A question about inclusion of $\mathcal L^p$ spaces for a general measure space

Question

Show that if $X=(0,\infty)$ then the function $f(x) =\frac{1}{x^{1/2}\left(1+|\log x|\right)}$ is in $\mathcal L^2(\mu)$ but not in $\mathcal L^p(\mu)$ for any $p\in (1,2) \cup (2,\infty)$ where $\mu$ is the Lebesgue measure on $(0,\infty).$

I could solve the case for $p=2$ but for $p\neq 2,$ if we calculate the integral $\int_1^a f^p(x)\ \mathrm{d}x$, as $x >1,$ we can write $x=1+t,$ then we know $\log(1+t)<t$ and so $\int_1^a f^p(x) dx$ $>$ $\int_0^{a-1}$ $ \frac{\mathrm{d}t}{(1+t)^{3p/2}}$ but then I can't conclude anything from this .

A very small hint is very much required at this moment, thanks in advance.

Answer 1

To get you started, verify that $x^\alpha(1+|\log x|)^\beta\to 0$ as $x\to 0^+$ for any $\alpha,\beta >0.$ Thus if $p>2,$

$$\frac{1}{x^{p/2}(1+|\log x|)^p}= \frac{1}{x[x^{p/2-1}(1+|\log x|)^p]} > \frac{1}{x}$$

for small $x>0.$

Answer 2

Another approach to this is to break the integral $I_p=\int^\infty_0 |f^p|$ in two pieces:

$$I_p=\int^1_0\frac{dx}{x^{p/2} (1-\log x)^p} + \int^\infty_1\frac{dx}{x^{p/2} (1+\log x)^p} $$

On $(0,1)$, $x\mapsto 1-\log x$ is monotone non increasing and so, \begin{align} \int^1_0\frac{dx}{x^{p/2} (1-\log x)^p}&=\sum_n\int^{e^{-n}}_{e^{-(n+1)}}\frac{dx}{x^{p/2} (1-\log x)^p} \\ &\geq \sum_{n\geq0}\frac{e^{-n}(1-e^{-1})}{e^{-np/2}(1+n+1)^p}\\ &=(1-e^{-1})\sum_{n\geq0} \frac{e^{n(p/2 - 1)}}{(2+n)^p} \end{align}

If $p>2$ then $\lim_{n\rightarrow\infty}\frac{e^{n(p/2 - 1)}}{(2+n)^p}=\infty$ and so the integral over $(0,1]$ diverges to $\infty$.

Similarly, over $(1,\infty)$, $x\mapsto 1+\log x$ is monotone non decreasing and so

\begin{align} \int^\infty_1\frac{dx}{x^{p/2} (1+\log x)^p}&=\sum_{n\geq0}\int^{e^{n+1}}_{e^n}\frac{dx}{x^{p/2} (1+\log x)^p}\\ &\geq\sum_{n\geq0}\frac{e^n(e-1)}{e^{(n+1)p/2}(2+n)^p}\\ &=\frac{e-1}{e^{{p/2}}}\sum_{n\geq0}\frac{e^{n(1-p/2)}}{(2+n)^p} \end{align} If $p<2$, the $\lim_{n\rightarrow\infty}\frac{e^{n(1-p/2)}}{(2+n)^p}=\infty$ and so, the integral over $[1,\infty)$ diverges to $\infty$.