Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
1  
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
add comment

1 Answer 1

up vote 3 down vote accepted

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 $f^{(n)}(0)$.

share|improve this answer
1  
MUCH more complicated than the question asked! –  Gerald Edgar Sep 4 '12 at 15:20
    
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
1  
It seems to be copied wrong. But yes, once you get the formula for scalar values, the same one holds when the outer function has vector values. But the thing to do is provide the proof. Use the scalar theorem to prove the vector theorem. –  Gerald Edgar Sep 4 '12 at 18:33
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.