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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?

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    $\begingroup$ For each $x$, the Taylor series has infinite radius of convergence and you can extend $f$ to a function on $\mathbb R \times \mathbb C$ which is holomorphic in the second variable. If this function is differentiable w.r.t. $x$ you can use Cauchy's formula $$f(x,z) = \frac{1}{2\pi i} \int_S \frac{f(x,\xi)}{\xi-z} d\xi$$ (where $S$ is a circle around $0$ with radius $>|z|$) to differentiate with respect to $x$ under the integral. But I do not think that the extended function is always differentiable w.r.t. $x$. Thm 1.2.6 of Hörmander (Anal. part. Diff. Op. I) ponits in this direction. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 20, 2012 at 13:27

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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.

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  • $\begingroup$ Doesn't one need something about the function $x\mapsto f(x,y)$ for $y\ in \mathbb C$ (or at least some open complex sets) to differentiate under the integral? Moreover, is it obvious that the analyticity of the lhs for all test functions $\phi$ implies the analyticity of $\frac{\partial f}{\partial x}$? Finally, are you sure about the reference to 2.3.1 ? (In my edition this is states that $\mathscr E'$ is the dual of $\mathscr C^\infty$.) $\endgroup$
    – Jochen Wengenroth
    Commented Aug 24, 2012 at 11:47
  • $\begingroup$ @Jochen I need indeed that for each $y$, $f(⋅,y)$ is a distribution on $\mathbb R^m_x$, then I can differentiate with the above rule. Using the $\overline{\partial}$ operator as suggested in my answer, you can prove a weak analyticity result. I doubt that much more could be proven: think about a smooth non-analytic function $f$ depending only on $x$. Sorry for the wrong reference for the variation on this topic: it is Theorem 2.1.3. – Bazin 4 hours ago $\endgroup$
    – Bazin
    Commented Aug 24, 2012 at 20:40
  • $\begingroup$ Just to be sure: So under mildly stronger assumptions than the ones I stated one can indeed conclude analyticity of $\frac{\partial f}{\partial x^i}$? Is there an easy example where one sees that my conditions don't suffice? $\endgroup$
    – Florian
    Commented Sep 2, 2012 at 12:51

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