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I am trying to find this limit. Let's conisder $\phi \in C_0(R)$. $$\int^{\infty}_{-\infty} \frac{sin^2(x/ {\epsilon})}{\pi x^2}\phi(x)dx$$ = $$\int^{\infty}_{-\infty} \frac{sin^2(y)}{\pi \epsilon y^2}\phi(\epsilon y)dy=I_{\epsilon}$$. I know that ansewer is $\delta(x)$. Then we have to proof that $lim_{\epsilon \to 0} I_{\epsilon}=\phi(0)$. $$|I_{\epsilon}| \leq \int_{supp \phi(\epsilon y)} \frac{sin^2(y)}{\pi \epsilon y^2}|\phi(\epsilon y)|dy$$. But this $\epsilon$ ruined all my integral estimates, and it is going to infinity. Please, give me some hints.

Sneach hcaens
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  • No, @OliverDiaz, that one isn't squared. But this one is. – md2perpe Jun 01 '20 at 16:41
  • @md2perpe, loot at integral carefully, i have $1/\epsilon$ and I can’t immediately go to the limit to get $\phi(0)$ – Sneach hcaens Jun 01 '20 at 17:08
  • I agree that something seems wrong here. Have you written down $f_\epsilon$ incorrectly from the exercise or is there an error in the exercise? – md2perpe Jun 01 '20 at 18:54
  • my apologies, I just removed my comment. In any case, that should be easier to get since the function you have is related to the Cesaro sums of Fourier transforms, which has an $L_1$ kernel. – Mittens Jun 01 '20 at 19:28
  • @md2perpe, it is absolutely correct from workbook. Moreover, in my exercise in test i had function $f_{\epsilon}(x) = \frac{sin^2(x/ {\epsilon})}{\pi x^2 \epsilon}$ – Sneach hcaens Jun 01 '20 at 21:08
  • Your "exercise in test"? Did you have $\frac{sin^2(x/ {\epsilon})}{\pi x^2 \epsilon}$ in an earlier exercise, and now have $\frac{sin^2(x/ {\epsilon})}{\pi x^2}$ in another exercise? – md2perpe Jun 01 '20 at 21:33
  • I had this example $f_{\epsilon}(x) = \frac{sin^2(x/ {\epsilon})}{\pi x^2 \epsilon}$ in test and didn't solve. Today i found example in workbook and asked about it instead of example in my test, because I know answer. I gues those examplese are same, but this one is simpler – Sneach hcaens Jun 01 '20 at 22:28

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