I am trying to show that

$$\int_0^\infty f_1(x) h_1(x) dx = \int_0^\infty f_2 (x) h_1(x) dx \\ \vdots\\ \int_0^\infty f_1(x) h_n(x) dx = \int_0^\infty f_2 (x) h_n(x) dx$$

if and only if $f_1(x) = f_2(x)$, $x \in \mathbb{R}^+$. I know that:

  • $h_j(x)$ is a real continuous function such that $h_j(x) > 0$ on the interval $(0, a_j)$, where $a_j > b > 0$ for $j = 1,...,n$.
  • $f_1(x)$ and $f_2(x)$ are real continuous functions such that $f_i(x) > 0$.
  • $f_i(x) h(x) \leq M \exp(-c x), \, i =1,2$ for some $M,\,c \in \mathbb{R}$.
  • $0 < \int_0^\infty f_i(x) h_j(x) dx < \infty, \, i = 1,2, \, j = 1,...,n$.
  • $0 < \int_0^\infty f_i(x) dx < \infty, \, i = 1,2$.

    Would anyone know if this result can possibly be shown? Thank you very much!


This seems hopeless without more constraints. Set $a_j = 2\pi$ for all $j$ and you've transformed this into a problem about Fourier transforms. We know the set of continuous functions cannot be spanned by a finite basis -- that is we can orthonormalize the $h_j$ and find entire subspaces of functions whose projections onto the $h_j$ all agree, but (would) disagree on extensions of this basis.

  • $\begingroup$ Hi Eric, Thanks a lot! That's a good point. Any idea what constraints on the functions $f_1(x)$ and $f_2(x)$ should be added to prove this result? $\endgroup$ – user304347 Mar 28 '16 at 4:04
  • $\begingroup$ @user304347 : An "unhelpful" answer is $f_1, f_2 \in \mathrm{span}\{h_1, h_2, \dots, h_n\}$. $\endgroup$ – Eric Towers Mar 29 '16 at 7:14

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