Function with compact support whose iterated antiderivatives also have compact supportNotation: If f : R → R is continuous, let us denote I f : R → R its indefinite integral from 0, i.e., ( I f ) ( x ) = ∫ 0 x f ( t ) d t, 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 ( I f ) ( 1 ) = 0.Main question: Does there exist a C ∞ 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 ≥ 0, or, equivalently, ( I k f ) ( 1 ) = 0 for all k ≥ 0?Equivalent formulation: Does there exists a sequence ( f k ) k ∈ Z of C ∞ 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 ∞ 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, ξ 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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Function with compact support whose iterated antiderivatives also have compact supportNotation: If   f  :      R    →      R   is continuous, let us denote   I  f  :      R    →      R   its indefinite integral from 0, i.e.,   (  I  f  )  (  x  )  =      ∫    0    x    f  (  t  )    d  t, 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   (  I  f  )  (  1  )  =  0.Main question: Does there exist a       C    ∞   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  ≥  0, or, equivalently,   (      I    k    f  )  (  1  )  =  0 for all   k  ≥  0?Equivalent formulation: Does there exists a sequence   (      f    k        )          k      ∈              Z             of       C    ∞   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    ∞   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,       ξ    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.

Mya Hurst · Accepted Answer

Step 1I think it is not possible even for f only measurable and bounded. Indeed,  (      I    k    f  )  (  x  )  =      ∫                  x        0            =      0      &amp;lt;              x        1            &amp;lt;      .      .      .      &amp;lt;              x        k            =      x        f  (      x    1    )  d      x    1    .  .  .  d      x    k    =      ∫    0    x    f  (      x    1    )      (          ∫                        x          1                &amp;lt;        .        .        .        &amp;lt;                  x          k                =        x              d          x      2        .    .    .    d          x      k        )    d      x    1  Step 2Now since       ∫          a      &amp;lt;              y        1            .      .      .      &amp;lt;              y        k            &amp;lt;      b        d      y    1    .  .  .  d      y    k    =  (  b  −  a      )    k        ∫          0      &amp;lt;              y        1            .      .      .      &amp;lt;              y        k            &amp;lt;      1        d      y    1    .  .  .  d      y    k    =            (      b      −      a              )        k                    k      !      we get:   (      I    k    f  )  (  1  )  =      ∫    0    1    f  (  y  )            (      1      −      y              )                  k          −          1                            (      k      −      1      )      !        d  y  .If this was vanishing for any k then for all polynomial P we would get:      ∫    0    1    f  (  y  )  P  (  y  )  d  y  =  0and therefore f is 0 almost everywhere.The hypothesis that f is bounded is probably not necessary (in fact if f is locally integrable,       I    1    f is bounded continuous and we can apply the previous argument to       I    1    f.)

Function with compact support whose iterated antiderivatives also have compact support Notation: If

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