When is the logarithm of this function square integrable?

Question

I was trying to prove $\log(1-|r(s)|^2)$ lies in $L^2$ when $r\in L^1\cap L^2$. How should I do this? Thank you!

Answer 1

You can use for instance that $|\log(1+\delta)|\leq2\delta$ for all $\delta$.

Answer 2

To begin with, did you mean $f_1\circ r$ or $f_2\circ r$, where $$f_1(x)=\log(1+|x|^2)\text{ and }f_2(x)=\log(1-x^2)?$$

Note that $f_1\circ r$ is defined everywhere, whereas the function $$f_2\circ r(s)=\log(1-|r(s)|^2)=\log(1+|r(s)|)+\log(1-|r(s)|)$$ has meaning only when $|r(s)|<1$ [*].

Now, $f_1$ and $f_2$ are continuous (Borel measurable will do) hence, under the restriction [*], $f_k\circ r$ is Lebesgue measurable $k=1,2$.

Hence to prove $f_k\circ r\in L^2$, it suffice to find some estimate of the form $$|f_k(x)|\leq C x\tag{1$_k$}$$

To reach (1$_1$) we use $1<1+x^2\leq1+2x+x^2=(1+x)^2$ for $x>0$, and hence $$0\leq f_1(x)=\log(1+x^2)=2\log(1+x)\leq 2x$$
where in the last step we used the (sort of) standard estimate $\log(1+x)\leq x$ (to prove that, consider the sign of the derivative of $x\mapsto\log(1+x)-x$ and check the value at $x=0$).

For (1$_2$), however, we have $1-x^2<0$ for $|x|<1$ and hence $$|f_2(x)|=\log(1+x)-\log(1-x)\leq x-\log(1-x)$$ and the function $$x\mapsto-\log(1-x)\tag{2}$$ blows up at $x=1$...

(2) suggest it is time to search for a counterexample, that is we look for $r\in L^1\cap L^2$ with say $0<r<1$ and $$\log(1-|r(x)|)\notin L^2$$ To construct such a function, a good start is $r(x)=1-\exp(-f(x))$ with $f(x)>0$, because then $0<r(x)<1$ and $$-\log(1-r(x))=f(x)$$
Now the idea is that

• $f$ should decay to $0$ slow enough so that $f\notin L^1\cap L^2$

• while in the same time $\exp(-f(x))$ is close enough to $1$ making $r\in L^1\cap L^2$.

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The choice $f(x)=1/x^2$ leads to $r(x)= 1-\exp(-1/x^2)$, this function stays very close to 1, around $x=0$, and $\log(1-r(x))=-1/x^2$ is not integrable. However, $r$ is continuous and has decays enough to be in $L^p,\,p\geq 1$.

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(If you hold the mouse pointer in the grey area above you will see a counterexample.)