In the book of Tao's nonlinear dispersive equations Exercise1.1, the author want us to prove $$\|\partial^{m}_{t}u(0)\|_{\mathcal{D}}\leq K^{m+1}m!$$ use the equation $\partial_{t}u=F(u(t))$ where $F:\mathcal{D}\to\mathcal{D}$ is analytic and $\mathcal{D}$ denote a(real or complex) finite dimensional vector space and we suppose initial data $t_{0}=0,u(0)=0$.
Let me make some calculation. For simplicity, we assume that $\mathcal{D}=\mathbb{R}^{1}$. Then we have $$\partial_{t}^{2}u=F^{(1)}(u)\partial_{t}u=F^{(1)}(u)F(u),$$ which contain $1=1!$ term$$\partial_{t}^{3}u=F^{(2)}(u)(F(u))^{2}+F^{(1)}(u)F^{(1)}(u)F(u),$$ which contains $2=2!$ terms$$\partial_{t}^{4}u=F^{(3)}(u)(F(u))^{3}+2F^{(2)}(u)F^{(1)}(u)F(u)+2F^{(1)}(u)F^{(2)}(u)(F(u))^{2}+(F^{(1)}(u))^{3}F(u)$$ which contains $6=3!$ terms $$\dots\dots$$
By induction, we know that $\partial^{m}_{t}u$ contain $m!$ terms
Observing the above equalities, and take $t=0$ so $u(0)=u_{0}=0$(this is a condition of exercise), can we get the above inequality? If we have $|F^{(l)}(0)|\leq K$ for all $l\in\mathbb{N}$, then we get Tao's answer, but in general, for analytic function $F$, we have $$|F^{(l)}(0)|\leq C(r)\frac{l!}{r^{l}}$$ where $r$ is some convergence radius value and $C(r)$ is depend on $r$.
If we have the condition that all the derivative of $F$ at $0$ is bounded, that is to say, there exists a $K>0$ large such that $|F^{(n)}(0)|\leq K,\forall n\in\mathbb{N}$, then we can really get the final estimate$|\partial_{t}^{m}u(0)|\leq m! K^{m}$, for example,$\partial^{4}_{t}u(0)$ have $6$ terms, and each term have $4$ entries $F^{(3)}(u)(F(u))^{3}$ has $3$ $F(u)$ and $1$ $F^{(3)}(u)$.
I have searched for some materials, there is "method of majorants" for Cauchy-Kowalevski theorem, but Tao's method is different from others,
Any hints are wellcome!!!!
I have known that the answer. This is the so called "the method of majorant".
According to the calculation in the above question, we know $\partial_{t}^{m}u=P_{m}(F^{m-1}(u),\cdots,F(u))$ where $P_{m}(\cdots)$ is a multivariable polynomial of degree $m$ with positive coefficients. Because $F(u)$ is analytic in $\mathcal{D}$, we take $r>0$ such that $$F(u)=\sum_{l=0}^{\infty}\frac{F^{(l)}(0)}{n!}u^{l}$$ convergence in $B(0,r)$ and there exist a constant $C$ such that $$|\frac{F^{(l)}(0)}{n!}r^{l}|\leq C\Leftrightarrow |F^{(l)}(0)|\leq \frac{Cn!}{r^{l}}$$ We construct $$\frac{Cl!}{r^{l}}=G^{(l)}(0)\geq|F^{(l)}(0)|.$$so $$G(v)=\sum_{l=0}^{\infty}\frac{G^{l}(0)}{l!}v^{l}=\sum_{l=0}^{\infty}C\frac{v^{l}}{r^{l}}=\frac{C}{1-\frac{v}{r}}$$ Then we let $$\frac{d}{dt}v=G(v),v(0)=0$$ We have $$\partial_{t}^{m}v(0)=P_{m}(G^{m-1}(0),\cdots,G(0))\geq P_{m}(F^{m-1}(0),\cdots,F(0))=\partial_{t}^{m}u(0)$$
and $$v(t)=r-\sqrt{r^{2}-2ctr}$$ which is analytic in a neighborhood of $t=0$, that is to say, there exist a $C,r_{0}>0$ such that $|v^{(l)}(0)|\leq C\frac{l!}{r^{l}}$.
Combined with $$|\partial_{t}^{m}u(0)|\leq \partial_{t}^{m}v(0)\leq\frac{m!}{r_{0}^{m}}$$