Simplified estimate for derivatives of $f\circ g$? Let $g:\mathbb R\to \mathbb R$ be a smooth compactly supported function, and let $f:\mathbb R\to \mathbb R$ be a smooth function with $f(0)=0$. Let $\|f\|_m := \|f\|_{C^m}=\sum_{0\le k\le m}\sup_{x\in\mathbb R} |f^{(k)}(x)|$ and let  $[f]_m:= \|  f^{(m)} \|_{0}$. I am able to prove the following estimates-
\begin{align}
[f \circ g]_{m} & \lesssim_m \sum_{i=1}^{m}[f]_{i}\|g\|_{0}^{i-1}[g]_{m}, \tag{$A_1$}\label{A1}\\
[f \circ g]_{m} & \lesssim_m \sum_{i=1}^{m}[f]_{i}[g]_{1}^{(i-1) \frac{m}{m-1}}[g]_{m}^{\frac{m-i}{m-1}},\tag{$A_2$}\label{A2}
\end{align}
($a\lesssim_m b$ means that $a\le Cb$ with a constant $C>0$ depending on $m$.) Their proofs are just Faà di Bruno's formula followed up by either of the interpolation inequalities
\begin{align}
[g]_i \lesssim \|g\|_{0}^{1-i/m} [g]_m^{i/m}, \quad\text{or}\quad [g]_i \lesssim [g]_{1}^{1-\frac{i-1}{m-1}} [g]_m^{\frac{i-1}{m-1}} . \quad(i\ge 1)
\end{align}
Question I'm wondering however if its possible to get the following simpler looking bounds?
\begin{align}
[f \circ g]_{m} & \overset{\color{red} ?}\lesssim_m  [f]_1[g]_{m} +\|f'\|_{m-1}\|g\|_0^{m-1}[g]_m, \tag{$B_1$}\label{B1}\\
[f \circ g]_{m} & \overset{\color{blue} ?}\lesssim_m  [f]_1[g]_m + \|f'\|_{m-1}[g]_1^m.\tag{$B_2$}\label{B2}
\end{align}
These are something like only taking the $i=1$ and $i=m$ terms in the sums of ($A_{1,2}$), but some seminorms are replaced with norms. I came across this in a certain Arxiv paper but I have found a number of other (minor) mistakes so my confidence in exactly this inequality is not the highest.
($B_{1,2}$) are obvious in the simplest case $f(x)=x$, and 'feel' like the above interpolation inequalities (specifically after further applying Young's inequality)
$ [g]_j \lesssim \|g\|_0 + [g]_m$, but I am unable to see how to estimate a general term of $(A_{1,2})$, because it seems $f$ and $g$ cannot be estimated separately without decoupling them, hence getting a worse estimate, like e.g.
$$ [f\circ g]_m \lesssim_m [f]_1 [g]_m + \|f'\|_{m-1}\|g\|^m_{m}.$$
I did manage to get a counterexample to a stronger estimate $[f \circ g]_{m}  \lesssim_m  [f]_1[g]_m + [f]_m[g]_1^m$, indicating that $\|Df\|_{m-1}$ cannot be replaced with $[f]_m$. (Just take $f(x)=x^2$, $g(x)=\lambda g_0(x)$ with $g_0\in C^\infty_c$, then send $\lambda\to\infty$.)
 A: It turns out, the proof is just one more application of the interpolation inequalities.
For B1, we start from A1 and use the interpolation inequality $[\Psi ]_{C^p} \lesssim \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \|D\Psi\|^{(p-1)/(m-1)}_{C^{m-1}}$:
\begin{align*}
     [ \Psi\circ u ]_m 
     &\lesssim    [u]_m \sum_{p=1}^m \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \|D\Psi\|^{(p-1)/(m-1)}_{C^{m-1}} \|u\|_{C^0}^{p-1} 
     \\&=   [u]_m \sum_{p=1}^m \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \left( \|u\|_{C^0}^{m-1}\|D\Psi\|_{C^{m-1}}\right)^{(p-1)/(m-1)} 
     \\&\lesssim_m  [\Psi]_{C^1}[u]_m + \|\nabla\Psi\|_{C^{m-1}}\|u\|_{C^0}^{m-1}[u]_{C^m},
\end{align*}
where we used Young's inequality for products $a^{\theta}b^{1-\theta} \lesssim a + b$ which is valid for all $\theta\in[0,1]$ (including the endpoints $\theta=0,1$), in particular for $\theta = \frac{p-1}{m-1}$ with $1-\theta=\frac{m-p}{m-1}$.
$$ \sum_{p=2}^m [\Psi]_{C^p} \|u\|_{C^0}^{p-1}  \le \max_{k\in[1,m-1]} \|u\|_{C^0}^{k-1} \|\nabla\Psi\|_{C^{m-1}} \lesssim (\|u\|_{C^0} + \|u\|_{C^0}^{m-1})\|\nabla\Psi\|_{C^{m-1}}.$$
For B2, we instead start from A2, and proceed similarly:
\begin{align*}
    [\Psi\circ u ]_{C^m} &\lesssim \sum_{p=1}^m \|D\Psi\|_{C^1}^{(m-p)/(m-1)} \|D\Psi\|^{(p-1)/(m-1)}_{C^{m-1}}  [u]_{C^1}^{(p-1)\frac{m}{m-1}} [u]_{C^m}^{\frac{m-p}{m-1}} 
   \\ &= \sum_{p=1}^m \left([\Psi]_{C^1} [u]_{C^m}\right)^{(m-p)/(m-1)}  \left( \|D\Psi\|_{C^{m-1}} [u]_{C^1}^m\right)^{(p-1)/(m-1)} 
   \\&\lesssim_m [\Psi]_{C^1} [u]_{C^m} + \|D\Psi\|_{C^{m-1}} [u]_{C^1}^m. 
\end{align*}
