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Let $K_z(x) \in \mathcal D'(\mathbb R)$ be a distribution kernel depending analytically on $z \in O \subset \mathbb C$, i.e. for $f \in \mathcal D(\mathbb R)$ fixed, $$ \int K_z(x) f(x) dx $$ is analytic for $z$ on an open set $O$.

Q: Is this still true if we multiply $K_z(x)$ with an analytic function $a(x,z)$ that is bounded on $O$? I.e., is $$ z \mapsto \int f(x) a(x,z) K_z(x) dx $$ still analytic in $O$?

Attempt:
Showing that the integral satisfies the Cauchy-Riemann equations would be a way to prove that it is analytic, but is it possible to

  1. Exchange integral and derivative $\frac{d}{dz} \int f(x) a(x,z) K_z(x) dx = \int \frac{d}{dz} f(x) a(x,z) K_z(x) dx$?
  2. Argue that the Kernel $K_z(x)$ satisfies the CR-equations and not just $\int f(x) K_z(x) dx$?

Is there another approach?


Update:
While the answer by @reuns provides some inside, it is still missing the crucial point, i.e. exchanging two limit operations (in this case an integral and a derivative). To make this point more obvious, one can also consider $$ (z,z') \mapsto \int f(x) a(x, z') K_z(x) dx $$ which is certainly analytic in $z$. It remains to show that it is analytic in $z'$. This boils down to the question whether distribution are linear in derivatives w.r.t. a paramter of the test function, i.e. $$ \partial_z \int a(x,z) K(x) dx = \int \partial_z a(x,z) K(x) dx $$ This is something I suspect to be true, but am unsure about.

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2 Answers 2

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Take $\phi \in D(\Bbb{R}),\int \phi = 1,\phi_n(x)=n \phi(nx)$, let

$$H(y,z)= \int_\Bbb{R} f(x-y) a(x-y,z) K_z(x)dx$$ $$h_n(z) = \int_\Bbb{R} \phi_n(y)H(y,z) = \int_\Bbb{R}f(y) a(y,z) (\int_\Bbb{R} \phi_n(x-y) K_z(x)dx)dy$$ where $y \to x-y$ was substituted.
$\int_\Bbb{R} \phi_n(x-y) K_z(x)dx$ is continuous in $y$ and analytic in $z$ so by Morera's theorem, considering the sequence of Riemman sums $$h_{n,M}(z)=\frac1M\sum_{m=-\infty}^\infty f(m/M) a(m/M,z) \int_\Bbb{R} \phi_n(m/M-x) K_z(x)dx$$ $h_n=\lim_{M\to \infty} h_{n,M}$ is analytic.

$H$ is continuous in $(y,z)$ so $h_n$ converges locally uniformly as $n\to \infty$.

Edit: not sure of the proof, it is somewhat the key step

Whence by Morera's theorem the limit $$h_\infty(z)=H(0,z)=\int_\Bbb{R} f(x) a(x,z) K_z(x)dx$$ is analytic.

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Here is a proof assuming $K(z,x),a(x,z)$ are analytic in $z$ and $f(x) K(x,z)$ is absolutely integrable in $z,x$. I am aware that the question wants for dsitribution kernel. But i thought this would be useful. Let $$g(z) = \int f(x) a(x,z) K(z,x) dx.$$ Now if $\int_{\gamma} g(z) dz = 0$ then by morera's theorem $g(z)$ is holomorphic. Now $$\int_{\gamma} g(z) dz = \int_{\gamma} \int f(x) a(x,z) K(z,x) dx dz.$$ Assuming the integral can be swapped, since $a(x,z) K(z,x)$ is analytic: $\int_{\gamma} a(x,z) K(z,x) dz = 0$. $$ = \int f(x) \int_{\gamma} a(x,z) K(z,x) dz dx = 0.$$ So the proof is complete if we can swap the two iterated integrals. Since we assume $f(x)K(z,x)$ is absolutely integrable in $x,z$ with $a(x,z)$ bounded uniformly in $x,z$, we can swap the integrals by Fubini's theorem.

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