An amazing inequality of the integration of two functions.

Question

Let $f:[0,1]\longrightarrow\mathbb{R}$ be measurable and $g\in L^1[0,1]$ such that for all $t>0$, $$ \int_{|f(x)|>t}|g(x)|~\mathrm{d}x\leq \frac{3}{t^2}. $$ Prove that for $1<p<2$, $$ \int_0^1 |f(x)|^p|g(x)|~\mathrm{d}x<\infty. $$

How to prove?

Answer 1

Rewrite the integral as follows $$\int_0^1 |f(x)|^p|g(x)|dx = \int_0^1 \int_0^{|f(x)|} pt^{p-1}|g(x)| dt dx$$ Switching the order of integration we obtain $$\int_0^\infty \int_{|f(x)|>t} pt^{p-1}|g(x)| dx dt = \int_0^\infty pt^{p-1}\int_{|f(x)|>t} |g(x)| dx dt$$ Now note that $\int_{|f(x)|>t} |g(x)| dx \leq \min(\frac{3}{t^2},|g|_{L^1})$ Hence the above integral is bounded by $$\int_0^\infty pt^{p-1}\min\left(\frac{3}{t^2},|g|_{L^1}\right) dt = p\int_0^\infty \min\left(\frac{3}{t^{2-p+1}},t^{p-1}|g|_{L^1}\right) dt$$ Now since $t^{p-1}|g|_{L^1}$ is bounded near $0$ and $2-p+1 > 1$ we have that the above integral converges which implies the desired result.