Let $f_k$ be a sequence probability density functions on the real line $\mathbb{R}$, is there a way to bound (uniformly in $k$) the following integral from below (for any small enough $h$) ? $$ \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x|^2}{h} + \log|x-y| \Big)f_k(x)f_k(y) ~dx dy. $$
Here $\log$ is the natural logarithm.
Edit my attempt : Recall $f_k$ is a probability density, note
\begin{align*} \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x|^2}{h} + \log|x-y| \Big)f_k(x)f_k(y) ~dx dy \geq \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x-y|^2}{Ch} + \log|x-y| \Big)f_k(x)f_k(y) ~dx dy \end{align*}
for some $C>0$. Next
\begin{align*} \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x-y|^2}{Ch} + \log|x-y| \Big)f_k(x)f_k(y) ~dx dy \geq& \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x-y|^2}{Ch} -\frac{1}{|x-y|} \Big)f_k(x)f_k(y) ~dx dy+1 \\ \geq & \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x-y|^3-Ch}{|x-y|} \Big)f_k(x)f_k(y) ~dx dy \end{align*}
Now I'm stuck...
No chance. Consider e.g.
$$f_k(x)=k^2\left(x+\frac{1}{k}\right)\chi_{[-1/k,0]}(x)+k^2\left(x-\frac{1}{k}\right)\chi_{(0,\frac{1}{k})}(x),$$
Then
$$\int_{\mathbb{R}} f_k = 2k^2\int_{-1/k}^0(x+\frac{1}{k}) dx = 2k^2 [ \frac{x^2}{2}+\frac{x}{k}]^0_{-\frac{1}{k}} = 2k^2(-1/(2k^2)+1/k^2) = 1.$$
Now for $h>0$ arbitrary and fixed there holds
$$\int_{\mathbb{R}^2} \frac{x^2}{h} f_k(x)f_k(y) dy dx = \frac{2k^2}{h} \int_{-1/k}^0(x+1/k) dx = 2k^2(-\frac{1}{4k^4} + \frac{1}{3k^4}) = \frac{1}{24hk^2}\to 0 $$
as $k\to\infty$. Moreover,
\begin{align*} \int_{\mathbb{R}^2} \log|x-y|f_k(x)f_k(y) dx dy &\le k^4\int_{-1/k}^0\int_{-1/k}^0 \log|x-y|(x+1/k)(y+1/k)dxdy\\ &\le k^4\int_{\frac{-1}{2k}}^0\int_{-\frac{1}{2k}}^0 \log (\frac{1}{k}) \frac{1}{2k}\frac{1}{2k}dx dy\\ &= \log(\frac{1}{k}) \frac{1}{4}\to - \infty \end{align*}
as $k\to\infty$, thus there is no (uniform in $h$ and $k$) lower bound.