product rule in banach algebra

**Sogan** · June 22nd 2011, 02:35 AM

Hello,

i asked myself, whether the product rule is correct for functions which have values in a banach algebra.
I think it is true, but i'm not so sure. Do you know it? Or can you check my "proof" for the product rule please?:
Let f,g:V->W be two differentiable maps, whereas W is some banachalgebra. Then we have:

$(f*g)(x+h)=(f(x+h))*(g(x+h))=( f(x)+df(x)[h]+o(\|h\|) )*( g(x)+dg(x)[h]+o(\|h\|) )<br /> =...=f(x)*g(x)+f(x)*dg(x)[h]+df(x)[h]*g(x)+o(\|h\|)$

therefore we have the derivative:
f(x)*dg(x)[h]+df(x)[h]*g(x) and the product rule is proved.

Of course we use in the calculation above, that the multiplication is distributive and also that $\|u*v\|<=\|u\|*\|v\|$

Thank you a lot!

Regards

**Mauritzvdworm** · June 22nd 2011, 11:14 PM

The product rule is defined for operator valued functions, and will then also be defined for functions whose values are in Banach algebras. It al depends on the definition of the derivative. The following definition I find particularly useful

$0=\lim_{h\rightarrow 0}\frac{\|f(x+h)-f(x)-f'(x)h \|}{\|h\|}$

where $f'(x)$ is then the derivative of $f$ at $x$ . The situation can be simplified slightly when we consider functions of the form

$f:\mathbb{R}\rightarrow \mathcal{B}$

with $\mathval{B}$ a Banach algebra. The above definition then reduces to

$0=\lim_{h\rightarrow 0}\|\frac{f(x+h)-f(x)}{h}-f'(x) \|$ .

Here the resemblance with the normal definition of the derivative is clear. So let us now consider the product rule

$\lim_{h\rightarrow 0}\frac{\|f(x+h)g(x+h)-f(x)g(x)-f'(x)g(x)h-f(x)g'(x)h \|}{\|h\|}$

$=\lim_{h\rightarrow 0}\frac{\|\left(f(x+h)-f(x)\right)g(x+h)+f(x)\left(g(x+h)-g(x)\right)-f'(x)g(x)h-f(x)g'(x)h \|}{\|h\|}$

$\leq\lim_{h\rightarrow 0}\frac{1}{\|h\|}\left(\|f(x+h)-f(x)-f'(x)h\|\|g(x+h)\|+\|g(x+h)-g(x)-g'(x)h\|\|f(x)\|+\|f'(x)\left(g(x+h)-g(x)\right)h\|\right)$

$\leq\lim_{h\rightarrow 0}\frac{1}{\|h\|}\|f'(x)\left(g(x+h)-g(x)\right)\|\|h\|$

$=\lim_{h\rightarrow 0}\|f'(x)\left(g(x+h)-g(x)\right)\|$

$=0$ .

Just check my algebra, but I think this should clear it up.

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