Given a radial symmetric function $f$, the estimate of |$\Delta ^ {m/2}f$| in $R^{2m}$ by induction

Question

This question might be a little strange; my order of Laplacian is related to the dimension of the space.

Actually, I’m reading a result which is obtained by induction; it is the absolute value of different order laplacian $|\Delta ^ {m/2}f|$ on $R^{2m}$.

The function we are dealing with is a radial function $f(r)$; we know that (since the dimension is $2m$.) $$\Delta f(r)=f^{\prime \prime}+(2m-1)f^{\prime}/r$$ and $$\nabla f=f^{\prime}(r) x/|x|.$$

The author gave a formula of |$\Delta ^ {m/2}f$| on $2m$ dimension; denote it as $g(r,m)$. I have no idea about how could we start the induction since the dimension is also changing.

We see that $|\nabla f|=|f^{\prime}(r)|$, my question is: starting from |$\Delta ^ {m/2}f$|=$g(r,m)$ on $2m$ dimension, then how to calculate $|\Delta ^ {(m+1)/2}f|$ on $2m+2$ dimension based on $g(r,m)$? So that I can verify if this in the form of $g(r,m+1)$·

Details: $f=w_\sigma(r):=\log \left(\frac{2 \sigma}{1+\sigma^2 r^2}\right)$ and the fomula $g(r,m)$ is $$ \left|\Delta^{\frac{m}{2}} w_\sigma\right|=2^m(m-1) ! \sigma^m \frac{\sigma^m r^m+p(\sigma r)}{\left(1+\sigma^2 r^2\right)^m} $$with $\operatorname{deg} p \leq m-1$. I don't know how to use induction because when computing $\Delta^2 f$ on dim=8, we can't use the result of $\Delta^{3/2}f$ since it's on dim=6. The background is in (14) of Martinazzi and Petrache.

Answer 1

The formula $$\left|\Delta^{\frac{m}{2}} \log \left(\frac{2 \sigma}{1+\sigma^2 r^2}\right)\right|=2^m(m-1) ! \sigma^m \frac{\sigma^m r^m+{\cal O}(r^{m-1})}{\left(1+\sigma^2 r^2\right)^m}$$ can be derived by recursion as follows. Note that this is equivalent to $$\left|\Delta^{\frac{m}{2}} \log \left(\frac{1}{1+r^2}\right)\right|=2^m(m-1) ! \frac{ r^m+{\cal O}(r^{m-1})}{\left(1+ r^2\right)^m}.$$ Consider first even $m$. The recursion starts at $$\Delta \log \left(\frac{1}{1+r^2}\right)=-4(m-1)\frac{r^2+{\cal O}(r^0)}{(1+r^2)^2}.$$ The step from $p$ to $p+1$ is $$\Delta\frac{r^{2p}}{(1+r^2)^{2p}}=C_{p,m}\frac{r^{2(p+1)}+{\cal O}(r^{2p})}{(1+r^2)^{2(p+1)}},$$ with $C_{p,m}=4p(1 - m + p)$. We thus arrive for $m$ even at $$\Delta^{\frac{m}{2}} \log \left(\frac{1}{1+r^2}\right)=-4(m-1)\left(\prod_{p=1}^{m/2-1}C_{p,m}\right)\frac{ r^m+{\cal O}(r^{m-1})}{\left(1+ r^2\right)^m}=$$ $$(-1)^{m/2} 2^m(m-1) ! \frac{ r^m+{\cal O}(r^{m-1})}{\left(1+ r^2\right)^m}.$$

For the case of $m=2s+1$ odd, I first calculate $$\Delta^{s} \log \left(\frac{1}{1+r^2}\right)=(-1)^{s} 2^{1+2s} (2s-1)! \frac{ r^{2s}+{\cal O}(r^{2s-1})}{\left(1+ r^2\right)^{2s}},$$ followed by one more derivative, $$\frac{d}{dr}\Delta^{(m-1)/2} \log \left(\frac{1}{1+r^2}\right)=(-1)^{(m+1)/2} 2^{m} (m-1)! \frac{ r^{m}+{\cal O}(r^{m-1})}{\left(1+ r^2\right)^{m}}.$$ And we're done.