TITLE: How prove this integral $\int_{0}^{\infty}f^{\alpha}(x)dx,\alpha>1$ is convergent
QUESTION [8 upvotes]: Question:

let the function $f(x)\ge 0$,and such $$f'(x)\le\dfrac{1}{2},\forall x\ge 0$$
  and this integral
  $\displaystyle\int_{0}^{\infty}f(x)dx$ is convergent.

show that:

$$\int_{0}^{\infty}f^{\alpha}(x)dx,\alpha>1$$ is convergent. And
  $$\int_{0}^{\infty}f^{\alpha}(x)dx\le\left(\int_{0}^{\infty}f(x)dx\right)^{\frac{\alpha-1}{2}}.$$

Maybe this problem use Cauchy-Schwarz inequality to solve it
$$\int_{0}^{\infty}f^2(x)dx\int_{0}^{\infty}g^2(x)dx\ge\left(\int_{0}^{\infty}f(x)g(x)dx\right)^2$$
maybe Holder inequality also is useful for this problem.
But I can't solve this problem
Thank you for you help

REPLY [3 votes]: The first part can be proven as follows:
Claim: There is an $M$ with $0\leq f(x)\leq 1$ for all $x\geq M$. Using this claim one has
$$\int_0^\infty f^\alpha(x)\ dx\leq\int_0^M f^\alpha(x)\ dx+\int_M^\infty f(x)\ dx<\infty$$
for all $\alpha>1$.
Proof of the Claim: Assume there is an increasing sequence $x_n\to\infty$ with $f(x_n)>1$ for all $n$. Put $n_0:=1$ and for each $k\geq 1$ choose $n_k>n_{k-1}$ such that $x_{n_k}>x_{n_{k-1}}+1$. It follows that
$$f(x)\geq{1\over2}\qquad (x_{n_k}-1\leq x\leq x_{n_k})$$
for all $k\geq1$, which is incompatible with $\int_0^\infty f(x)\ dx<\infty$.