Fix a small $\delta>0$ and let $p,q>1$. Consider the integral
$$I(p,q):=\int\limits_{1-\delta}^{1+\delta} \int\limits_{y/2}^{2y}\frac{1}{|y-x|^{\frac{1}{p}}|1-x|^{\frac{1}{q}}} \,\mathrm{d}x\,\mathrm{d}y. $$
I am trying to show that $I(p,q)$ diverges if $\frac{1}{p}+\frac{1}{q}\geq 1$. I am not sure this is even the case ? Any hints on how to handle this?
Remark: This seems to be related to the failure of the Hardy-Littlewood-Sobolev inequality (HLS) at the endpoint $p=1$. HLS reads:
If $1<p,q<\infty$, $f\in L^p$ and
$$Tf(x):=\int_{\mathbb{R}^n} \frac{f(y)}{|x-y|^{\gamma}}dy$$
Then $$\|Tf\|_q\leq \|f\|_p$$ if and only if $$\frac{1}{p}-\frac{1}{q}=1-\frac{\gamma}{n}.$$
Many thanks.
Split the $x$ integral into two
$$ \int_{y/2}^{2y} = \int_{y/2}^{(1+y)/2} + \int_{(1+y)/2}^{2y} $$
If $y < 1$:
$$ \int_{y/2}^{(1+y)/2} \lesssim |y/2|^{1-1/p} |1-y|^{-1/q} $$
and
$$ \int_{(1+y)/2}^{2y} \lesssim |1-y|^{-1/p} |2y-1|^{1-1/q} $$
If $y > 1 $:
$$ \int_{y/2}^{(1+y)/2} \lesssim |1 - y|^{-1/p} |1-y/2|^{1-1/q} $$
and
$$ \int_{(1+y)/2}^{2y} \lesssim |y|^{1-1/p} |1-y|^{-1/q} $$
Note: Each of these integrals put the singular term in $L^1$ and the other term in $L^\infty$. E.g. for the first one, since $y < 1$ we have $(1+y)/2 < 1$ so $|1-x| > |1-y|/2$. So the integral becomes $$ \leq 2^{1/q}|1-y|^{-1/q} \int_{y/2}^{(1+y)/2} |x-y|^{-1/p}~ dx $$ and the interior integral can be evaluated explicitly. The other terms are treated similarly.
So the integral is bounded by
$$ \int_{1-\delta}^{1+\delta} |y|^{1-1/p} |1-y|^{-1/q} + |2y-1|^{1-1/q} |1-y|^{-1/p} ~dy $$
and so your $I(p,q)$ should converge.