2

Is there any nontrivial function, say $f\in C_0^\infty([0,\infty))$ such that $$\int_0^\infty f(x)\,dx=\int_0^\infty e^{-x}f(x)\,dx=0?$$ My guess is no (i.e. $f\equiv 0$), but am not really sure. I could show there is no such $f$ within a subset in which $f$ change its sign only once, but for general $f\in C^\infty([0,\infty))$ (of course in this case we will need a further assumption that $f\in L^1$) it is unclear.

Jingeon An-Lacroix
  • 1,635
  • 1
    @SineoftheTime It is the space of infinitely differentiable functions with compact support, with some topology structure (but not normable in a useful sense), but this setting is just to describe some "nice" function with sufficient decay in this context. – Jingeon An-Lacroix Dec 20 '23 at 12:33

1 Answers1

3

$V:=C_0^\infty([0,\infty))$ is an infinite dimensional vector space and $$ \phi: V \to \mathbb{R}, ~ \phi(f)=\int_0^\infty f(x)dx, \quad \psi: V \to \mathbb{R}, ~ \psi(f)=\int_0^\infty e^{-x}f(x)dx $$ are linear functionals. Choose any three linear independent functions $g_1,g_2,g_3 \in V$. Then there is a nontrivial linear combination $f= \sum_{k=1}^3 \alpha_kg_k$ such that $\phi(f)=\psi(f)=0$.

Edit: Note that the vectors $(\phi(g_1),\psi(g_1)),(\phi(g_2),\psi(g_2)),(\phi(g_3),\psi(g_3)) \in \mathbb{R}^2$ are linear dependent, so you find $(\alpha_1,\alpha_2,\alpha_3) \in \mathbb{R}^3\setminus\{(0,0,0)\}$ such that $$ \sum_{k=1}^3 \alpha_k(\phi(g_k),\psi(g_k)) = (0,0). $$.

Gerd
  • 7,034
  • 1
    I edited the answer. It is clearer now? – Gerd Dec 20 '23 at 13:20
  • In a vector space of dimension at least $3$ the intersection of the kernels of two linear functionals is never trivial. – Gerd Dec 20 '23 at 13:28
  • Great, thank you! – Jingeon An-Lacroix Dec 20 '23 at 13:29
  • Now it make sense. For example, you can consider two atoms $f,g$ i.e. $\int f=\int g=0$ with $\int e^{-x}f\neq \int e^{-x}g$ with their support does not overlap (so they are linearly independent), then one can find a linear combination of $f$ and $g$ satisfying the given identities. Thank you for your help, this kinda shows that the approach I was about to take was completely invalid, so I save lots of times :P – Jingeon An-Lacroix Dec 20 '23 at 13:33
  • 1
    Glad I could help. – Gerd Dec 20 '23 at 13:36