Timeline for Smoothness of finite-dimensional functional calculus
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2019 at 15:25 | comment | added | Mizar | The formula, in turn, is easy to prove by induction: we can subtract $f(\lambda_0)$ from each numerator in the left-hand side (thanks to this: math.stackexchange.com/questions/104262/…) and obtain $LHS=\sum_{j=1}^k\int_0^1\frac{f'(t_0\lambda_0+(1-t_0)\lambda_j)}{\prod_{\ell\neq j,\ell>0}(\lambda_j-\lambda_\ell)}$. We are done applying induction with $g:=f'(t_0\lambda_0+(1-t_0)z)$ in place of $f(z)$ and noticing $g^{(k-1)}(z)=(1-t_0)^{k-1}f^{(k)}(t_0\lambda_0+(1-t_0)z)$. | |
| Apr 7, 2019 at 15:21 | comment | added | Mizar | Higher derivative estimates follow from the formula $\sum_{j=0}^k\frac{f(\lambda_j)}{\prod_{\ell\neq j}(\lambda_j-\lambda_\ell)}=\frac{1}{k!|\Delta_k|}\int_{\Delta_k}f^{(k)}(\sum_jt_j\lambda_j)\,dt_0\cdots dt_k$, $\Delta_k$ being the standard simplex $\{t_j\ge 0,\sum t_j=1\}$ (assuming wlog the $\lambda_j$'s are distinct). | |
| Apr 6, 2019 at 18:15 | comment | added | Mizar | Excellent! I see how to generalize it to higher derivatives. I am amazed by the speed of your answer :-) | |
| Apr 6, 2019 at 18:13 | vote | accept | Mizar | ||
| Apr 6, 2019 at 17:25 | history | answered | Dap | CC BY-SA 4.0 |