Suppose an extented real valued function $f$ defined on $\mathbb{R}^n$ satisfies the following two properties:
a) There is a $p$, $1\leq p < \infty$ such that $f\in L^p(I)$, for every bounded interval $I\in \mathbb{R}^n$ and
b) $|\int_I f|^p\leq c|I|^{p-1} \int_I |f|^p$ for every bounded interval $I\in \mathbb{R}^n$ and a constant $0<c<1$ independent of $I$.
Show that $f=0$ a.e.
My Work:
We must prove that $\int_{\mathbb{R}^n} |f|=0$. By Holder's inequality, $|\int_I f|\leq \int_I |f|=\int_I |f\chi_I|\leq (\int_I |f|^p)^\frac{1}{p}(\int_I |\chi_I|^\frac{p}{p-1})^\frac{p-1}{p}$. Hence $|\int_I f|^p\leq |I|^{p-1} \int_I |f|^p$. Since then I was stuck. What else I can do here to proceed? Can anybody please help me?
Suppose (a) holds and that it is not the case that $f$ is $0$ a.e.. Let $x$ be a Lebesgue point of $f$ and of $|f|^p$ such that $f(x)\neq 0$. Then
$$|f(x)|^p=\left|\lim\limits_{I\to x}\frac{1}{|I|}\int_I f\right|^p$$
and
$$|f(x)|^p=\lim\limits_{I\to x}\frac{1}{|I|}\int_I |f|^p,$$
so
$$\lim\limits_{I\to x}\left|\frac{1}{|I|}\int_I f\right|^p=\lim\limits_{I\to x}\frac{1}{|I|}\int_I |f|^p,$$
and recall these are nonzero limits. Thus given $c$ with $0<c<1$, there exists $I$ sufficiently small centered at $x$ such that $\left|\frac{1}{|I|}\int_I f\right|^p>c\frac{1}{|I|}\int_I |f|^p$. Therefore (b) does not hold.