
Title: The Delta Distribution Is Not Regular

Series: Distributions

YouTubeTitle: Distributions 6  The Delta Distribution Is Not Regular

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Subtitle on GitHub: dt06_sub_eng.srt

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Subtitle in English
1 00:00:00,270 –> 00:00:02,049 Hello and welcome back
2 00:00:02,059 –> 00:00:03,559 to distributions.
3 00:00:03,960 –> 00:00:05,369 And first, as always, I want
4 00:00:05,380 –> 00:00:06,849 to thank all the nice people
5 00:00:06,860 –> 00:00:07,909 that support this channel
6 00:00:07,920 –> 00:00:09,289 on Steady or paypal.
7 00:00:09,670 –> 00:00:11,130 In today’s part six, we will
8 00:00:11,140 –> 00:00:12,409 show that the delta
9 00:00:12,420 –> 00:00:14,189 distribution is not a
10 00:00:14,199 –> 00:00:15,560 regular distribution.
11 00:00:16,059 –> 00:00:17,940 So usually we say it’s a
12 00:00:17,950 –> 00:00:19,329 singular distribution
13 00:00:19,850 –> 00:00:21,309 more concretely, this means
14 00:00:21,319 –> 00:00:22,709 there’s no locally
15 00:00:22,719 –> 00:00:24,370 integrable function F
16 00:00:25,149 –> 00:00:26,819 defined on RN with
17 00:00:26,829 –> 00:00:28,600 values in R or C as
18 00:00:28,610 –> 00:00:30,090 always with the
19 00:00:30,100 –> 00:00:31,670 property that delta of
20 00:00:31,680 –> 00:00:33,290 Phi is the same as
21 00:00:33,299 –> 00:00:34,680 TF of Phi.
22 00:00:35,349 –> 00:00:36,830 And of course, this property
23 00:00:36,840 –> 00:00:38,529 should hold for all test
24 00:00:38,540 –> 00:00:40,509 functions Phi, OK.
25 00:00:40,520 –> 00:00:42,439 Now, as a reminder TF of
26 00:00:42,450 –> 00:00:44,310 Phi was defined by the
27 00:00:44,319 –> 00:00:45,130 integral.
28 00:00:45,520 –> 00:00:47,200 So we integrate FX
29 00:00:47,209 –> 00:00:49,130 times Phi X and
30 00:00:49,139 –> 00:00:50,130 we already showed in the
31 00:00:50,139 –> 00:00:51,869 last video that for locally
32 00:00:51,880 –> 00:00:53,360 integral with function F,
33 00:00:53,369 –> 00:00:54,689 this always defines a
34 00:00:54,700 –> 00:00:55,650 distribution.
35 00:00:56,119 –> 00:00:57,389 On the other hand, if we
36 00:00:57,400 –> 00:00:58,810 put Phi into the delta
37 00:00:58,819 –> 00:01:00,349 distribution, we get out
38 00:01:00,360 –> 00:01:01,759 the value of Phi at
39 00:01:01,770 –> 00:01:02,409 zero.
40 00:01:02,959 –> 00:01:04,650 Now, our claim here is for
41 00:01:04,660 –> 00:01:06,459 no F this equality is
42 00:01:06,470 –> 00:01:07,180 possible.
43 00:01:07,730 –> 00:01:09,220 Indeed, this is a standard
44 00:01:09,230 –> 00:01:10,199 exercise.
45 00:01:10,230 –> 00:01:11,760 And here I want to show you
46 00:01:11,769 –> 00:01:12,739 the proof of it.
47 00:01:13,269 –> 00:01:14,459 Let’s do it with a proof
48 00:01:14,470 –> 00:01:15,720 by contradiction.
49 00:01:16,339 –> 00:01:17,690 Therefore, let’s assume that
50 00:01:17,699 –> 00:01:19,230 we find such a function
51 00:01:19,239 –> 00:01:20,190 F.
52 00:01:20,199 –> 00:01:21,449 So it’s a locally integrable
53 00:01:21,750 –> 00:01:23,059 function where we usually
54 00:01:23,069 –> 00:01:24,389 use this symbol here.
55 00:01:24,870 –> 00:01:26,559 And now this property here
56 00:01:26,569 –> 00:01:28,269 will lead to a contradiction.
57 00:01:28,919 –> 00:01:29,290 OK.
58 00:01:29,300 –> 00:01:30,739 Now, in the first step, I
59 00:01:30,750 –> 00:01:32,129 just want to integrate the
60 00:01:32,139 –> 00:01:33,489 function F itself.
