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Let $\varphi$ be a test function ($\varphi$ is smooth and has compact support - is zero outside some bounded interval).

I know that the following

$$ \lim_{\epsilon \to 0_+} \frac{1}{\pi}\int_{-\frac{R}{\epsilon}}^{\frac{R}{\epsilon}}\frac{\sin(z)}{z}(\varphi(\epsilon z)-\varphi(0)) = 0$$

is easy to prove assuming the Riemann-Lebesgue lemma, because $$\frac{\varphi(x)-\varphi(0)}{x} \in L^1 (\mathbb{R})$$ if and only if is in $L^1$ on supp $\varphi$. But can this be also proved in without using the R-L lemma? Any hints appreciated!

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1 Answer 1

It does not really avoid the use of Riemann Lebesgue lemma (because I will use the ideas to prove it).

Use the substitution $s=t\varepsilon$ and $u=-s$ to be reduced to show that $$\lim_{\varepsilon\to 0}\int_0^R\frac{\varphi(t)-\varphi(-t)}t\cdot \sin\left(\frac t\varepsilon\right)dt=0.$$ Notice that $\frac{\varphi(t)-\varphi(-t)}t=\int_{-1}^1\varphi'(tu)du$. Hence we can switch the integrals then integrate by parts in the inner integral.

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