A limit of an Integral

Question

Consider the following limit $$K=\lim_{x\rightarrow \infty}\frac{1}{x(1-x)}\left(1-\int_{\mathbb{R}}g(y;x)^x f(y)^{1-x}\mathrm{d}y\right)$$ where $f$ and $g$ are any continuous probability density functions on $\mathbb{R}$, $(f\neq g)$. Additionally, $g$ is parametrized by $x$ as well.

Question $1$: Show that if $g(y;x)=g(y)$ for all $x$, then K$\in\{0,\infty\}$.

Question $2$: Prove or disprove if there exists $g$, s.t. $\lim_{x\rightarrow \infty} g(y;x)\neq f(y)$ on at least some (one) interval $y\in I\subset\mathbb{R}$ and $0<K<\infty$.

Added: According to the examples that I did, for question $1$, the limit was always $\infty$. I suspect that $K=0$ is never a solution..

Answer 1

Considering $f$ and $g$ some smoothed versions of the functions $\bar f=\mathbf 1_{(0,1)}$ and $\bar g=2\mathbf 1_{(0,1/2)}$, say $f=\bar f\ast h$ and $g=\bar g\ast h$ where $h$ is a gaussian density of vanishingly small variance, one sees that the conclusion of Question 1 does not necessarily hold.

Answer 2

For question $1$ the only solution is $\infty$. Here is the explanation:

if $g(y)=g(y;x)\forall y$ I just replace it with $g(y)$ and I have

$$K=\lim_{x\rightarrow \infty}\frac{1}{x(1-x)}\left(1-\int_{\mathbb{R}}g(y)^x f(y)^{1-x}\mathrm{d}y\right)$$

Here $h_1(x)=\int_{\mathbb{R}}g(y)^x f(y)^{1-x}\mathrm{d}y$ is a convex function of $x$, i.e., $h_1^{''}>0$. Since $h_1>0$ for all $x>0$, then

$$\lim_{x\rightarrow \infty}\int_{\mathbb{R}}g(y)^x f(y)^{1-x}\mathrm{d}y=\infty\quad\Longrightarrow \quad K=\frac{\infty}{\infty}$$ Using L'Hospital rule twice we get

$$K=\lim_{x\rightarrow \infty}\frac{\int_{\mathbb{R}}\log^2(g(y)/f(y))g(y)^x f(y)^{1-x}\mathrm{d}y}{2}$$

The function $h_2(x)=\int_{\mathbb{R}}\log^2(g(y)/f(y))g(y)^x f(y)^{1-x}\mathrm{d}y$ is also positive and convex therefore $K=\infty$ when $x\rightarrow \infty$.

For question $2$:

I assume that there exists a $g$ s.t. $\lim g(y;x)=g(y)\neq f(y)$ at least in some interval. Then considering this $g$ as above gives me $K=\infty$ for $x\rightarrow \infty$. In other words, $g$ should approach $f$ at some rate ($c x(1-x)$ for some constant $c$) which will eventually makes $K$ a constant.