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I recently met this problem from Folland's real analysis second edition involving a specific question on distributions (exercise 19 page 299) which reads as follows:

On $ R $ let $ F_0 = PV(\frac{1}{x}) $ where PV stands for "Principle Value" and defined as follows: $ \langle PV(f),\phi\rangle = \lim_{\epsilon \to 0^+} \int_{|x|>\epsilon} f(x)\phi(x) \, dx $ for all $ \phi \in C_C^\infty $. Also for $ \epsilon > 0 $ we define $ F_\epsilon(x) = x(x^2+\epsilon^2)^{-1} $, $ G_\epsilon^\pm(x)=(x \pm i\epsilon)^{-1} $ and $ S_\epsilon(x) = \operatorname{sgn}(x) e^{-2 \pi \epsilon |x|} $

a. We are to prove $ \lim_{\epsilon \to 0} F_\epsilon=F_0 $ in the weak topology* on $ \mathcal{S}' $ (distributions on Schwartz class of functions where we define the weak topology in the usual point-wise convergence sense). As hint we are told to use the theorem below the question with a=0.

b. We are to prove that $ \lim_{\epsilon \to 0} G_\epsilon^\pm = F_0 \mp i \pi \delta $ (Hint : $(x \pm i\epsilon )^{-1} = (x \mp i\epsilon)(x^2+\epsilon ^2)^{-1} $).

c. We are to prove that $ \widehat{S}_\epsilon = (i\pi)^{-1} F_\epsilon $ and hence $ \widehat{\operatorname{sgn}} = (i\pi)^{-1}F_0 $.

d. From part c it follows that $ \widehat{F}_0 = -i\pi \operatorname{sgn} $. We are to prove this directly by showing $ \lim_{\epsilon \to 0 , N \to \infty } H_{\epsilon,N} = F_0 $ where we define $ H_{\epsilon,N} $ to be $ \frac{1}{x} $ if $ \epsilon < |x| < N $ and 0 otherwise, and via the exercise at the bottom.

e. We are to compute $ \widehat{\chi}_{(0,\infty)} $ (i) By writing $\chi = \frac{1}{2} \operatorname{sgn} + \frac{1}{2} $ and by using part c (ii) By using $ \chi(x) = \lim_{\epsilon \to 0} e^{-x\epsilon} \chi_{(0,\infty )} $ and by using b.

The theorem instructed to use (Notation used here is for $ \phi \in \mathbb R^n $ we define $ \phi_t(x) = t^{-n} \phi(t^{-1}x) $) : enter image description here

The exercise instructed to use:

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Here are where my problems are: I cannot seem to tackle any of parts a,b,c,d and also part e, as simple as it might sound, I tried doing but always ended up getting some close result but with something wrong. So I really need the help on this in order to do it, I realize it is a long question but I tried asking two people I know and they could not help me either, and of course I appreciate the help on this. Thanks all helpers.

*********** I am sorry I have just added notation for the theorem I brought here

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    $\begingroup$ Note that $\langle PV(f),\phi\rangle$ is standard usage and $< PV(f),\phi >$ is not. (I changed it.) ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 10 '15 at 17:33
  • $\begingroup$ @MichaelHardy : please forgive my faulty typing $\endgroup$ – kroner Oct 10 '15 at 17:39
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For part a. I didn't use the theorem; I did the following: Suppose $g$ is bounded and continuous on $\mathbb {R}$ with $g(0)=0.$ Then $$\tag 1 \int_0^\infty \frac{x}{x^2 + \epsilon^2}g(x)\,dx - \int_\epsilon^\infty \frac{1}{x}g(x)\,dx \to 0$$ as $\epsilon\to 0.$ This proves the result we're after: For $\phi \in S,$ we look at $$\int_{-\infty}^\infty \frac{x}{x^2 + \epsilon^2}\phi(x)\,dx - \int_{|x|>\epsilon} \frac{1}{x}\phi(x)\,dx $$ $$= \int_{0}^\infty \frac{x}{x^2 + \epsilon^2}(\phi(x)-\phi(-x))\,dx - \int_\epsilon^\infty \frac{1}{x}(\phi(x)-\phi(-x))\,dx,$$ which $\to 0$ by $(1).$

To prove $(1),$ first observe that $$ \int_0^\epsilon \frac{x}{x^2 + \epsilon^2}g(x)\,dx \to 0.$$ See that by letting $x=\epsilon y.$ Then use DCT and the continuity of $g$ at $0$ with $g(0)=0.$ So we need to look at

$$\int_\epsilon^\infty \frac{x}{x^2 + \epsilon^2}g(x)\,dx - \int_\epsilon^\infty \frac{1}{x}g(x)\,dx = \int_\epsilon^\infty \frac{-\epsilon^2}{x(x^2 + \epsilon^2)}g(x)\,dx = - \int_1^\infty \frac{-1}{y(y^2 + 1)}g(\epsilon y)\,dy.$$ As above, the DCT and the assumptions on $g$ show this $\to 0.$


b. $$\int_{\mathbb {R}} \frac{1}{x + i\epsilon}\phi(x)\,dx = \int_{\mathbb {R}} \frac{x-i\epsilon}{x^2 + \epsilon^2}\phi(x)\,dx = \int_{\mathbb {R}} \frac{x}{x^2 + \epsilon^2}\phi(x)\,dx - i\int_{\mathbb {R}} \frac{\epsilon}{x^2 + \epsilon^2}\phi(x)\,dx.$$

We know how the first integral on the right behaves from a., and I'm guessing you know the second integral $\to \pi \phi(0).$


Basic ideas for d. Step 1, Lemma: $$\lim_{h\to 0^+} \int_h^{1/h} \sin(xt)/x \, dx = (\pi/2)\text {sgn} (t).$$ For the proof of the lemma, let $x=y/t$ and use the exercise.

Step 2: For $h>0$ define $F_h(\phi) = \int_{h<|x|<1/h}\phi(x)/x \, dx.$ Verify that $F_h \to F_0$ as $h\to 0^+.$ Hence $F_0(\hat{\phi}) = \lim_{h\to 0^+} F_h(\hat{\phi}).$

Step 3: We can evaluate $F_h(\hat{\phi})$ by using the definition of $\hat{\phi}$ and reversing the order of integration. We get $$F_h(\hat{\phi}) = \int_{\mathbb {R}}\phi (t) \int_{h<|x|<1/h}\frac{e^{-ixt}}{x} \, dx = \int_{\mathbb {R}}\phi (t) (-2i)\int_h^{1/h}\frac{\sin(xt)}{x} \, dx\,dt.$$

Step 4: Verify the the inner integrals are uniformly bounded for all $h,t.$ Use the lemma and the DCT to let $h\to 0^+$ to get the desired result.

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  • $\begingroup$ Thanks for part a I managed to understand your solution completely and it makes sense to me now. Might I please ask you to provide help with the rest of the parts as well (b,c,d) now they are more accessible but still puzzle me? $\endgroup$ – kroner Oct 10 '15 at 17:42
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    $\begingroup$ I added part b. $\endgroup$ – zhw. Oct 10 '15 at 18:12
  • $\begingroup$ Ok I think part c I can do simply computing directly but parts d and e (especially d) I still cannot do, would you mind please adding those? $\endgroup$ – kroner Oct 10 '15 at 18:19
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    $\begingroup$ Made a minor correction to b. $\endgroup$ – zhw. Oct 10 '15 at 18:27
  • $\begingroup$ I noticed it thanks for your meticulousness $\endgroup$ – kroner Oct 10 '15 at 18:30

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