Mollifiers: Derivative

Asked Jan 09 '15 at 16:10

Active Feb 09 '15 at 22:35

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This thread is meant as lemma for: Semigroups & Generators: Entire Elements: Construction

Given a smooth mollifier: $\varphi\in\mathcal{L}(\mathbb{R}): \varphi'\in\mathcal{L}(\mathbb{R})$

Do the derivatives exist in the sense: $$f\in\mathcal{C}(\mathbb{R}):\quad\int_{-\infty}^{\infty}\frac{1}{h}\left\{\varphi(\hat{x}+h)-\varphi(\hat{x})\right\}f(\tfrac{1}{n}\hat{x})\mathrm{d}\hat{x}\to\int_{-\infty}^{\infty}\varphi'(\hat{x})f(\tfrac{1}{n}\hat{x})\mathrm{d}\hat{x}\quad(\|f\|_\infty<\infty)$$ (Note the unbounded domain of integration.)

edited Jun 12 '20 at 10:38

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asked Jan 09 '15 at 16:10

C-star-W-star

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Do you really need that every such $\phi$ satisfies this or are you just in the need of some $\phi$ with this property? The latter is rather easy to find. – Thomas Jan 09 '15 at 16:15
@Thomas: Preferable for any such mollifier. Specifically for the Gaussian. – C-star-W-star Jan 09 '15 at 16:25
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Hint: What is the value of $\int_{-\infty}^{\infty}{\varphi(x+h)-\varphi(x)}dx$? – Disintegrating By Parts Jan 09 '15 at 23:27
Do you mean: $\int_a^b\varphi(x+h)-\varphi(x)}\mathrm{d}x=\Phi(b+h)-\Phi(b)-\Phi(a+h)+ \Phi(a)$? But then I need to interchange some limits, don't I? – C-star-W-star Jan 09 '15 at 23:57
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Hint: The value is 0! – Disintegrating By Parts Jan 10 '15 at 00:45
@T.A.E.: Do you mean: $\lim_h\lim_c\frac{1}{h}{\Phi(c+h)-\Phi(c)}=\lim_h\frac{1}{h}0=0$? – C-star-W-star Jan 10 '15 at 00:54
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I mean that $\int_{-\infty}^{\infty}\varphi(x+h)dx=\int_{-\infty}^{\infty}\varphi(x)dx$ for all $h$. – Disintegrating By Parts Jan 10 '15 at 00:56
Ah that's also way the subtlty arises with the Gaussian integral that all derivatives vanish. Ok in fact it was then important explicitely involve a test function: $\int_{-\infty}^{\infty}\frac{1}{h}{\varphi(\hat{s}-h)-\varphi(\hat{s})} f(\frac{1}{n}\hat{s})\mathrm{d}\hat{s}$ – C-star-W-star Jan 10 '15 at 01:00
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I don't think this is what you intended, but it's what you have in the statement of the problem. Did you mean to integrate $\varphi$ against another function? – Disintegrating By Parts Jan 10 '15 at 01:03
@T.A.E.: Yes, I edited the problem. – C-star-W-star Jan 10 '15 at 01:10

Mollifiers: Derivative

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