Let $f: \mathbb{R} \rightarrow X$, where $X$ is a real Banach space, $g: \mathbb{R}\rightarrow \mathbb{R}$.


Is it then true the Faa di Bruno's formula on $(f\circ g)^{(n})(x)$ ? 



Faa di Bruno's formula for vector valued functions
For $g:B\_1\rightarrow B\_2$, $f:B\_2\rightarrow B\_3$, $B\_j$ Banach spaces , $g,f$ smooth,
we have for $n\ge 1$
$$
\frac{(f\circ g)^{(n)}}{n!}=\sum\_{n\_1+\dots+n\_r=n\atop
r\ge 1, n\_j\ge 1} \frac{f^{(r)}\circ g}{r!}\frac{g^{(n\_1)}}{n\_1!}\dots
\frac{g^{(n\_r)}}{n\_r!},\tag{FB}
$$
where the symmetric $n$-multilinear form $(f\circ g)^{(n)}$ is characterized by
$$T\in B\_1,\quad
\frac{(f\circ g)^{(n)}T^n}{n!}=\sum\_{n\_1+\dots+n\_r=n\atop
r\ge 1, n\_j\ge 1} \frac{(f^{(r)}\circ g)}{r!}\Bigl(\frac{g^{(n\_1)}T^{n\_1}}{n\_1!},\dots,
\frac{g^{(n\_r)}T^{n\_r}}{n\_r!}\Bigr).
$$
Note that a symmetric $n$-multilinear form is completely determined by its values on $T^n$: this follows from the polarization formula
$$
T\_1T\_2\dots T\_k=\frac{1}{2^k k!}\sum\_{\epsilon\_j=\pm 1} \epsilon\_1\dots\epsilon\_k
(\epsilon\_1T\_1
+\dots+\epsilon\_kT\_k)^k.
$$
The proof of (FB) is not different from the 1D proof: assuming $g(0)=0$
$$
(f\circ g)(x)=\sum\_r\frac{(f^{(r)}\circ g)(0)}{r!} g(x)^r=\sum\_r\frac{(f^{(r)}\circ g)}{r!} \bigl(\sum\_n \frac{g^{(n)}(0)x^{n}}{n!}\bigr)^r
$$
$$
=\sum\_{n\_1+\dots+n\_r=n}\frac{(f^{(r)}\circ g)}{r!} \frac{g^{(n\_1)}(0)x^{n\_1}}{n\_1!}\dots
\frac{g^{(n\_r)}(0)x^{n\_r}}{n\_r!},
$$
providing the sought expression for $f^{(n)}(0)$.


