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

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    it is taken from Robert Stichartz, A Guide to Distribution Theory and Fourier Transforms, page 17.2012-07-16

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(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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    @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...2012-07-16