Prove that the improper integral is convergent

Question

Let $f\colon\mathbb R\to \mathbb R$ be a non-negative function. Assume that

$\int_{-\infty}^{+\infty}x^kf(x)=u,\int_{-\infty}^{+\infty}f(x)=1$, please prove that $\int_{-\infty}^{+\infty}x^{k-1}f(x)(k\gt1)$ is convergent.

This is a little hard for me, could anyone help me? Thanks in advance!

Answer 1

If $|x|\lt 1$, then $x^{k-1}f(x)\leqslant f(x)$ and if $|x|\geqslant 1$, then $x^{k-1}f(x)\leqslant x^kf(x)$, hence $$\forall x\in\mathbb R,\quad 0\leqslant x^{k-1}f(x)\leqslant f(x)+x^kf(x).$$

Answer 2

In particular, using the inequality $$ \frac{a^p}{p}+\frac{b^q}{q}\ge ab, $$ whenever $a,b\ge 0$ and $\frac{1}{p}+\frac{1}{q}=1$ ($p,q>1$), we obtain that ($p=k,q=\frac{k}{k-1}$): $$ \frac{1}{k}\left(1+(k-1)x^k\right) \ge x^{k-1}, $$ and hence $$ \int_{-\infty}^\infty x^{k-1}f(x)\,dx\le \frac{1}{k}+\frac{(k-1)u}{k}. $$