Let $f\in L^1(\mathbb R^n)\cap L^{\infty}(\mathbb R^n) $.
Let $g(x):=\frac{1}{\vert x\vert^{n-2}}\,,n\gt3$ and $n$ is integer.
Then $g\in L^1_{loc}(\mathbb R^n)$ and is bounded outside the ball $B_r(0)$ for any given $r\gt 0$, where the notation "$L^1_{loc}(\mathbb R^n)$" means $g$ is integrable in any compact sets in $\mathbb R^n$.
Let $e_i :=(0,0,...,1,0,...,0)$, where $1$ is at the i-th term.
My problem is how to show: $$\lim_{r\to 0}\lim_{h\to 0}\int_{B_r(x)}\frac{g(x+he_i-y)-g(x-y)}{h}\,f(y)dy=0$$
Attempt $1$: I note that for each fixed $h\gt0$, $$\lim_{r\to 0}\int_{B_r(x)}\frac{g(x+he_i-y)-g(x-y)}{h}\,f(y)dy=0.$$ So if we can exchange the limits, then we are done. But I don't know if we can exchange.
Attempt $2$: Also I note that, if we can exchange the limit and integral, i.e. $\displaystyle\lim_{h\to0}$ and $\int$, then we are also done. But I can't find a dominated function to use the Lebesgue Dominated Convergence Theorem.
I spent one day on thinking this estimate, but I can't figure it out. I am frustrated now...
Any help will be appreciated. Thanks!
Note that $f$ is in $L^{\infty}(\mathbb R^n)$ too, so$$\left|\frac{g(x+he_i-y)-g(x-y)}{h}\,f(y)\right|\leq\|f\|_{L^\infty(\mathbb R^n)}\left|\frac{g(x+he_i-y)-g(x-y)}{h}\right|\\\leq 2\|f\|_{L^\infty(\mathbb R^n)}\left|\frac{\partial g}{\partial x_i}(x-y)\right|$$ uniformly for $h$ small enough, so $2\|f\|_{L^\infty(\mathbb R^n)}\left|\frac{\partial g}{\partial x_i}\right|$ would be a dominating function (which is clearly integrable) for the case $x\neq 0$. Uniformly, because we can make the ball $B_r(x)$ compact (for $r$ small enough) and take the closure and note that $\left|\frac{g(x+he_i-y)-g(x-y)}{h}\right|$ is decreasing as $h$ approaches $0$, so uniform convergence follows from Dini's theorem and thus the inequality holds uniformly for $h$ small enough.
For the case $x=0$ (actually for all $x$), the monotone convergence theorem can be used, since $\left|\frac{g(x+he_i-y)-g(x-y)}{h}\right|$ is decreasing as $h$ approaches $0$.