How prove this integral $\int_{0}^{\infty}f^{\alpha}(x)dx,\alpha>1$ is convergent

Question

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

Answer 1

Answer Changed

It is assumed that:

• $f$ is differentiable on $ \mathbb{R}^+$
• $f'\le M$ where $ M $ is a constant.
• $f$ is integrable on $\mathbb{R}^+$

We assume that $f$ does not converge to $0$ at infinity and lead to a contradiction.

Let $\epsilon>0$,since $f$ is integrable on $\mathbb{R}^+$, there exist $z_1\geq 0$ such that $f(z_1)\leq \epsilon /2$.

If $f$ does not converge to $0$ at infinity. Then there exist $x_1\geq z_1$ such that $|f(x_1)|\geq \epsilon$.

1. $f(x_1)\geq \epsilon$

Denote $y_1=sup\{ t\in [z_1,x_1]\quad /\quad f(t)\leq \epsilon/2\}$

The IVT implies that $y_1$ is defined, the continuity of $f$ implies that $f(y_1)=\epsilon/2$. The IVT tell us $f(t)\geq \epsilon/2$ for $t\in [y_1,x_1]$.

Once again by the IVT, there exist $t_1\in [y_1,x_1]$ such that $f(t_1)=\epsilon$.

Then,

$$\int_{y_1}^{x_1}f(u)du\geq \int_{y_1}^{t_1}f(u)du \geq \frac{\epsilon}{2}(t_1-y_1)\geq \frac{\epsilon}{4M}$$

As $\frac{\epsilon}{2} = f(t_1)-f(y_1) \geq M (t_1-x_1)$

And repeat the operation by taking:

$z_2\geq x_1$ such that $f(z_2)\leq \epsilon /2$ and $x_2\geq z_2$ such that $|f(x_2)|\geq \epsilon$.

1. $f(x_1)\leq -\epsilon$ The reasoning is broadly the same as in 1.

Let $y_1=inf\{ t\in [x_1,z_1]/ f(t)\geq -\epsilon/2\}$.

We find then :

$$\int_{x_1}^{y_1}f(u)du\leq-\frac{\epsilon}{4M}$$

Therefore, $f$ is not integrable, contradiction/

Then,$\lim_{x\rightarrow \infty}f(x)=0$,

$$ \exists A / x>A \implies 0\le f(x) \le 1 $$ Thus, $$ \forall x>A 0\le f^\alpha(x) \le f(x) $$ for $p \ge 2$, for $p=1$ there is nothing to prove.QED

Answer 2

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 $\int_0^\infty f(x)\ dx<\infty$.