Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

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 $\delta'(1)=0$.

share|improve this answer

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

share|improve this answer
    
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 $n$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.