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I was reading the a lecture note online about distribution theory and it said: The Dirac delta distribution $\delta \in D'$ is defined as $\delta(\varphi)= \varphi(0) $, and there's no locally integrable function $f$ such that $T_f=\delta$, namely, $\int_{R^n}=f(x)\varphi(x)ds=\varphi(0)$, for all $\varphi \in D$.

I was trying to construct a proof about it but failed... Can anyone help ?

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1 Answer 1

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Suppose there exists $f \in L^1_{\text{loc}}$ such that $T_f = \delta$. Define

$$\varphi_k(x) := \begin{cases} \exp \left(- \frac{1}{1-|kx|^2} \right) & |x| < \frac{1}{k} \\0 & \text{otherwise} \end{cases}.$$

for $k \in \mathbb{N}$. As $\varphi_k \in D$, we get

$$\int \varphi_k(x) f(x) \, dx = T_f(\varphi_k) = \delta(\varphi_k) = \varphi_k(0)= e^{-1}$$

for all $k \in \mathbb{N}$. Hence,

$$e^{-1} = \left| \int \varphi_k(x) f(x) \, dx \right| \leq \underbrace{\|\varphi_k\|_{\infty}}_{\leq e^{-1}} \int_{B(0,1/k)} |f(x)| \,dx.$$

Since $f$ is locally integrable, the dominated convergence theorem yields

$$e^{-1} \leq e^{-1} \inf_{k \in \mathbb{N}} \int_{B(0,1/k)} |f(x)| \, dx =0.$$

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