Let $f:\mathbb{R}\to \mathbb{R}$ be a measurable function such that:
1) there exists $p\in (1,\infty)$ such that $f\in L^p(I)$ for all bounded interval $I$.
2) there exists $\theta \in (0,1)$ such that $$ \left| \int_I f\; dm \right|^p \leq \theta \left(m(I) \right)^{p-1} \int_I |f|^p \; dm $$
Prove that $f=0$ almost everywhere.
Any hint?? I was trying to manipulate the intervals to show that $f$ is $0$ almost everywhere in certain covering of intervals, but that got me nowwhere.
Fix $x$ and $r>0$, put $I=B_{r}(x)$, we have \begin{align*} \left|\dfrac{1}{m(I)}\int_{I}f(y)dm(y)\right|^{p}\leq\theta\cdot\dfrac{1}{m(I)}\int_{I}|f(y)|^{p}dm(y), \end{align*} taking $r\rightarrow 0^{+}$, by Lebesgue Differentiation Theorem, we have for a.e. $x$ that \begin{align*} |f(x)|^{p}\leq\theta\cdot|f(x)|^{p}, \end{align*} if $f(x)\ne 0$, cancelling each side we have $\theta\geq 1$, this cannot happen by assumption, so $f(x)=0$, this shows that $f(x)=0$ a.e.
$\newcommand{\eps}{\varepsilon}$ Assume we even have $$ \left| \int_A f dm \right|^p \le \theta m(A)^{p-1} \int_A |f|^p $$ for any measurable set $A \subset \mathbb R$.
Choose any $\nu \gt 1$ such that $\theta \nu^p \lt 1$. (This is possible, as $\theta \in (0, 1)$.) For any $\eps \gt 0$ set $$ A_{\eps} := |f|^{-1}([\eps, \nu \eps]) $$ (As $f$ is measurable, $|f|$ is also measurable such that $A_\eps$ is measurable.)
Then $$ \eps^p m(A_\eps)^p \le \left| \int_{A_\eps} f \right|^p \le \theta m(A_\eps)^{p-1} \int_{A_\eps} |f|^p \le \theta \nu^p \eps^p m(A_\eps)^{p-1+1}, $$ which can only be true if $m(A_\eps) = 0$, since $\theta \nu^p \lt 1$.
Finally note that $$ m(\{f \neq 0\}) = m\left(\bigcup_{\eps \in \mathbb Q^+} A_\eps\right) \le \sum_{\eps \in \mathbb Q^+} 0 = 0. $$
