Math-3739 · mathematics · Analysis

Math-3739

Research Problems in Function Theory — Problem 7.61

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Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

For $n\ge1$ let $\mathcal E=\mathcal O(\mathbb C^n)$ be all entire functions. For a multi-index $\alpha=(\alpha_1,\ldots,\alpha_n)\in\{0,1,\ldots\}^n$, put $z^\alpha=\prod z_j^{\alpha_j}$ and $D^\alpha=\prod(\partial/\partial z_j)^{\alpha_j}$. If $P(z)=\sum_\alpha p_\alpha z^\alpha$ is a polynomial, define $P(D)=\sum_\alpha p_\alpha D^\alpha$ and $T_{P,Q}f=P(D)(Qf)$. Entire exponential type means $|f(z)|\le A e^{B|z|}$ for some finite positive $A,B$.Define $P^*(z)=\overline{P(\bar z)}=\sum_\alpha\overline{p_\alpha}z^\alpha$. The zero polynomial makes $T_{P^*,P}$ the zero operator; the substantive bijection question is for nonzero $P$, including nonzero constants.

Statement recorded in the supplied source.
The original asks whether $T_{P^*,P}:\mathcal E\to\mathcal E$ is bijective for every polynomial $P$. For $P=0$ the answer is negative, since the operator is zero. For every nonzero complex polynomial $P$, is this operator bijective? Also consider the source's proposed sufficient route: for every multi-index $\alpha$, does \[P^*(D)(Pf)=z^\alpha \] admit an entire solution of exponential type? The source records the homogeneous case as known and notes the injectivity claim separately; neither replaces the full nonhomogeneous bijection question. [S2]

Short English statement

Determine bijectivity of polynomial multiplication followed by the differential operator with conjugate coefficients, and whether exponential-type solutions exist for every monomial right-hand side.

Sources

Catalogue entry

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