61 00:01:33,980 –> 00:01:35,790 So we already know this gives
62 00:01:35,800 –> 00:01:37,279 us a finite number by
63 00:01:37,290 –> 00:01:39,250 assumption when we integrate
64 00:01:39,260 –> 00:01:40,690 over a compact set.
65 00:01:41,339 –> 00:01:42,730 And for this, let’s simply
66 00:01:42,739 –> 00:01:44,400 choose the unit ball.
67 00:01:44,800 –> 00:01:46,330 One possibility to describe
68 00:01:46,339 –> 00:01:47,839 this would be simply to write
69 00:01:47,849 –> 00:01:49,080 that the Euclidean norm of
70 00:01:49,089 –> 00:01:50,830 X is less or equal than
71 00:01:50,839 –> 00:01:51,279 one.
72 00:01:51,830 –> 00:01:53,389 And in order to get a positive
73 00:01:53,400 –> 00:01:55,150 number out here, we also
74 00:01:55,160 –> 00:01:56,959 use the absolute value inside.
75 00:01:57,709 –> 00:01:59,150 Then we know we get out a
76 00:01:59,160 –> 00:02:00,519 number A which is a
77 00:02:00,529 –> 00:02:02,250 finite positive number.
78 00:02:02,930 –> 00:02:04,069 Now to get an idea of what
79 00:02:04,080 –> 00:02:05,779 we do here, let’s sketch
80 00:02:05,790 –> 00:02:07,069 the region where we integrate
81 00:02:07,080 –> 00:02:08,429 over in the case that the
82 00:02:08,440 –> 00:02:09,669 dimension is two
83 00:02:10,399 –> 00:02:11,429 there, we just have this
84 00:02:11,440 –> 00:02:13,139 circle where we integrate
85 00:02:13,149 –> 00:02:14,539 over the whole disc.
86 00:02:15,059 –> 00:02:16,419 Now, what I want to do is
87 00:02:16,429 –> 00:02:17,660 to go closer to the
88 00:02:17,669 –> 00:02:19,600 0.0 to the origin
89 00:02:19,610 –> 00:02:21,250 because this is the one we
90 00:02:21,259 –> 00:02:22,139 are interested in.
91 00:02:22,850 –> 00:02:24,520 Therefore, we can take smaller
92 00:02:24,529 –> 00:02:25,970 and smaller circles here
93 00:02:26,029 –> 00:02:27,339 and can split up the whole
94 00:02:27,350 –> 00:02:27,750 region.
95 00:02:28,419 –> 00:02:29,880 Of course, we do this with
96 00:02:29,889 –> 00:02:31,580 infinitely many circles
97 00:02:31,589 –> 00:02:33,169 where we choose one over
98 00:02:33,179 –> 00:02:34,660 K as the radius.
99 00:02:35,050 –> 00:02:36,860 So we start with 1/1, then
100 00:02:36,869 –> 00:02:38,199 comes 1/2
101 00:02:38,210 –> 00:02:39,979 1/3 and so on.
102 00:02:40,419 –> 00:02:41,800 And then our idea is that
103 00:02:41,809 –> 00:02:43,600 we can simply take the union
104 00:02:43,610 –> 00:02:45,470 of all these rings here.
105 00:02:45,949 –> 00:02:47,619 And then this infinite union
106 00:02:47,630 –> 00:02:49,580 is again the whole disc.
107 00:02:50,160 –> 00:02:50,559 OK.
108 00:02:50,570 –> 00:02:51,639 I don’t think we need to
109 00:02:51,649 –> 00:02:53,360 formalize this even more
110 00:02:53,479 –> 00:02:54,830 because the idea should be
111 00:02:54,839 –> 00:02:55,320 clear.
112 00:02:55,929 –> 00:02:57,119 The important thing we get
113 00:02:57,130 –> 00:02:58,399 is that here at the
114 00:02:58,410 –> 00:03:00,369 integral, we have a countable
115 00:03:00,380 –> 00:03:01,000 union.
116 00:03:01,699 –> 00:03:02,929 And now when you know some
117 00:03:02,940 –> 00:03:04,619 measure theory or integration
118 00:03:04,630 –> 00:03:06,149 theory, you immediately know
119 00:03:06,160 –> 00:03:08,070 what to do because here
120 00:03:08,080 –> 00:03:09,699 we have a disjoint union,
121 00:03:10,259 –> 00:03:11,839 we can simply write the whole
122 00:03:11,850 –> 00:03:13,669 integral as an infinite sum
123 00:03:13,679 –> 00:03:15,380 over the single integrals
124 00:03:15,389 –> 00:03:16,380 over the rings.
