1 00:00:00,400 –> 00:00:02,170 Hello and welcome back to

2 00:00:02,180 –> 00:00:03,089 distributions.

3 00:00:03,200 –> 00:00:04,659 And as always many, many

4 00:00:04,670 –> 00:00:06,219 thanks to all the nice people

5 00:00:06,230 –> 00:00:07,530 that support this channel

6 00:00:07,539 –> 00:00:08,890 on Steady or paypal.

7 00:00:09,390 –> 00:00:10,550 This is part five.

8 00:00:10,560 –> 00:00:11,750 And today I want to talk

9 00:00:11,760 –> 00:00:13,359 about so called regular

10 00:00:13,369 –> 00:00:14,260 distributions.

11 00:00:15,079 –> 00:00:16,680 However, before we do this,

12 00:00:16,739 –> 00:00:18,010 I first want to show you

13 00:00:18,020 –> 00:00:19,569 a useful characterization

14 00:00:19,579 –> 00:00:20,590 for distributions.

15 00:00:21,540 –> 00:00:23,200 Of course, you already know

16 00:00:23,209 –> 00:00:24,799 a distribution is a linear

17 00:00:24,809 –> 00:00:26,770 map T from the space of

18 00:00:26,780 –> 00:00:27,840 test functions

19 00:00:28,790 –> 00:00:30,600 to the real or complex numbers.

20 00:00:31,500 –> 00:00:32,939 And you also know it has

21 00:00:32,950 –> 00:00:34,529 to be continuous in the

22 00:00:34,540 –> 00:00:35,840 sense as we define it.

23 00:00:35,849 –> 00:00:37,740 In the last video, we

24 00:00:37,750 –> 00:00:39,099 use sequences for this.

25 00:00:39,110 –> 00:00:40,189 But there are also a lot

26 00:00:40,200 –> 00:00:41,459 of other equivalent

27 00:00:41,470 –> 00:00:42,759 characterizations for it.

28 00:00:43,919 –> 00:00:45,139 Hence, we can write it in

29 00:00:45,150 –> 00:00:47,040 the following way T is a

30 00:00:47,049 –> 00:00:49,009 distribution if and only

31 00:00:49,020 –> 00:00:50,819 if for all

32 00:00:50,830 –> 00:00:52,650 compact sets in RN,

33 00:00:53,479 –> 00:00:55,189 there exists a natural number

34 00:00:55,200 –> 00:00:56,979 M which could be zero

35 00:00:58,110 –> 00:00:59,770 and there exists a positive

36 00:00:59,779 –> 00:01:00,959 constant C

37 00:01:01,729 –> 00:01:03,270 such that for all test

38 00:01:03,279 –> 00:01:04,319 functions, Phi

39 00:01:05,319 –> 00:01:06,870 we have that if the

40 00:01:06,879 –> 00:01:08,349 support of Phi

41 00:01:08,360 –> 00:01:10,269 lies in the compact set

42 00:01:10,279 –> 00:01:12,260 K, then this

43 00:01:12,269 –> 00:01:14,010 implies that the value

44 00:01:14,029 –> 00:01:15,620 T of Phi is

45 00:01:15,629 –> 00:01:16,809 bounded from above,

46 00:01:17,660 –> 00:01:19,160 namely we have the constant

47 00:01:19,169 –> 00:01:20,809 C times the sum

48 00:01:20,819 –> 00:01:22,529 over all multi indices

49 00:01:22,540 –> 00:01:24,480 with degree less or equal

50 00:01:24,489 –> 00:01:25,599 than the constant M.

51 00:01:26,260 –> 00:01:27,589 And then we look at all the

52 00:01:27,599 –> 00:01:29,440 derivatives of Phi in the

53 00:01:29,449 –> 00:01:30,400 supremum norm.

54 00:01:31,269 –> 00:01:33,010 Please note this means that

55 00:01:33,019 –> 00:01:34,449 whenever you can write down

56 00:01:34,459 –> 00:01:36,330 such an estimate, you immediately

57 00:01:36,339 –> 00:01:38,330 have the continuity and therefore

58 00:01:38,339 –> 00:01:39,809 you have a distribution T

59 00:01:40,650 –> 00:01:41,959 since at first, it looks

60 00:01:41,970 –> 00:01:43,620 complicated, we really should

61 00:01:43,629 –> 00:01:44,769 write down a proof for it.

