Bound between difference of logarithms and difference of functions Let $f,g$ be two functions such that $f(x)>0$ and $g(x)>0$ for all $x$ and $\int f(x)dx=1$ and $\int g(x)dx=1$, and $f,g\in L_2$. The goal is to find $h$, if it exists, such that:
If $\int \log f(x) -\log g(x) dx < \epsilon$, then $\int (f(x)-g(x))^2dx<h(\epsilon)$, for $\epsilon>0$.
You may assume $f,g$ are continuous if that helps. Also, if not possible to find an exact expression $h$, an upper bound would be great.
This is what I have thought so far.
My first thought was to use Jensen's inequality as follows:
$$\exp(\epsilon)>\exp\Bigg(\int \log f-\log  dx\Bigg)\leq \int\exp \log\dfrac{f}{g}dx=\int \dfrac{f}{g}dx$$
Unfortunately, that doesn't lead anywhere because $\exp(\epsilon)$ needs to be an upper bound for $\int\frac{f}{g}dx$, but we can't say anything about that from the above equation.
Another thought was to expand $\int (f-g)^2dx=\int f^2-2fg +g^2 dx$. We know that since $f,g\in L_2$, both integrals with squares are bounded, but there is nothing we can say about $\int fgdx$, and moreover, that is not connected with $\epsilon$. I don't have anything else.
I would appreciate help with this.
Edit: this isn't a homework problem where I know what the solution should be, or even if it exists.
 A: First of all we will prove the following property :
Property :
If $\,\gamma>1\,,\,$ then $\;\big|\log z\big|\geqslant\dfrac{(z-1)^2}{\gamma^2}\quad\forall\,z\in\big]0,\gamma\big]\,.\quad\color{blue}{(*)}$
Proof :
Let $\,\gamma\,$ be any real number greater than $\,1\;(\gamma>1)\,.$
Let $\,\varphi:\big]0,\gamma\big]\to\mathbb R\,$ be the function defined as
$\varphi(z)=\log z-\dfrac{z-1}\gamma\;\;\quad\forall\,z\in\big]0,\gamma\big]\,.$
The function $\,\varphi\,$ is differentiable on $\,\big]0,\gamma\big]\,$ and it results that
$\varphi(z)=\dfrac1z-\dfrac1\gamma>0\qquad\forall\,z\in\big]0,\gamma\big[\,.$
Consequently the function $\,\varphi\,$ is monotonically increasing on $\,\big]0,\gamma\big]\,$ and it results that
$\varphi(z)<\varphi(1)=0\qquad\forall\,z\in\big]0,1\big[\;,\quad\color{blue}{(1)}$
$\varphi(z)>\varphi(1)=0\qquad\forall\,z\in\big]1,\gamma\big]\;.\quad\color{blue}{(2)}$
Moreover ,
$0<\dfrac{1-z}\gamma<\dfrac1\gamma<1\qquad\forall\,z\in\big]0,1\big[\;,\quad\color{blue}{(3)}$
$0<\dfrac{z-1}\gamma\leqslant\dfrac{\gamma-1}\gamma<1\qquad\forall\,z\in\big]1,\gamma\big]\;.\quad\color{blue}{(4)}$
From the inequalities $\,(1)\,$ and $\,(3)\,,\,$ it follows that
$\begin{align}
\big|\log z\big|&=-\log z=-\varphi(z)-\dfrac{z-1}\gamma>-\dfrac{z-1}\gamma=\\
&=\!\!\!\!\!\underset{\text{it belongs to }\left]0,1\right[}{\underbrace{\dfrac{1-z}\gamma}}\!\!\!\!\!>\dfrac{(1-z)^2}{\gamma^2}=\dfrac{(z-1)^2}{\gamma^2}\;\quad\forall\,z\in\big]0,1\big[\,.
\end{align}$
From the inequalities $\,(2)\,$ and $\,(4)\,,\,$ it follows that
$\begin{align}
\big|\log z\big|&=\log z=\varphi(z)+\dfrac{z-1}\gamma>\dfrac{z-1}\gamma=\\
&=\!\!\!\!\!\underset{\text{it belongs to }\left]0,1\right[}{\underbrace{\dfrac{z-1}\gamma}}\!\!\!\!\!>\dfrac{(z-1)^2}{\gamma^2}\qquad\forall\,z\in\big]1,\gamma\big]\,.