125 00:03:16,740 –> 00:03:18,320 Of course, this whole procedure
126 00:03:18,330 –> 00:03:19,710 here is very natural.
127 00:03:19,720 –> 00:03:20,869 But it’s the thing we have
128 00:03:20,880 –> 00:03:22,520 to put in for the integrals.
129 00:03:23,309 –> 00:03:24,169 In the case, you want to
130 00:03:24,179 –> 00:03:25,660 learn more about the details
131 00:03:25,669 –> 00:03:27,169 here, you can simply watch
132 00:03:27,179 –> 00:03:28,860 my measure theory videos.
133 00:03:29,490 –> 00:03:31,059 Now the result we get here
134 00:03:31,070 –> 00:03:32,860 is an infinite sum a series
135 00:03:32,869 –> 00:03:34,619 which has a finite value.
136 00:03:35,309 –> 00:03:36,800 Therefore, we can conclude
137 00:03:36,809 –> 00:03:38,360 that the sequence given by
138 00:03:38,369 –> 00:03:40,139 these integrals here is a
139 00:03:40,149 –> 00:03:41,520 sequence that converges to
140 00:03:41,529 –> 00:03:42,139 zero.
141 00:03:42,639 –> 00:03:44,220 And we know even more if
142 00:03:44,229 –> 00:03:45,619 we shift the starting point
143 00:03:45,630 –> 00:03:47,479 here, we can make this series
144 00:03:47,490 –> 00:03:48,889 as small as we want.
145 00:03:49,250 –> 00:03:50,679 For example, we could make
146 00:03:50,690 –> 00:03:52,539 it smaller than one half
147 00:03:53,070 –> 00:03:54,929 and I take one half because
148 00:03:54,940 –> 00:03:56,300 it’s a number that is smaller
149 00:03:56,309 –> 00:03:57,259 than one.
150 00:03:57,389 –> 00:03:58,800 And this is what we need
151 00:03:58,809 –> 00:03:59,860 for the contradiction.
152 00:04:00,490 –> 00:04:00,869 OK.
153 00:04:00,880 –> 00:04:02,380 And now we can just summarize
154 00:04:02,389 –> 00:04:03,210 the whole thing.
155 00:04:03,270 –> 00:04:04,839 I only did it to show
156 00:04:04,850 –> 00:04:06,770 you that we find an epsilon
157 00:04:06,779 –> 00:04:07,949 greater than zero.
158 00:04:08,380 –> 00:04:09,979 Such that the integral
159 00:04:09,990 –> 00:04:11,830 over the epsilon ball is
160 00:04:11,839 –> 00:04:13,809 a number B which is smaller
161 00:04:13,820 –> 00:04:14,789 than one half.
162 00:04:15,350 –> 00:04:16,630 So the whole explanation
163 00:04:16,640 –> 00:04:18,410 told you if we make the
164 00:04:18,420 –> 00:04:20,190 epsilon ball small enough,
165 00:04:20,200 –> 00:04:21,910 we can make the whole integral
166 00:04:21,920 –> 00:04:23,350 as small as we want.
167 00:04:23,970 –> 00:04:25,380 Of course, this makes totally
168 00:04:25,390 –> 00:04:26,750 sense if you just think of
169 00:04:26,760 –> 00:04:28,369 some ordinary functions.
170 00:04:28,690 –> 00:04:30,549 However, it also holds for
171 00:04:30,559 –> 00:04:32,359 a locally integrable function.
172 00:04:32,519 –> 00:04:34,029 Now, this was just a rough
173 00:04:34,040 –> 00:04:35,369 idea, but I don’t think you
174 00:04:35,380 –> 00:04:37,019 will have any problems filling
175 00:04:37,029 –> 00:04:37,950 in the details.
176 00:04:38,119 –> 00:04:39,670 For example, one detail I
177 00:04:39,679 –> 00:04:41,149 skipped here is that in this
178 00:04:41,160 –> 00:04:42,709 union, we will not
179 00:04:42,720 –> 00:04:44,510 include zero as a point.
180 00:04:44,760 –> 00:04:45,839 However, since the whole
181 00:04:45,850 –> 00:04:47,470 integral here does not care
182 00:04:47,480 –> 00:04:49,049 about a single point, the
183 00:04:49,059 –> 00:04:50,170 whole calculation is still
184 00:04:50,179 –> 00:04:50,790 correct.
