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Notation: If $f\colon\mathbb{R}\to\mathbb{R}$ is continuous, let us denote $If\colon\mathbb{R}\to\mathbb{R}$ its indefinite integral from $0$, i.e., $(If)(x) = \int_0^x f(t)\,dt$, and iteratively $I^{k+1}f = I(I^k f)$.

Remark: If $f$ is a continuous function with support contained in the open interval $]0,1[$ then $If$ has support contained in $]0,1[$ iff $(If)(1) = 0$.

Main question: Does there exist a $C^\infty$ function $f$ with support contained in the open interval $]0,1[$ such that $I^k f$ has support contained in $]0,1[$ for every $k\geq 0$, or, equivalently, $(I^k f)(1) = 0$ for all $k\geq 0$?

Equivalent formulation: Does there exists a sequence $(f_k)_{k\in\mathbb{Z}}$ of $C^\infty$ functions each with support contained in the open interval $]0,1[$, such that $f_{k-1}$ is the derivative of $f_k$?

Weaker question: Does there at least exist a continuous function $f$ with the properties demanded in the main question?

Stronger question: Does there exist a $C^\infty$ function $f$ with compact support, whose Fourier transform vanishes identically on a nontrivial interval?

(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in $]0,1[$, multiply it appropriately so its Fourier transform vanishes in a neighborhood of $0$, and observe that the Fourier transform of $I^k f$ is, up to constants, $\xi^k$ times that of $f$.)

Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

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    $\begingroup$ I think the stronger assertion should be false by Paley-Wiener, $\endgroup$ – Paul K Apr 4 at 12:39
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    $\begingroup$ @PaulK: Something along the lines of “if a function has compact support, its Fourier transform is analytic, so it cannot vanish identically on a nontrivial interval without vanishing identically”? Indeed, this seems to work. It might even answer the original question by imposing all derivatives of the Fourier transform to vanish at the origin… $\endgroup$ – Gro-Tsen Apr 4 at 12:54
  • $\begingroup$ You might be right that this could also works for your original question! $\endgroup$ – Paul K Apr 4 at 12:57
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I think it is not possible even for $f$ only measurable and bounded. Indeed, $$(I^kf)(x)=\int_{x_0=0<x_1<...<x_k=x}f(x_1)dx_1...dx_k=\int_0^xf(x_1)\left(\int_{x_1<...<x_k=x}dx_2...dx_k\right)dx_1$$ Now since $$\int_{a<y_1...<y_k<b}dy_1...dy_k=(b-a)^k\int_{0<y_1...<y_k<1}dy_1...dy_k=\frac{(b-a)^k}{k!}$$ we get : $$(I^kf)(1)=\int_0^1f(y)\frac{(1-y)^{k-1}}{(k-1)!}dy.$$ If this was vanishing for any $k$ then for all polynomial $P$ we would get : $$\int_0^1f(y)P(y)dy=0$$ and therefore $f$ is $0$ almost everywhere.

The hypothesis that $f$ is bounded is probably not necessary (in fact if $f$ is locally integrable, $I^1f$ is bounded continuous and we can apply the previous argument to $I^1f$.)

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