Function in $L^p$ but not in $L^{\infty}$

Question

Show that if

$$f(x) = \ln\left({1\over x}\right),\quad 0<x\le 1$$

then $f\in L^p((0, 1])$ for all $1 \le p < \infty$ but $f\not\in L^{\infty}((0, 1])$.

Intuitively I know I have to show that $\int_{[0,1]}|f|^p <\infty $ but $f$ is not essentialy bounded, that means $\exists M >0$, s.t. $|f(x)|>M$ but I forget how to compute the integral and I am not sure how to find $M$.

Any help, Thanks.

Answer 1

Make a simple change of variables, $u=x^{-1}$ so that $du = -x^{-2}dx\iff -u^{-2}du = dx$

Then you get

$$\int_\infty^1 -{\log u\over u^2}\,du=\int_1^\infty {\log u\over u^2}\,du$$

And this is easily verified to be integrable by directly comparing with

$$\int_1^\infty {du\over u^{3/2}}$$

Since we know

$$\lim_{x\to\infty} {\log x\over \sqrt{x}}=0$$

So there is $N>0$ so that for all $x>N$ we have $\log x<x^{1/2}$. Hence

$$\int_1^\infty{\log u\over u^2}\,du\le \int_1^N{\log x\over x^2}\,dx +\int_N^\infty {u^{1/2}\over u^2}.$$

For $p^{th}$ powers this is as simple, since given $\epsilon>0, p>1$ we have ${\log^p x\over x^{p\epsilon}}\to 0$ as $x\to\infty$ by the same result as before, so you just do

$$\int_1^N \left|{\log^p x\over x^{2p}}\right|\,dx +\int_N^\infty {x^{p\epsilon}\over x^p}\,dx$$

and we need only choose $\epsilon$ small enough so that $p(\epsilon -1)<-1$ since then we have the second integral is

$$\int_N^\infty {dx\over x^{p(1-\epsilon)}}.$$

And this converges exactly when $p(1-\epsilon)>1$.

For the second part, it's simple: select $x<e^{-N}\iff {1\over x} > e^N$ then

$$\log\left({1\over x}\right) >\log e^N = N$$

so the function is not essentially bounded, since $|f(x)|>N$ on the set $(0, e^{-N})$ which has measure $e^{-N}>0$.

Answer 2

For any $a>0,\lim_{x\to 0^+}x^a\ln (1/x) =0.$ So given $0<p<\infty,$ we have for small positive $x$

$$x^{1/(2p)}\ln(1/x) < 1 \implies x^{1/2}(\ln(1/x))^p < 1 \implies (\ln(1/x))^p < x^{-1/2}.$$

Since $\int_0^1 x^{-1/2}\, dx < \infty,$ we have the result.

Answer 3

You can do this change of variables : u = $\log{1/x} \implies x = e^{-u}$

So, to calculate the p-norm, consider the following integral

$\int_0^{1}\log(\frac{1}{x})^pdx = \int_{0}^{\infty} u^p e^{-u} du$

The integral in the r.h.s is the well-known gamma function $\Gamma(p+1)$, which converges for the values of p that we are interested.

For the $L^{\infty}$ part, consider $M \in [0,\infty)$

So, we have that $\log(\frac{1}{x})> M \iff x < e^{-M} $

Now, consider the interval $A = (0, e^{-M}) $. We have that, for all $x \in A, f(x) > M$ because the function is strictly decreasing. Also, f is positive in $(0,1)$, so we can throw away the modulus.

So we found a measurable set where the function is greater than M, for all M, and have positive measure.This implies that $ f \notin L^{\infty} $ and we are done.