185 00:04:51,320 –> 00:04:51,670 OK.
186 00:04:51,679 –> 00:04:53,290 Now, with this result from
187 00:04:53,299 –> 00:04:54,829 the first step, we are ready
188 00:04:54,839 –> 00:04:56,130 to finish the proof
189 00:04:56,679 –> 00:04:58,250 because we look at the delta
190 00:04:58,260 –> 00:04:59,609 distribution, it would be
191 00:04:59,619 –> 00:05:01,410 good to have test functions
192 00:05:01,420 –> 00:05:02,609 that are concentrated at
193 00:05:02,619 –> 00:05:03,209 zero.
194 00:05:03,600 –> 00:05:04,910 Indeed, we already know a
195 00:05:04,920 –> 00:05:06,769 nice one from part two of
196 00:05:06,779 –> 00:05:08,769 the video series, it was
197 00:05:08,779 –> 00:05:10,750 set to zero outside of
198 00:05:10,760 –> 00:05:12,410 the unit ball and
199 00:05:12,420 –> 00:05:14,109 inside the unit ball, it
200 00:05:14,119 –> 00:05:15,769 was given by an exponential
201 00:05:15,779 –> 00:05:16,209 function.
202 00:05:16,910 –> 00:05:18,660 This is a nice test function
203 00:05:18,670 –> 00:05:20,510 and the one dimensional visualization
204 00:05:20,519 –> 00:05:22,399 would be just such a bump
205 00:05:23,109 –> 00:05:24,410 where the maximum is
206 00:05:24,420 –> 00:05:25,989 exactly at zero.
207 00:05:26,450 –> 00:05:26,790 OK.
208 00:05:26,799 –> 00:05:27,980 Now you already know from
209 00:05:27,989 –> 00:05:29,420 the first part, the unit
210 00:05:29,429 –> 00:05:30,850 ball might be not small
211 00:05:30,859 –> 00:05:31,529 enough.
212 00:05:31,540 –> 00:05:32,670 Therefore, we should take
213 00:05:32,679 –> 00:05:34,200 this epsilon ball here.
214 00:05:34,649 –> 00:05:35,809 Therefore, I would say let’s
215 00:05:35,820 –> 00:05:37,329 change the function phi a
216 00:05:37,339 –> 00:05:39,130 little bit with this epsilon,
217 00:05:39,549 –> 00:05:41,309 it should be zero outside
218 00:05:41,320 –> 00:05:43,290 of the epsilon ball and
219 00:05:43,299 –> 00:05:45,010 inside it should be the same
220 00:05:45,019 –> 00:05:46,170 exponential function.
221 00:05:47,059 –> 00:05:48,790 Therefore, also inside the
222 00:05:48,799 –> 00:05:50,489 exponential function we have
223 00:05:50,500 –> 00:05:52,070 to divide by epsilon.
224 00:05:52,579 –> 00:05:53,970 And then you see this was
225 00:05:53,980 –> 00:05:55,540 just scaling the whole test
226 00:05:55,549 –> 00:05:56,070 function.
227 00:05:56,779 –> 00:05:58,010 Everything looks exactly
228 00:05:58,019 –> 00:05:59,839 the same just zoom to
229 00:05:59,850 –> 00:06:01,160 the epsilon ball instead of
230 00:06:01,170 –> 00:06:02,070 the unit ball.
231 00:06:02,440 –> 00:06:02,670 OK.
232 00:06:02,679 –> 00:06:04,450 Then let’s calculate with
233 00:06:04,459 –> 00:06:05,480 this test function
234 00:06:06,209 –> 00:06:07,869 by assumption this integral
235 00:06:07,880 –> 00:06:09,420 here is given by the delta
236 00:06:09,429 –> 00:06:11,029 distribution which means
237 00:06:11,040 –> 00:06:12,380 it gives us the value of
238 00:06:12,390 –> 00:06:13,940 the test function at zero.
239 00:06:14,500 –> 00:06:15,700 It’s a positive number.
240 00:06:15,709 –> 00:06:17,299 Hence, we don’t change anything
241 00:06:17,309 –> 00:06:18,869 when we use absolute values
242 00:06:18,880 –> 00:06:20,420 here, it makes everything
243 00:06:20,429 –> 00:06:21,790 easier because we can just
244 00:06:21,799 –> 00:06:23,640 apply the triangle inequality
245 00:06:23,649 –> 00:06:24,799 for integrals.
246 00:06:25,269 –> 00:06:26,470 So now we have the absolute
247 00:06:26,480 –> 00:06:27,929 value inside the
248 00:06:27,940 –> 00:06:28,640 integral.
