# Derivative of delta distribution

I'm reading Reed & Simon's book on Functional Analysis. In the chapter of locally convex spaces they say: "consider the tempered distribution $\delta'(f)=-f'(0)$, which doesn't come from a measure". Why is that true? I've tried to prove that claim but it's been unsuccessful.

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A measure cannot depend on the value of the derivatives of its test functions. –  Giuseppe Negro Feb 12 '13 at 21:24
What if you try the constant function ? –  Damien L Feb 12 '13 at 21:26
I get the idea, but I can't find a rigorous framework to prove it. –  user43014 Feb 12 '13 at 21:27

The support of the $\delta'$ is $\{0\}$; for any $f$ that is $0$ on a neighborhood of $\{0\}$, $\delta'(f)=0$.

If $\delta'$ were described by a measure, $\mu$, then that measure must also be supported on $\{0\}$. Then $$\mu(f)=\int f\,\mathrm{d}\mu\tag{1}$$ and $(1)$ is dependent only on the value of $f$ on $\{0\}$. Both $1+x$ and $1$ are functions which on $\{0\}$ have the same value, but $\delta'(1+x)=1$ and $\delta'(1)=0$.

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Suppose that $\delta'$ is given by a mesure $\mu$. Then we have for any compact $\mathrm K$ $$\int_{\mathrm K} d\mu = 0$$

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That's not really an objection since the measure could be signed. –  Giuseppe Negro Feb 12 '13 at 21:37
The constant function is not in $S$, it doesn't "decay faster than polynomials". –  user43014 Feb 12 '13 at 21:37
@user I edited. –  Damien L Feb 12 '13 at 21:41
This still doesn't work, since the function $\chi_K$ is not smooth. However this is morally correct for sure. –  Giuseppe Negro Feb 12 '13 at 21:54
Ho, you are right ! So we have to work with non negative functions with compact support instead, that would be flat at $0$. –  Damien L Feb 12 '13 at 21:57

According to Riesz's representation theorem, a measure is a continuous linear functional on the Fréchet space of continuous functions with compact support. The functional you have here is not even well-defined on that space.

A possible objection is that the definition should be considered on the subspace of sufficiently regular functions and then extended by density. But the assignment $$f\in C^1(\mathbb{R})\to f'(0)$$ is not continuous with respect to the topology of $C(\mathbb{R})$. For example, the sequence $$f_n(x)=\frac{\sin(nx)}{n}\zeta(x),$$ where $\zeta$ is a smooth cutoff function, is such that $f_n \to 0$ in $C(\mathbb{R})$ (that is, uniformly on compact sets) but $f'_n(0)=1$ for all $n$.

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To continue with your solution, where can I see the continuity of the measure with respect to the functions of compact support? I'm not familiar with that version of the Riesz's representation theorem. –  user43014 Feb 12 '13 at 22:12
@user43014: If $\mu$ is a Radon measure, then the assignment $$f \to \int f\, d\mu$$ is linear and continuous with respect to uniform convergence on compact sets. This is a consequence of the Lebesgue dominated convergence theorem. The Riesz representation theorem says that all (signed) Radon measures arise in that way. So to define a signed measure we only need to give a linear functional on the space of continuous functions and verify that it is continuous with respect to the mentioned notion of convergence. –  Giuseppe Negro Feb 13 '13 at 1:29