# Faa di Bruno's formula for vector valued functions

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 $(f\circ g)^{(n})(x)$ ?

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What did you try? –  Gerald Edgar Sep 4 '12 at 12:23
Why doesn't the same proof work? –  Deane Yang Sep 4 '12 at 14:16

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 $f^{(n)}(0)$.
In Schwartz, Analysis, vol.I, is the following formula for scalar functions: $$(f\circ g)^{(n)}(x)=\sum_{k_1+k_2+...+k_m=m} \frac{m!}{k_1! k_2! \ldots k_m! (1!)^{k_1} (2!)^{k_2} \ldots (m!)^{k_m} }$$ $$g^{(k_1+k_2+\ldots +k_m)}(f(a)) (f')^{k_1}(a) (f'')^{k_2}(a) \ldots (f^{(m)})^{k_m}(a).$$ (the same is in Wikipedia) Is this formula true if $f,g$ are as in my question? –  M-S Sep 4 '12 at 16:27
Elaborating Gerald's comment: A way to verify a formula in a Banach space is to use the Hahn-Banach theorem: $x=y$ holds if and only if $\phi(x)=\phi(y)$ for all continuous linear functionals $\phi$. This leads to a reduction to a scalar case and you do not even need to know the scalar proof. –  Jochen Wengenroth Sep 5 '12 at 8:24