}{|B_i|}\int_{B_i}f(y)dy\geq C\min_{0\leq j\leq M}\frac{1}{|B_j|}\int_{B_j}f(y)dy>0 $$ since $f$ is positive over $\mathbb R$.
Counterexample when $N=2:$ Let $\Omega$ be the open triangle with vertices $(0,0), (1,0),(1,1).$ Define $f(x,y)=x.$ Suppose $(x,y)\in \Omega.$ Consider a ball $B=B((a,b),r) \subset \Omega,$ with $(x,y) \in B$ and $a\ge x.$ Then a little geometry shows $r\le a/2.$ Thus $a-x \le a/2 \implies a/2 \le x.$ It follows that the largest value $f$ can take in $B$ is $\le a+r \le 3a/2 \le 3x/2.$ Therefore the average of $f$ over this ball is $\le 3x/2.$ If we took $B$ to have its center to the left of $x,$ then it's obvious the average of $f$ over $B$ is $\le x.$ It follows that $Mf(x,y)\le 3x/2 = (3/2)f(x,y).$ But clearly $\inf_{\Omega} f = 0,$ so we have a counterexample.