\end{align}$
Consequently, it results that
$\big|\log z\big|>\dfrac{(z-1)^2}{\gamma^2}\qquad\forall\,z\in\big]0,\gamma\big]\setminus\big\{1\big\}\,.$
Since $\,\big|\log z\big|=\dfrac{(z-1)^2}{\gamma^2}\;$ for $\,z=1\,,\,$ we get that
$\big|\log z\big|\geqslant\dfrac{(z-1)^2}{\gamma^2}\quad\forall\,z\in\big]0,\gamma\big]\,.$

$\text{ }\\ \text{Let }\,f,g:\mathbb R\to\mathbb R^+\,\text{ be two positive bounded functions such}\\ \text{that }\displaystyle\int_\mathbb R f(x)dx=1\,,\;\displaystyle\int_\mathbb R g(x)dx=1\,\text{ and }\,f,g\in L_2\,.$
We will find a function $\,h:\mathbb R^+\to\mathbb R\,$ such that:
if $\!\!\displaystyle\int_\mathbb R\!\!\big|\log f(x)\!-\!\log g(x)\big|dx\!<\!\varepsilon,\,$ then $\!\!\displaystyle\int_\mathbb R\!\!\big(\!f(x)\!-\!g(x)\!\big)\!^2\!dx\!<\!h(\varepsilon).$
$ $
Let $\,F=\sup_\limits{x\in\mathbb R} f(x)\;$ and $\;G=\sup_\limits{x\in\mathbb R} g(x)\,.$
Since the functions $\,f\,$ and $\,g\,$ are positive and bounded for hypothesis, then $\,F\,$ and $\,G\,$ are positive real numbers $\,\big(F>0\,,\,G>0\big)\,.$
Let $\,X\!=\!\left\{\!x\in\mathbb R\,\bigg|\,\dfrac{f(x)}{g(x)}\!>\!\sqrt e\!\right\},\;Y\!=\!\left\{\!x\in\mathbb R\,\bigg|\,\dfrac{f(x)}{g(x)}\!\leqslant\!\sqrt e\!\right\}.$
It results that $\;X\cup Y=\mathbb R\;$ and $\;X\cap Y=\emptyset\,.$
Moreover $\;Y\neq\emptyset\;,\;$ otherwise $\;\dfrac{f(x)}{g(x)}>\sqrt e\quad\forall\,x\in\mathbb R\;$ and
$\displaystyle\int_\mathbb R\!\big|\log f(x)\!-\!\log g(x)\big|dx=\!\int_\mathbb R\!\left|\log\frac{f(x)}{g(x)}\right|dx\geqslant\!\!\int_\mathbb R\!\dfrac12 dx=+\infty$
but it would contradict the hypothesis.
For any $\;x\in Y\;$ it results that
$\big(f(x)\!-\!g(x)\big)^2\!=e\,g^2(x)\dfrac{\left(\dfrac{f(x)}{g(x)}\!-\!1\right)^2}{\big(\sqrt e\big)^2}\!\!\!\!\!\!\!\!\underset{\text{from }(*)\text{ with }\gamma=\sqrt e\text{ and }z=\frac{f(x)}{g(x)}\in\left]0,\sqrt e\right]}{\underbrace{\leqslant e\,g^2(x)\left|\log\dfrac{f(x)}{g(x)}\right|}}\!\!\!\!\!\!\!\!\!\!=$
$=e\,g^2(x)\big|\log f(x)-\log g(x)\big|\leqslant e\,G^2\big|\log f(x)-\log g(x)\big|\,.$
Therefore ,
$\begin{align}
\displaystyle\int_Y\big(f(x)-g(x)\big)^2dx&\leqslant e\,G^2\int_Y\big|\log f(x)-\log g(x)\big|dx\leqslant\\
&\leqslant e\,G^2\int_\mathbb R\big|\log f(x)-\log g(x)\big|dx<\\
&<e\,G^2\varepsilon\,.
\end{align}$
There are two possible cases :
$1)\;\;X=\emptyset\;,$
$2)\;\;X\neq\emptyset\;.$
If $\,X=\emptyset\,,\,$ then $\,Y=\mathbb R\,$ and
$\displaystyle\int_\mathbb R\big(f(x)-g(x)\big)^2dx=\int_Y\big(f(x)-g(x)\big)^2dx<e\,G^2\varepsilon\,.$
If $\,X\ne\emptyset\,,\,$ then we will proceed as follows :
$\,f(x)>\sqrt e\,g(x)>g(x)\qquad\forall\,x\in X\;,\;\;\;$ hence ,
$0<f(x)-g(x)<f(x)\leqslant F\qquad\forall\,x\in X\;,$
$\big(f(x)-g(x)\big)^2<F^2\qquad\forall\,x\in X\;.\quad\color{blue}{(5)}$
Since $\,\dfrac{f(x)}{g(x)}>\sqrt e\quad\forall\,x\in X\;,\;$ then $\;\log\dfrac{f(x)}{g(x)}\!>\!\dfrac12\!>\!0\;$ and
$\displaystyle\dfrac12m(X)\leqslant\int_X\log\dfrac{f(x)}{g(x)}dx=\int_X\left|\log\dfrac{f(x)}{g(x)}\right|dx\leqslant$
$\displaystyle\leqslant\int_\mathbb R\left|\log\dfrac{f(x)}{g(x)}\right|dx=\int_\mathbb R\big|\log f(x)-\log g(x)\big|dx<\varepsilon\;,$
hence ,$\;\;m(X)<2\varepsilon\;.\quad\color{blue}{(6)}$
From $\,(5)\,$ and $\,(6)\,,\,$ it follows that
$\displaystyle\int_X\big(f(x)-g(x)\big)^2dx\leqslant\int_X F^2dx=F^2m(X)<2F^2\varepsilon\;.$
Hence ,
$\begin{align}
\displaystyle\int_\mathbb R\!\!\big(f(x)\!-\!g(x)\big)^2\!dx&=\!\!\int_X\!\!\big(f(x)\!-\!g(x)\big)^2\!dx+\!\!\int_Y\!\!\big(f(x)\!-\!g(x)\big)^2\!dx\!<\\
&<\left(2F^2+eG^2\right)\varepsilon\;.
\end{align}$
In any case, it results that
$\displaystyle\int_\mathbb R\!\!\big(f(x)\!-\!g(x)\big)^2\!dx<\left(2F^2+eG^2\right)\varepsilon\;.$
Consequently, the function $\,h:\mathbb R^+\to\mathbb R\,$ is the following one :
$h(\varepsilon)=\left(2F^2+eG^2\right)\varepsilon\qquad\forall\,\varepsilon\in\mathbb R^+\;,$
where $\;F=\sup_\limits{x\in\mathbb R} f(x)\;$ and $\;G=\sup_\limits{x\in\mathbb R} g(x)\,.$
