Radial-derivative expansion of the Euler–Poisson–Darboux transform and its zero-radius limit

Base case. For $k=1$ and $f\in C^0((0,\mathrm{\infty }))$ one has $D_r^0(r^{2k-1}f)=rf={\alpha }_{1,0}{r}^1{f}^{(0)}$ with ${\alpha }_{1,0}=1=(2\cdot 1-1)!!$; under the limit hypothesis, $r^{-1}{D}_r^0(rf)=f(r)\to f(0)$ as $r\downarrow 0$.

Induction hypothesis. Fix $k\ge 1$ and assume that for every $h\in C^{k-1}((0,\mathrm{\infty }))$ there are real constants ${\alpha }_{k,0},\dots ,{\alpha }_{k,k-1}$ such that $D_r^{k-1}(r^{2k-1}h)={\sum }_{j=0}^{k-1}{\alpha }_{k,j}{r}^{j+1}{h}^{(j)}$ and ${\alpha }_{k,0}=(2k-1)!!$.

Shape of the $(k+1)$-st transform. Let $f\in C^k((0,\mathrm{\infty }))$ and put $h:=r^{2}f$, so that $h\in C^k((0,\mathrm{\infty }))\subseteq C^{k-1}((0,\mathrm{\infty }))$ and $r^{2k+1}f=r^{2k-1}h$. The induction hypothesis gives $D_r^{k-1}(r^{2k+1}f)={\sum }_{j=0}^{k-1}{\alpha }_{k,j}{r}^{j+1}{h}^{(j)}$, and the product rule gives $h^{(j)}=r^2{f}^{(j)}+2jr{f}^{(j-1)}+j(j-1)f^{(j-2)}$ for $0\le j\le k-1$, where the last two terms are read as $0$ for $j=0$ and $j=1$.

Applying $D_r$ once to the finitely many resulting terms, and using $D_r(r^{m}v)=m{r}^{m-2}v+r^{m-1}{v}^{\mathrm{\prime }}$, produces a finite sum ${\sum }_{i=0}^k{\beta }_i{r}^{i+1}{f}^{(i)}$ with constants ${\beta }_i$ independent of $f$: the term ${\alpha }_{k,j}{r}^{j+1}\cdot r^2{f}^{(j)}$ contributes $(j+3){\alpha }_{k,j}{r}^{j+1}{f}^{(j)}$ and ${\alpha }_{k,j}{r}^{j+2}{f}^{(j+1)}$; the term $2j{\alpha }_{k,j}{r}^{j+2}{f}^{(j-1)}$ contributes $2j(j+2){\alpha }_{k,j}{r}^j{f}^{(j-1)}$ and $2j{\alpha }_{k,j}{r}^{j+1}{f}^{(j)}$; and the term $j(j-1){\alpha }_{k,j}{r}^{j+1}{f}^{(j-2)}$ contributes $j(j-1)(j+1){\alpha }_{k,j}{r}^{j-1}{f}^{(j-2)}$ and $j(j-1){\alpha }_{k,j}{r}^j{f}^{(j-1)}$. Every displayed monomial $r^m{f}^{(i)}$ has $m=i+1$ with $0\le i\le k$, and negative powers do not occur because the terms with $j=0,1$ have one or both of the last two summands read as zero.

Leading coefficient. Evaluating the identity of the previous step at the constant function $f\equiv 1$, which lies in $C^k$, gives ${\beta }_{0}r=D_r^k(r^{2k+1})=[(2k+1)(2k-1)\cdots 3]r=(2k+1)!!r$: indeed $D_r(r^{2k+1})=(2k+1)r^{2k-1}$, each further application of $D_r$ lowers the exponent by $2$ and multiplies the coefficient by the previous exponent, and $k$ applications leave the exponent $1$. Hence ${\beta }_0=(2k+1)!!$, and setting ${\alpha }_{k+1,i}:={\beta }_i$ for $0\le i\le k$ extends the conclusion of the induction hypothesis from $k$ to $k+1$.

Conclusion. By the base case and the induction step, the expansion $D_r^{k-1}(r^{2k-1}f)={\sum }_{j=0}^{k-1}{\alpha }_{k,j}{r}^{j+1}{f}^{(j)}$ with ${\alpha }_{k,0}=(2k-1)!!$ holds for every $k\ge 1$ and every $f\in C^{k-1}((0,\mathrm{\infty }))$. If $f(0):={\lim }_{r\downarrow 0}f(r)$ is finite and each $f^{(j)}$ with $1\le j\le k-1$ is bounded on $(0,1]$, then dividing by $r$ gives $r^{-1}{D}_r^{k-1}(r^{2k-1}f)(r)={\sum }_{j=0}^{k-1}{\alpha }_{k,j}{r}^j{f}^{(j)}(r)\to {\alpha }_{k,0}f(0)=(2k-1)!!f(0)$ as $r\downarrow 0$: the $j=0$ term converges by the assumed limit, and each term with $j\ge 1$ is bounded by $\mathrm{\mid }{\alpha }_{k,j}\mathrm{\mid }{r}^j{\sup }_{(0,1]}\mathrm{\mid }{f}^{(j)}\mathrm{\mid }\to 0$.