Suppose that $0<p<1$. Let $h:(0,1]\to\mathbb C$ be a Lebesgue-measurable function such that $$\int_{x}^1|h(t)|\,\mathrm dt<\infty\quad\forall x\in(0,1].$$ Suppose also that $$\int_0^1t\cdot|h(t)|^p\,\mathrm dt<\infty.$$
Claim: $$\lim_{x\downarrow0}\left\{x^{2/p-1}\cdot\int_x^1|h(t)|\,\mathrm dt\right\}=0.$$
I have a proof for the case in which $p\in[1,2)$, but the trick here is that $L^p$ is not a normed space for $p\in(0,1)$, so the usual tricks (e.g., Hölder's inequality) fail to work here. Somewhat related to my earlier question.
I tried to come up with some heuristic counterexamples, to no avail. My conjecture is that the claim is true but the proof has eluded me so far. Thanks in advance for any hints.
This is not true. To clarify the issue, it helps to decompose $(0,1]$ into dyadic intervals $I_k = (2^{-k},2^{1-k}]$. Then your assumption is $$\sum_k 2^{-k} \int_{I_k}|h|^p<\infty \tag1$$ while the desired conclusion is $$\sum_{k=1}^m \int_{I_k}|h| = o \left(2^{m(2/p-1)}\right) \tag2$$ Now if we recall that $\int_{I_k}|h|^p$ provides no upper bound on $\int_{I_k}|h|$ whatsoever (since $p<1$), a counterexample appears: define $h$ on each $I_k$ so that $\int_{I_k}|h|^p=1$ and $\int_{I_k}|h| = 2^{2^{2^{2^k}}}$ or whatever you want.
E.g., $h(x) = (x-2^{-k})^{-1}$ on $I_k$, slightly truncated and normalized.