0
$\begingroup$

Let $f: \mathbb R_{\geq 0} \to \mathbb R$ measurable such that there is $\alpha \in (0,1)$ with $|f(t)| \leq \dfrac{t^{\alpha}}{1+t}$ for all $t \geq 0$. Let $G: \mathbb R_{\geq 0} \times \mathbb R_{\geq 0} \to \mathbb R$ be defined as $$G(x,t)=e^{-xt}f(t)$$ Prove that $G$ is integrable.

It is sufficient to show that $|G|$ is integrable. By Fubini-Tonelli's theorem, we have $$\int_{(\mathbb R_{\geq 0})^2}|G(x,t)|=\int_{ \mathbb R_{\geq 0}}(\int_{ \mathbb R_{\geq 0}}|f(t)|e^{-xt}dx)dt$$ $$=\int_{ \mathbb R_{\geq 0}}|f(t)|(\int_{ \mathbb R_{\geq 0}} e^{-xt}dx)dt \space(*)$$

If for each $t>0$ I define $f_{n,t}(x)=e^{-xt}\chi _{[0,n]}$, it is clear that $\lim_n f_{n,t}(x)=f(x)$ with $(f_{n,t})_{n}$ an increasing sequence.

The Riemann-Integral $$\int_0^n f_{n,t}(x)dx=\dfrac{1}{t}-\dfrac{e^{-nt}}{t},$$ which coincides with the Lebesgue Integral. So by this and by the monotone convergence theorem, the last expression becomes $$(*)=\int_{\mathbb R_{\geq 0}}|f(t)|\dfrac{1}{t}dt\leq \int_{\mathbb R_{\geq 0}}\dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt$$

I would like to show that this last integral is finite, but I don't know how to prove this. I would appreciate if someone could help me to complete the answer and to correct me if necessary.

$\endgroup$
1
$\begingroup$

So consider the integrals $\int_0^1 t^{- \alpha} \textrm{d}t $ if $\alpha \in (0,1)$ and $\int_1^\infty t^{-\beta} \textrm{d}t $ with $ \beta \in (1 ,2)$.
Then we must bound $\frac{1}{1+t}$ by $1$ on $t \in [0,1]$ and by $\frac{1}{t}$ for $t \in [1, \infty)$.

$\endgroup$
  • $\begingroup$ Since $1+t>t$, then $\dfrac{1}{1+t} <\dfrac{1}{t}$, so $(*) \leq \int_0^{\infty} t^{\alpha-2}dt$. Since $0<\alpha<1$, then $-2<\alpha-2<-1$, so $2>2-\alpha>1$, this last inequality shows that the improper integral is finite. From here it follows that the original Lebesgue integral is finite. Is this correct? $\endgroup$ – user156441 Oct 2 '15 at 8:59
  • 1
    $\begingroup$ No, since $t^{-\frac{3}{2}}$ is not integrable as is goes to fast to infinity at $0$. And $t$ is not a constant, so perhaps try $1+t \ge 1$. $\endgroup$ – Hetebrij Oct 2 '15 at 12:16
  • 1
    $\begingroup$ You're right, so maybe I could break up the integral into $\int_0^1 \dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt+\int_1^{\infty} \dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt \leq \int_0^1 t^{\alpha}\dfrac{1}{t}dt+\int_1^{\infty} \dfrac{t^{\alpha}}{t}\dfrac{1}{t}dt$. The last expression equals to $ \int_0^1 \dfrac{1}{t^{1-\alpha}}dt+\int_1^{\infty} \dfrac{1}{t^{2-\alpha}}dt$, we have $0<1-\alpha<1, 1<2-\alpha<2$, which means these two integrals are convergent. $\endgroup$ – user156441 Oct 2 '15 at 16:19
  • 1
    $\begingroup$ You're also right, as I was not considering the limit as $t \to \infty$. I will update my answer. $\endgroup$ – Hetebrij Oct 2 '15 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.