Show $f(x)=\frac{1}{|x|^{1/4}}\in H^{1/2}(B)$ by definition.

Question

Let $B=B(0,1)\subset \mathbb{R}^2$ the unit ball. I am trying to prove that $f(x):B\rightarrow \mathbb{R}$, $f(x)=\frac{1}{|x|^{1/4}}\in H^{1/2}(B)$ by definition. This is:

$$\int_{B}\int_{B} \frac{|f(x)-f(y)|^2}{|x-y|^3}dxdy <+\infty $$

Obviuosly the problem here is in the region near $y=x$.

I know that $\int_{B} \frac{1}{|x|^p}<+\infty$ if $p<n=2$. So what I have tried to expand the numerator as a Taylor series:

$$f(x)-f(y)\sim \frac{(x-y)}{|y|^{5/4}}$$

when $x$ is near $y$. So I think the integral will converge if

$$\int_{B}\int_{B} \frac{1}{|x-y||y|^{5/2}}dxdy <+\infty $$

Is it true? Moreover, is there any condition on $\alpha,\beta$ for

$$\int_{B}\int_{B} \frac{1}{|x-y|^{\alpha}|y|^{\beta}}dxdy $$

to converge when $B$ is the unit ball in $\mathbb{R}^n$?

Answer 1

Intuition. Formally, we have $|D^{1/2}f(x)|\sim |x|^{-1/4-1/2}\sim|x|^{-3/4}$. Therefore, the fact that $f\in H^{1/2}(B)$ essentially comes from $|x|^{-3/4}\in L^2(B)$, i.e., $\int_B|x|^{-3/2}\,dx<+\infty$. As we will see in the rigorous proof below, the key point is indeed what we've mentioned.

Rigorous proof. Let $\phi(u)=u^{-\frac14}$ for $u>0$, then $f(x)=\phi(|x|)$. If $0<u<v$, then we have $$\left|\phi(u)-\phi(v)\right|\leq |u-v|\int_u^v|\phi'(t)|\,dt\leq Cu^{-5/4}|u-v|.\tag{1}$$ We show that $$I=\int\int_{x,y\in B, |x|\leq|y|}\frac{|f(x)-f(y)|^2}{|x-y|^3}\,dxdy <+\infty.\tag{$*$}$$ We split the integral $I$ into two parts $I=I_1+I_2$, where $$I_1=\int\int_{x,y\in B, |x|\leq|y|\leq\frac32|x|}\frac{|f(x)-f(y)|^2}{|x-y|^3}\,dxdy,$$ $$I_2=\int\int_{x,y\in B, |y|\geq\frac32|x|}\frac{|f(x)-f(y)|^2}{|x-y|^3}\,dxdy.$$ By $(1)$, for $|x|\leq |y|$ we have $$|f(x)-f(y)|\leq C|x|^{-5/4}||x|-|y||\leq C|x|^{-5/4}|x-y|,$$ hence \begin{align*} I_1&\leq C\int\int_{x,y\in B, |x|\leq|y|\leq\frac32|x|}|x|^{-5/2}|x-y|^{-1}\,dxdy\\ &\leq C\int\int_{|x|\leq1, |x-y|\leq \frac52|x|}|x|^{-5/2}|x-y|^{-1}\,dxdy\\ &=C\int_{|x|\leq 1}|x|^{-5/2}\left(\int_{|x-y|\leq \frac52|x|}|x-y|^{-1}\,dy\right)\,dx\\ &\leq C\int_{|x|\leq 1}|x|^{-3/2}\,dx<+\infty. \end{align*} As for $I_2$, we note that if $|y|\geq \frac32|x|$, then $|f(x)-f(y)|\leq f(x)+f(y)=|x|^{-1/4}+|y|^{-1/4}\leq C|x|^{-1/4}$ and we also have $|x-y|\geq|y|-|x|\geq\frac12|x|$. Hence \begin{align*} I_2&\leq C\int\int_{x,y\in B, |y|\geq\frac32|x|}|x|^{-1/2}|x-y|^{-3}\,dxdy\\ &\leq C\int\int_{|x|\leq1, |y-x|\geq\frac12|x|}|x|^{-1/2}|x-y|^{-3}\,dxdy\\ &=C\int_{|x|\le 1}|x|^{-1/2}\left(\int_{|y-x|\geq\frac12|x|}|x-y|^{-3}\,dy\right)\,dx\\ &\leq C\int_{|x|\leq 1}|x|^{-3/2}\,dx<+\infty. \end{align*}

Therefore, we have shown the validity of $(*)$ and by symmetry we have $f\in H^{1/2}(B)$.

Remark. Here we note that $|f(x)-f(y)|$ can be approximate well by the derivative only when $x$ and $y$ are near to each other; if they are far from each other, we use the $L^\infty$ estimate.

Note. Your wish of $$\int_{B}\int_{B} \frac{1}{|x-y||y|^{5/2}}dxdy <+\infty $$ is NOT correct: We have \begin{align*} \int_{B}\int_{B} \frac{1}{|x-y||y|^{5/2}}dxdy&\geq \int_{|y|\leq 1}\left(\int_{|x-y|\leq1-|y|}|x-y|^{-1}\,dx\right)\,dy\\ &\geq C\int_{|y|\leq 1}(1-|y|)|y|^{-5/2}\,dy=+\infty. \end{align*}

Using similar ideas, we may show that $$\int_{B}\int_{B} \frac{1}{|x-y|^{\alpha}|y|^{\beta}}dxdy<+\infty $$ if and only if $\alpha<2, \beta<2$.