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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?

Emptymind
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3 Answers3

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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 $f \notin L^\infty(0,1]$.

mechanodroid
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  • I do not understand from where your last line came ... why for every $\epsilon > 0$ we have $f \geq \epsilon$? and why specifically on this set $(0, e^{-\epsilon}]$? could you provide more details please? – Idonotknow Nov 26 '19 at 16:17
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    @Idonotknow Let $\varepsilon > 0$. If $x \in (0,e^{-\varepsilon}]$, then $$0 < x \le e^{-\varepsilon} \implies \frac1x \ge e^{\varepsilon} \implies \ln\frac1x \ge \varepsilon \implies f(x) \ge \varepsilon.$$ – mechanodroid Nov 26 '19 at 16:20
  • Also, I can not see why you used the substitution $t = 1/x$. is this because the integral is difficult to integrate? –  Nov 29 '19 at 03:01
  • Is not the first integral is an improper integral as $x$ can not take the value $0$? –  Nov 29 '19 at 15:19
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    @Mathstupid Ok, perhaps it would work without substitution as well. Let $t_0 \ge 1$ be chosen as above. Then for $0 < x < \frac1{t_0}$ we have $\ln\frac1x \le \sqrt[2p]{\frac1x}$ so $$\int_0^1 \left(\ln\frac1x\right)^p,dx \le \int_{0}^{\frac1{t_0}} \frac{dx}{\sqrt{x}} + \int_{\frac1{t_0}}^1 \left(\ln\frac1x\right),dx < +\infty$$ Whichever suits you better. I interpreted the integral as a Lebesgue integral over $(0,1]$ but yes, it can also be calculated as an improper Riemann integral. – mechanodroid Nov 30 '19 at 08:54
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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.

Claudio Moneo
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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.

Aweygan
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