smooth families of analytic functions - MathOverflow

smooth families of analytic functions

2012-06-20T11:54:31Z

My question is essentially whether taking partial derivatives of a smooth family of analytic functions yields again a smooth family of analytic functions.

The precise question is the following:

Let $f: \mathbb{R}^m\oplus \mathbb{R}^n \to \mathbb{R}$ be a smooth function in two variables $x\in \mathbb{R}^m$ and $y\in\mathbb{R}^n$, such that for each fixed x, the function

$$ y \mapsto f(x.y)$$

is globally analytic, i.e. its Taylor expansion around y=0 converges to f(x,y) for all y.

My question is: do the partial derivatives with respect to one of the x-coordinates $\frac{\partial f}{\partial x^i}$ have the same property, i.e. is $\frac{\partial f}{\partial x^i}$ globally analytic in the above sense for each fixed x?

And what about the case when the Taylor expansion around y=0 converges to f(x,y) only if y lies in some open interval of 0?

Answer by Bazin for smooth families of analytic functions

2012-08-23T16:32:54Z

For $\phi\in C^\infty_c(\mathbb R^m)$ and duality products, we have $$ \langle\frac{\partial f}{\partial x}(x,y),\phi(x)\rangle_{\mathscr D'(\mathbb R^m),\mathscr D(\mathbb R^m)}=- \langle f(x,y),\phi'(x)\rangle. $$ The rhs is an analytic function of $y$, so the same holds for the lhs. There are several variations on this topic, and one example is Theorem 2.3.1 in the first volume of L. Hörmander ALPDO. Here to prove analyticity, the simplest way is certainly to extend the function $f$ to a bit of complex flesh in the $y$ variable (which can depend on the support of $\phi$) and apply the $\overline\partial$ operator in the complex variable $y$. This type of result amounts to differentiating under the integral sign, where that integral is in fact replaced by a bracket of duality.