Showing this to $1 \leq p < \infty$ Show that if $f(x) = \ln{1/x}$ for $x \in (0,1],$ then $f$ belongs to $L^{p}(0,1]$ for all $1 \leq p < \infty$ but not to $L^{\infty}(0,1]$.
I can show it to $L^{1}$ but how can I show it to the rest?
 A: Let $1 \le p < +\infty$. Using the substitution $t = \frac1x$ we get
\begin{align}
\int_0^1 \left(\ln \frac1x\right)^p\,dx &= \begin{bmatrix} t = \frac1x \\ dx = -\frac1{t^2}dt\end{bmatrix}\\
&= \int_1^{+\infty} \frac{(\ln t)^p}{t^2}\,dt
\end{align}
Now notice that for large enough $t_0 \ge 1$ we have $t \ge t_0 \implies \ln t \le \sqrt[2p]{t}$ so the above integral can be bounded by
$$\int_1^{t_0}\frac{(\ln t)^p}{t^2}\,dt + \int_{t_0}^{+\infty}\frac{dt}{t^{3/2}} < +\infty$$
which is finite since $\frac{(\ln t)^p}{t^2}$ is bounded on $[1,t_0]$ and $t \mapsto \frac1{t^{3/2}}$ is integrable on $[1, +\infty)$.
We conclude that $f \in L^p(0,1]$.
On the other hand, for every $\varepsilon > 0$ we have $f \ge \varepsilon$ on the set $(0,e^{-\varepsilon}]$ which has positive measure so $f \notin L^\infty(0,1]$.
A: It´s sufficient to prove that $(log(x))^{p}$ is in $L^{1}(0,1]$ for any $p \in [1,\infty)$ because $log(1/x) = -log(x)$. As a hint, consider the substitution $x = exp(t)$ and remember that the exponential function converges faster to zero as $t \rightarrow - \infty$ than any polynomial function. For $f \notin L_{\infty}((0,1])$, note that $f$ is not bounded.
A: To show $f\in L^p(0,1]$ for $1\leq p<\infty$, it suffices to show that $|f|^n\in L^1(0,1]$ for all $n\in\mathbb N$ (since $(0,1]$ has finite measure, so $L^q(0,1]\subset L^p(0,1]$ whenever $p\leq q$).  Proceed by induction.
