2
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ What did you try? $\endgroup$ Sep 4, 2012 at 12:23
  • $\begingroup$ Why doesn't the same proof work? $\endgroup$
    – Deane Yang
    Sep 4, 2012 at 14:16

1 Answer 1

4
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ MUCH more complicated than the question asked! $\endgroup$ Sep 4, 2012 at 15:20
  • 1
    $\begingroup$ 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? $\endgroup$
    – M-S
    Sep 4, 2012 at 16:27
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 4, 2012 at 18:33
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 5, 2012 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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