Prove that there is a continuous function $f:(0,a)\rightarrow\mathbb{R}$ ($a>0$) such that $f\notin\mathcal{L}_q(\lambda)$ for all $q\in (0,+\infty)$.

Question

Prove that there is a continuous function $f:(0,a)\rightarrow\mathbb{R}$ ($a>0$) such that $f\notin\mathcal{L}_q(\lambda)$ for all $q\in (0,+\infty)$.

I think $f(x)=e^{1/x}$ works but I don't know how to prove it. I have to demonstrate that $\int |f|^q\,d\lambda=+\infty$ but I don't know how to do it. My only idea is to define the sequence of functions $(f_n)$:

$$ f_n(x)= \left\{ \begin{array}{ll} e^{n/a}&\mbox{if }x<a/n\\ e^{1/x}&\mbox{if }x\geq a/n \end{array} \right. $$

So we have that $\lim_{n\uparrow+\infty}f_n=f=|f|$ and that $\lim_{n\uparrow+\infty}f_n^q=|f|^q$.

Thus, $$\int |f|^q\,d\lambda\geq\int f_n^q\,d\lambda\geq\int_{(0,a/n)}f_n^q\,d\lambda=(a/n)e^{qn/a}=q\frac{e^{qn/a}}{qn/a}\xrightarrow{n \to+\infty}+\infty$$

This is what I have but I am not really sure that my demonstration is correct. Can someone help me?

Answer 1

$\frac {e^{qy}} y \to \infty$ as $ y \to \infty$ (for any $q >0$). Hence there exist $M$ such that $\frac {e^{qy}} y >2$ if $y >M$. Taking $x=\frac 1y $ see that This gives $e^{q/x} >\frac 2 x$ for $0< x<\frac 1M$. Now it is easy to show that $\int_0^{a} e^{q/x} dx=\infty$.

Answer 2

Exercise: For any $q>0,$ $\dfrac{e^{nq}}{n}\to \infty$ as $n\to \infty.$

So let $q>0.$ Then for any $n\in \mathbb N,$

$$\int_0^1 (e^{1/t})^q\,dt =\int_0^1 e^{q/t}\,dt $$ $$\ge\int_0^{1/n} e^{q/t}\,dt > e^{nq}/n.$$

By the exercise, $\int_0^1 (e^{1/t})^q\,dt= \infty.$