# Only two parts left : Problem on Fourier Transform and convergence of Tempered Distributions

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 $$\phi_t(x) = t^{-n} \phi(t^{-1}x)$$) :

The exercise instructed to use:

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

• Note that $\langle PV(f),\phi\rangle$ is standard usage and $< PV(f),\phi >$ is not. (I changed it.) ${}\qquad{}$ – Michael Hardy Oct 10 '15 at 17:33
• @MichaelHardy : please forgive my faulty typing – kroner Oct 10 '15 at 17:39

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.

• 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? – kroner Oct 10 '15 at 17:42
• I added part b. – zhw. Oct 10 '15 at 18:12
• 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? – kroner Oct 10 '15 at 18:19
• Made a minor correction to b. – zhw. Oct 10 '15 at 18:27
• I noticed it thanks for your meticulousness – kroner Oct 10 '15 at 18:30