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Might be simple, but i don't get it. Why is the integral in the last line approximately equal to $n(\varphi(\frac{-1}{2n}) - \varphi(\frac{1}{2n}))$?

enter image description here

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Please mention the name of the book the next time you ask about something you see in a book... – J. M. Jul 16 '12 at 13:48
it is taken from Robert Stichartz, A Guide to Distribution Theory and Fourier Transforms, page 17. – Stefan Jul 16 '12 at 14:14

(Some pages of Strichartz's book are available here).

There is probably a typo at the left of the approximation. $\phi_n$ is supposed to have the limit $\delta$ as $n \to \infty$ so that $\int \phi_n(x)\phi(x)\;dx$ should go to $\int \delta(x)\phi(x)\;dx=\phi(0)$ and not to $-\phi'(0)$ as $n \to \infty$.

A derivative is clearly missing somewhere in the integral (I should say on the first term from the sign obtained) and the approximation should read in agreement with the last picture : $$\int \phi'_n(x)\phi(x)\;dx\approx n\left(\phi\left(\frac{-1}{2n}\right)-\phi\left(\frac 1{2n}\right)\right)\to -\phi'(0)\ \text{as}\ n\to\infty$$

It seems that the (kind of) 'bump function' $\phi_n$ was chosen such that its derivative looks like $n\phi_n\left(2x+\frac 1n\right)-n\phi_n\left(2x-\frac 1n\right)$ (other half derivative of bump function could look less symmetrical!). The approximation was obtained by replacing the two $\phi_n$ by their limit $\delta$.

Note that the choice of the special function $\phi_1$ (for example) bounded in $(-1,1)$ and verifying $\phi'_1(x)=\phi_1(2x+1)-\phi_1(2x-1)$ (a not easy 'delay differential equation') is not primordial since we replace it by delta functions on both sides anyway. The derivative of $\delta$ is sometimes used to describe an electrical dipole ('unit doublet') and perhaps that the derivation proposed in (1.11) of this file will help you more. You may observe that the two 'bumps' of the derivative shown in Fig 1.6 are not similar to the starting test function, for the rectangular test function you would get directly two delta functions as derivative.

Hoping this clarified a little,

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ah ok, its a typo, but your last paragraph still is mysterious for me, what you mean by "The approximation was obtained by replacing the two $\phi_n$ by their limit", in the first approximation there is still the parameter $n$ so no limit process was made? – Stefan Jul 16 '12 at 17:51
@Stefan: yes and that is why it is an 'approximation'! The approximation is to replace $\phi_n$ by $\delta$ for $n$ 'large enough' fixed. The limit as $n\to \infty$ will give you $-\phi'(0)$ (if you want remove the 'as $n\to \infty$' from my sentence). – Raymond Manzoni Jul 16 '12 at 17:57
and how you came to $n\phi_n(x+\frac{1}{2n})-n\phi_n(x-\frac{1}{2n})$, and what about the $x$, at what point you make the approximation? – Stefan Jul 16 '12 at 18:18
@Stefan: it was 'deduced' from the last picture. The author probably choose $\phi(n)$ so that this worked (that's why the initial function is 'rounded' the same way at bottom and at top) (the $n$ factor is a more general feature of derivatives of functions with limit $\delta$). – Raymond Manzoni Jul 16 '12 at 18:24
@Stefan: concerning $x$ there is a scaling error in my $\phi_n$ formula (the 'bumps' of the derivative should be 2 times less large than the original...). I'll correct that after some verifications... – Raymond Manzoni Jul 16 '12 at 19:42

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