non-decreasing by young's inequality

Question

Suppose that $\int_0^{\infty}f(x)dx=1 $ and $\int_0^{\infty}e^{kx}f(x)dx<\infty$ for some $k>0 $. Prove $(\int_0^{\infty}x^tf(x)dx)^{\frac{1}{t}}$ is non-decreasing in t (Hint: Use young's inequality).I want to follow the hint,but i don't know how to apply young's inequality.Can someone give more hint or a complete proof?

Answer 1

Young's inequality says $ab \leq \frac {a^{p}} p + \frac {b^{p}} q$ if $a,b \geq 0$,$1<p<\infty $ and $\frac 1 p + \frac 1 q =1$. Let $0<t<s$ and denote by $c$ the number $(\int_0^{\infty } x^{s}f(x)\, dx)^{t/s}$. Let $p=\frac s t$, $q=\frac s {s-t}$. Apply Young's inequality with $a=\frac {x^{t}} c$ and $b=1$. You will get $\frac {x^{t}} c \leq \frac {x^{s}} {pc^{p}}+\frac 1 q$. Mulitiply by $f(x)$ and integrate to get $ \frac {\int_0^{\infty}x^{t}f(x)\, dx } c \leq \frac 1 p +\frac 1 q =1$ (where we used the fact that $c^{p}$ in the denominator cancels with $\int_0^{\infty } x^{s} f(x)\, dx$). Hence $ {\int_0^{\infty}x^{t}f(x)\, dx } \leq c=(\int_0^{\infty } x^{s}f(x)\, dx)^{t/s}$. This gives $ ({\int_0^{\infty}x^{t}f(x)\, dx })^{1/t} \leq c=(\int_0^{\infty } x^{s}f(x)\, dx)^{1/s}$ which is what we want to prove.