# Function with compact support whose iterated antiderivatives also have compact support

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.

• I think the stronger assertion should be false by Paley-Wiener, – Paul K Apr 4 at 12:39
• @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… – Gro-Tsen Apr 4 at 12:54
• You might be right that this could also works for your original question! – Paul K Apr 4 at 12:57

I think it is not possible even for $$f$$ only measurable and bounded. Indeed, $$(I^kf)(x)=\int_{x_0=0 Now since $$\int_{a 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 $$I^1f$$.)