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Prove that there is no integrable function $\delta:[0,1]\to \mathbb{R}$ with the property that $\int_0^1\delta(x)f(x)\mathrm{d}x=f(0)$ for every continuous function $f:[0,1]\to\mathbb{R}$.

I have no idea how to prove this.

Zhang Yifeng
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  • This would be the actual $\delta$ "function". Lookup any book/notes on distributions. – PierreCarre May 17 '20 at 09:25
  • Maybe you can look at https://math.stackexchange.com/questions/3106973/lebesgue-integrability-of-the-dirac-delta-function?rq=1. – EDX May 17 '20 at 09:41

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For $0<a<1$, define $$f_a(x)= \left\{ \begin{array}{cl}\exp\left(-\dfrac{a^2}{a^2-x^2}\right), & 0\leq x < a\\ 0, & x \ge a \end{array}\right.$$ These are continuous functions on $[0,1]$ and you have that $$ e^{-1} = f_a(0) = \int_0^{1} \delta(x) f_a(x) dx \leq e^{-1} \int_0^a |\delta(x)| dx $$

Since the RHS goes to zero as $a\to 0$, you get the contradiction $e^{-1}\leq 0$.

PierreCarre
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    A function $\delta:[0,1]\to \Bbb R$ can be integrable and yet $\int_0^a|\delta(x)|dx=\infty $ for all $a\in (0,1] . $ So can you show that the constraints of the $\delta$ of the Q don't allow that? – DanielWainfleet May 17 '20 at 11:17
  • @DanielWainfleet If the property indeed holds, we have that $\int_0^1 \delta(x) dx =1$. – PierreCarre May 17 '20 at 13:28
  • @DanielWainfleet What definition of "integrable" are you thinking of? – zhw. May 17 '20 at 16:38
  • Why are you able to say $\int_0^{1} \delta(x), f_a(x),dx \leq e^{-1} \int_0^a |\delta(x)|,dx$? – gen-ℤ ready to perish May 18 '20 at 03:15
  • @zhw. Suppose $g:[0,1]\to \Bbb R$ is continuous on $(0,1]$ with $g(x)>0$ when $1/2n<x<1/(2n-1),$ and $g(x)<0$ when $1/(2n+1)<x< 1/2n,$ for $n\in \Bbb N.$ And $\int_{1/(m+1)}^{1/m}g(x)dx=(-1)^{m+1}/m$ for $m\in \Bbb N.$ And $g(0)=0$... Is $g $ integrable? – DanielWainfleet May 18 '20 at 05:13
  • @DanielWainfleet It's not Lebesgue integrable and it's not Riemann integrable. The improper Riemann/Lebesgue integrals do exist however – zhw. May 18 '20 at 05:24
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    @gen-zreadytoperish When both integrals exist, $\int_I g(x) dx \leq \int_I |g(x)| dx$, so $\int f_a(x) \delta (x) dx \leq \int |f_a(x)| |\delta(x)| dx$. Finally, $f_a(x) = |f_a(x)| \leq f_a(0) = e^{-1}$. – PierreCarre May 18 '20 at 13:09