Riemann Integrable Dirac Function does not Exist

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\delta : \R \to \R$ be a real function.

Let $\phi : \R \to \R$ be a smooth function vanishing outside $\closedint a b$.

Let $a \in \R_{> 0}$ be a real number.

Suppose $\delta$ is Riemann integrable on $\closedint {-a} a$.

Suppose for every $\phi$ we have that:

$\ds \int_{-a}^a \map \delta x \map \phi x = \map \phi 0$


Then $\delta$ does not exist.


Proof

Aiming for a contradiction, suppose $\delta$ exists.

Let $\phi$ be a test function of the following form:

$\map \phi x = \begin {cases}

\map \exp {\dfrac 1 {x^2 - 1} } & : \size x < 1 \\ 0 & : \size x \ge 1 \end {cases}$

Let $n \in \N$ be a natural number.

Let $\phi_n : \R \to \R$ be a real function such that:

$\forall n \in N : \forall x \in \R : \map {\phi_n} x := \map \phi {n x}$

Then $\map \phi {nx}$ is smooth, vanishes outside $\ds \closedint {- \frac 1 n} {\frac 1 n}$ and:

$\forall x \in \R : 0 \le \map \phi {nx} \le 1$

Hence:

\(\ds \frac 1 e\) \(=\) \(\ds \map {\phi_n} 0\)
\(\ds \) \(=\) \(\ds \int_{-\frac 1 n}^{\frac 1 n} \map \delta x \map \phi {nx} \rd x\)
\(\ds \) \(\le\) \(\ds \int_{-\frac 1 n}^{\frac 1 n} \paren {\sup_{x \mathop \in \closedint {-\frac 1 n} {\frac 1 n} } \map \delta x} \cdot 1 \rd x\) Riemann Integrable Function is Bounded
\(\ds \) \(=\) \(\ds \paren {\sup_{x \mathop \in \closedint {-\frac 1 n} {\frac 1 n} } \map \delta x} \frac 2 n\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 e\) \(\le\) \(\ds \paren {\sup_{x \mathop \in \closedint {-1} 1} \map \delta x} \lim_{n \mathop \to \infty} \frac 2 n\) taking the limit $n \to \infty$ of both sides
\(\ds \) \(=\) \(\ds 0\)

This is a contradiction.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Riemann_Integrable_Dirac_Function_does_not_Exist&oldid=536951"