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Consider the delta dirac distribution $\delta (\varphi) = \varphi (0), \varphi \in \mathcal{S}(\mathbb{R}^n)$ (the Schwartz space). I know that $\delta ^{'} (\varphi) = - {\varphi }^{'} (0)$. How can I prove $\delta^{'}$ is not given by a measure, that is , doesn't exists a measure $\mu$ such that

$$\delta^{'} (\varphi) = \displaystyle\int_{R^n} \varphi (x)\,d \mu (x) $$

I have no idea how to proceed. Someone can give me a hint ?

Thanks in advance

Xiang Yu
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math student
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1 Answers1

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Choose a test function $\varphi$ with $\frac{\partial \varphi}{\partial x_i}(0) = 1$. For $n \in \mathbb{Z}^+$, let

$$f_n(x) = \frac1n\varphi\left(nx\right).$$

For every measure $\mu$, we have

$$\lim_{n\to\infty} \int f_n(x)\,d\mu = 0$$

by the dominated convergence theorem. But

$$\frac{\partial\delta}{\partial x_i}[f_n] = - \frac{\partial f_n}{\partial x_i}(0) = -1$$

for all $n$.

Daniel Fischer
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  • Is trivial to give a explicit form to your test function ? – math student Nov 06 '13 at 00:05
  • You could take $\varphi(x) = x_i\cdot \psi(x)$ where $\psi$ is a cutoff function, say $\psi(x) = \psi_0(\lVert x\rVert)$ and $\psi_0\colon \mathbb{R}\to\mathbb{R}$ is a function with $\psi_0(x) = 1$ for $x \leqslant 1$, $\psi_0(x) = 0$ for $x \geqslant 2$ constructed using $\exp \left(-\frac{1}{x^2}\right)$. I wouldn't call it trivial, but you can construct an explicit example. – Daniel Fischer Nov 06 '13 at 00:12
  • I understand that the sequence $f_n$ is dominated by $\frac{1}{n} || \varphi||_{\infty}$. But I have dominate by a Integrable function . Which integrable functions do you take ? (my english is terrible, sorry) – math student Nov 06 '13 at 00:23
  • Take a cutoff function $\eta$ that is $\equiv 1$ on a ball containing the support of $\varphi$. Since $\operatorname{supp} f_n = \frac1n\operatorname{supp}\varphi$, the sequence $f_n$ is dominated by $\lVert\varphi\rVert_\infty\cdot \eta$. – Daniel Fischer Nov 06 '13 at 00:26
  • I understand your construction of the test function , your last coment and the solution for my problem . Thank you !!! . Nice solution for my problem! – math student Nov 06 '13 at 01:02
  • I think you need the measure to be locally finite, that is $\mu(K)<\infty$ for every compact set $K$. – Xiang Yu Feb 29 '16 at 08:31
  • @XiangYu Yes (well, for the argument it suffices that $\lvert \mu\rvert(B) < \infty$ for some ball containing the support of $\varphi$, it doesn't matter how $\mu$ behaves outside $B$). But we're looking at tempered distributions, and all measures that define tempered distributions are locally finite. – Daniel Fischer Feb 29 '16 at 11:44