Bounding integral for a function in $L_1$

Question

Consider $(\mathbb{R}, \mathbb{B}, \lambda) $If $f$ is non-negative and in $L_1(\mathbb{R}, \mathbb{B}, \lambda)$ is it true that for some $\alpha < -1$ $$ \int_{[1,\infty)} f d\lambda \leq \int_{[1,\infty)} x^\alpha $$
I was thinking about this but was not sure on a rigorous proof, if it is even true that is.

Answer 1

For the sake of contradiction, assume there exists a function $f^*\in L^1([1,\infty))$ such that $$\int_{[1,\infty)} fd\lambda\le \int_{[1,\infty)} f^*d\lambda \tag1 $$ for all $f\in L^1([1,\infty))$. Note that since (e.g.) $x\mapsto x^{-2}\in L^1([1,\infty))$, this implies $$ \int_{[1,\infty)} f^*d\lambda \ge \int_{[1,\infty)} x^{-2}\ d\lambda(x)>0 \tag 2 $$

But now consider $f\equiv 2f^*$ : clearly $f\in L^1([1,\infty))$ as multiplying by a scalar doesn't affect integrability. Furthemore, because we assumed that $(1)$ is true we get $$ \int_{[1,\infty)} fd\lambda:= \int_{[1,\infty)} 2f^*d\lambda = 2\int_{[1,\infty)} f^*d\lambda\le \int_{[1,\infty)} f^*d\lambda\tag 3 $$ Substracting $\int_{[1,\infty)} f^*d\lambda $ from both sides of the inequality $(3)$ we find $$\int_{[1,\infty)} f^*d\lambda \le 0 \tag 4 $$ This is a contradiction with $(2)$. Hence such a function $f^*$ can't exist.