# Prove or disprove that the given expression is “always” positive

I have previously asked a question and I tried to solve it by my own and it led to the question below:

Prove or disprove that

$$\small\int_{\mathbb{R}}l(y)^xf_0(y)\mathrm{d}y\int_{\mathbb{R}}f_0(y)l(y)^x\ln (l(y)^x)\ln (l(y))\mathrm{d}y-\int_{\mathbb{R}}l(y)^xf_0(y)\ln (l(y))\mathrm{d}y\int_{\mathbb{R}}f_0(y)l(y)^x\ln (l(y)^x)\mathrm{d}y$$

is greater than $0$.

Given:

$\rightarrow f_0$ and $f_1$ are some density functions

$\rightarrow l(y)=\frac{f_1(y)}{f_0(y)}$ is an increasing function.

$\rightarrow x\in(0,1)$

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You can drop the denominator... – 1015 Mar 19 '13 at 14:12
@julien thanks, done! – Seyhmus Güngören Mar 19 '13 at 14:15
The first thing to do would be to consider the convergence of $\int \ell(y)^xf_0(y)dy$. – 1015 Mar 19 '13 at 14:19
@julien yes it must be convergent as $\int f_1(y)^xf_0(y)^{1-x}$ is just some weighted multiplication of densities. – Seyhmus Güngören Mar 19 '13 at 14:22
Yes, that's by Holder. And that's a start. – 1015 Mar 19 '13 at 14:30

This is an application of Cauchy-Schwarz in disguise. It suffices to find the appropriate Hilbert space. And then show that the relevant functions actually belong to it. The latter is the tedious part of this exercise.

Consider the positive function $w(y)=\ell(y)^xf_0(y)=f_1(y)^xf_0(y)^{1-x}$. By Holder, it is integrable over $\mathbb{R}$. Now consider the weighted $L^2$ space $L^2_w(\mathbb{R})$ equipped with the inner product

$$(g,h)_w:=\frac{\int_\mathbb{R}g(y)h(y)w(y)dy}{\int_\mathbb{R}w(y)dy}$$

and the resulting norm $\|g\|_w=\sqrt{(g,g)_w}$. Note that $g$ is in $L^2_w$ if, by definition, $g^2w$ is integrable over $\mathbb{R}$.

Observing $\ln(\ell(y)^x=x\ln(\ell(y))$, what you want to show boils down to $$\left(\int_\mathbb{R} \ln(\ell(y))w(y)dy\right)^2< \left(\int_\mathbb{R} (\ln(\ell(y)))^2w(y)dy\right) \left(\int_\mathbb{R} (w(y)dy\right).$$ Dividing by $\left(\int_\mathbb{R} (w(y)dy\right)^2$, this reads

$$(g,1)_w^2< \|g\|_w^2=\|g\|_w^2\|1\|_w^2\qquad\mbox{with}\;\;g(y)=\ln(\ell(y)).$$

This follows from Cauchy-Schwarz. Note that this will be a strict inequality as $g$ and $1$ are linearly independent, since $\ell$ is increasing.

It only remains to prove that $g$ belongs to $L^2_w$, i.e. $\int_\mathbb{R} g(y)^2w(y)dy<\infty$. This is where the assumption $\ell$ increasing comes into play.

If $g$ is bounded, the claim is obvious. If not, since $\ell$ is increasing and positive, this means that either $\lim_{+\infty} \ell=+\infty$ or $\lim_{-\infty} \ell=0$. So assume, to begin with, that $\lim_{+\infty} \ell=+\infty$. Now write $$(\ln \ell(y))^2w(y)=(\ln \ell(y))^2\ell(y)^xf_0(y) =\frac{(\ln \ell(y))^2}{\ell(y)^\frac{1-x}{2}}\ell(y)^\frac{1+x}{2}f_0(y)=\frac{(\ln \ell(y))^2}{\ell(y)^\frac{1-x}{2}}f_1(y)^\frac{1+x}{2}f_0(y)^\frac{1-x}{2}.$$ By Holder, the function $f_1(y)^\frac{1+x}{2}f_0(y)^\frac{1-x}{2}$ is integrable. Now $$\lim_{+\infty} \frac{(\ln \ell(y))^2}{\ell(y)^\frac{1-x}{2}}=0.$$ So $g(y)^2w(y)$ is integrable over $[0,+\infty)$ by comparison with $f_1(y)^\frac{1+x}{2}f_0(y)^\frac{1-x}{2}$. If $\lim_{-\infty}\ell>0$, then $g$ is bounded on $(-\infty,0]$ and integrability over $(-\infty,0]$ follows. If $\lim_{-\infty}\ell=0$, then write $$(\ln \ell(y))^2w(y)=(\ln \ell(y))^2\ell(y)^xf_0(y).$$ Now $$\lim_{-\infty}(\ln \ell(y))^2\ell(y)^x=0$$ hence integrability over $(-\infty,0]$ follows by comparison with $f_0$. n any case, we have proven that $g$ belongs to $L^2_w$, which completes the argument.

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Ok just checking, one min. – Seyhmus Güngören Mar 19 '13 at 15:35
what is $h$ in $(g,h)_w$, perhaps $f$? – Seyhmus Güngören Mar 19 '13 at 15:44
@SeyhmusGüngören ? No I just define $(g,h)_w$ for arbitrary functions such that $g^2w$ and $h^2w$ are integrable. This defines the inner product structure on $L^2_w$, the set of all $g$ such that $g^2w$ is integrable. I chose the letters $g,h$ randomly. Use $f$ if you prefer. – 1015 Mar 19 '13 at 15:48
I see. I am not familiar with $L_w^2$, I am currently reading your wiki post. I only checked the right and left side of the equation, i saw the term $h$ on the left side but not on the right side, thats the reason I asked, perhaps it was $f$. – Seyhmus Güngören Mar 19 '13 at 15:52
@SeyhmusGüngören Right, thanks and sorry, this is a typo. Fixed. – 1015 Mar 19 '13 at 15:54