249 00:06:28,920 –> 00:06:30,239 This means that we can just
250 00:06:30,250 –> 00:06:32,019 pull out the supreme norm
251 00:06:32,029 –> 00:06:33,119 of the test function.
252 00:06:33,579 –> 00:06:34,970 However, before we do that,
253 00:06:34,980 –> 00:06:36,260 we should change the region
254 00:06:36,269 –> 00:06:38,040 where we integrate over to
255 00:06:38,049 –> 00:06:39,049 the epsilon ball.
256 00:06:39,459 –> 00:06:41,170 Because outside of the epsilon
257 00:06:41,179 –> 00:06:42,609 ball, everything is zero
258 00:06:42,619 –> 00:06:43,890 by the test function.
259 00:06:44,239 –> 00:06:45,559 And maybe now you already
260 00:06:45,570 –> 00:06:47,239 see why we care so much about
261 00:06:47,250 –> 00:06:48,160 the epsilon ball.
262 00:06:48,260 –> 00:06:49,829 First, let’s pull out the
263 00:06:49,839 –> 00:06:51,130 supremum norm of the test
264 00:06:51,140 –> 00:06:51,779 function.
265 00:06:52,109 –> 00:06:53,630 And then you see only this
266 00:06:53,640 –> 00:06:55,250 integral here remains
267 00:06:55,420 –> 00:06:57,279 first, the supremum here
268 00:06:57,290 –> 00:06:58,660 we already know it’s the
269 00:06:58,670 –> 00:07:00,149 value at zero more
270 00:07:00,160 –> 00:07:01,690 concretely the same as the
271 00:07:01,700 –> 00:07:02,609 left hand side.
272 00:07:03,140 –> 00:07:04,660 And the second part here
273 00:07:04,670 –> 00:07:06,079 is equal to B
274 00:07:06,380 –> 00:07:08,339 however, B was smaller
275 00:07:08,350 –> 00:07:10,339 than one half, which
276 00:07:10,350 –> 00:07:12,220 means we have an inequality
277 00:07:12,239 –> 00:07:14,000 which can’t be satisfied.
278 00:07:14,410 –> 00:07:16,019 So we have a positive number
279 00:07:16,029 –> 00:07:17,820 which does not get smaller
280 00:07:17,829 –> 00:07:19,059 when we multiply it with
281 00:07:19,070 –> 00:07:19,799 one half.
282 00:07:20,290 –> 00:07:21,609 And such a positive real
283 00:07:21,619 –> 00:07:23,369 number does not exist.
284 00:07:24,029 –> 00:07:25,769 And this is indeed our wanted
285 00:07:25,779 –> 00:07:27,440 contradiction in
286 00:07:27,450 –> 00:07:29,190 conclusion such a locally
287 00:07:29,200 –> 00:07:30,869 integral function F for the
288 00:07:30,880 –> 00:07:32,559 delta distribution does not
289 00:07:32,570 –> 00:07:33,239 exist.
290 00:07:33,760 –> 00:07:35,570 Therefore, the delta distribution
291 00:07:35,579 –> 00:07:36,769 is not a regular
292 00:07:36,779 –> 00:07:37,670 distribution.
293 00:07:38,019 –> 00:07:38,320 OK.
294 00:07:38,329 –> 00:07:39,480 Then in the next video, I
295 00:07:39,489 –> 00:07:40,609 will show you what we can
296 00:07:40,619 –> 00:07:42,220 really do with distributions
297 00:07:42,230 –> 00:07:43,399 and how to calculate with
298 00:07:43,410 –> 00:07:43,980 them.
299 00:07:43,989 –> 00:07:45,149 Therefore, I hope I see you
300 00:07:45,160 –> 00:07:46,480 there and have a nice day.
301 00:07:46,489 –> 00:07:47,160 Bye.

Quiz Content
Q1: Is the delta distribution a regular distribution?
A1: Yes.
A2: No.
Q2: Is there a locally integrable function $f$ such that $\delta = T_f$?
A1: Yes.
A2: No.
Q3: Is the function $\varphi_\varepsilon \in \mathcal{D}(\mathbb{R}^n)$ given by $$\varphi_\varepsilon(x) = \begin{cases} 0 &,~  x  \geq \varepsilon \ \exp\left(  \frac{1}{1  x/\varepsilon ^2 } \right) &,~  x  < \varepsilon \end{cases}$$ a welldefined test function for $\varepsilon > 0$.
A1: Yes!
A2: No!