62 00:01:45,650 –> 00:01:46,800 So let’s start with this

63 00:01:46,809 –> 00:01:47,449 direction.

64 00:01:47,459 –> 00:01:49,099 So when we have the estimate,

65 00:01:49,110 –> 00:01:50,519 what can we say about

66 00:01:50,529 –> 00:01:52,489 continuity in

67 00:01:52,500 –> 00:01:53,680 order to show this, let’s

68 00:01:53,690 –> 00:01:55,260 choose test functions Phi

69 00:01:55,269 –> 00:01:57,220 K and Phi such that we have

70 00:01:57,230 –> 00:01:59,129 our D convergence for the

71 00:01:59,139 –> 00:02:00,779 sequence Phi K to

72 00:02:00,790 –> 00:02:02,769 Phi, then we know by

73 00:02:02,779 –> 00:02:04,660 definition there is a compact

74 00:02:04,669 –> 00:02:06,379 or bounded set K in

75 00:02:06,389 –> 00:02:08,330 RN such that the

76 00:02:08,339 –> 00:02:10,020 support of all phi K

77 00:02:10,080 –> 00:02:11,929 lies in this compact set

78 00:02:11,940 –> 00:02:12,330 K.

79 00:02:13,470 –> 00:02:14,919 And we also know that for

80 00:02:14,929 –> 00:02:16,729 all multi indices alpha,

81 00:02:16,740 –> 00:02:18,570 we have that the divers

82 00:02:18,660 –> 00:02:20,039 D alpha Phi K

83 00:02:20,050 –> 00:02:21,759 converge in the suprema

84 00:02:21,919 –> 00:02:22,330 norm.

85 00:02:22,990 –> 00:02:24,300 Please keep in mind this

86 00:02:24,309 –> 00:02:25,839 was just the explanation

87 00:02:25,850 –> 00:02:27,479 what the D-convergence here

88 00:02:27,490 –> 00:02:27,970 means.

89 00:02:28,630 –> 00:02:29,169 OK.

90 00:02:29,210 –> 00:02:30,630 And now we just have to check

91 00:02:30,639 –> 00:02:32,529 if T of Phi K is

92 00:02:32,539 –> 00:02:34,529 convergent to T of Phi.

93 00:02:35,389 –> 00:02:36,679 In other words, we want that

94 00:02:36,690 –> 00:02:38,679 this term goes to zero.

95 00:02:38,690 –> 00:02:40,320 When K goes to infinity.

96 00:02:41,279 –> 00:02:42,899 By using the linearity, we

97 00:02:42,910 –> 00:02:44,889 can rewrite that as T of

98 00:02:44,899 –> 00:02:46,860 Phi K minus Phi.

99 00:02:47,649 –> 00:02:49,149 Of course, we know this is

100 00:02:49,160 –> 00:02:50,500 again a test function with

101 00:02:50,509 –> 00:02:51,639 support at K.

102 00:02:51,699 –> 00:02:53,119 So we know they exist in

103 00:02:53,130 –> 00:02:54,940 M and in C such that this

104 00:02:54,949 –> 00:02:55,949 estimate holds.

105 00:02:56,679 –> 00:02:57,889 And there you see, we have

106 00:02:57,899 –> 00:02:59,300 what we want because this

107 00:02:59,309 –> 00:03:01,179 term goes to zero when K

108 00:03:01,190 –> 00:03:02,179 goes to infinity.

109 00:03:03,050 –> 00:03:04,339 And please don’t forget we

110 00:03:04,350 –> 00:03:05,669 have a finite sum here.

111 00:03:06,449 –> 00:03:07,770 So this is all we needed

112 00:03:07,779 –> 00:03:09,089 to show for the first part.

113 00:03:09,100 –> 00:03:10,429 Now let’s go from the left

114 00:03:10,440 –> 00:03:11,669 hand side to the right hand

115 00:03:11,679 –> 00:03:12,020 side.

116 00:03:12,800 –> 00:03:14,059 Here, I want to do a proof

117 00:03:14,070 –> 00:03:15,100 by contraposition.

118 00:03:15,229 –> 00:03:16,910 So we start with the negation

119 00:03:16,919 –> 00:03:18,059 of the right hand side.

120 00:03:18,809 –> 00:03:20,369 This means that we just have

121 00:03:20,380 –> 00:03:21,860 to exchange the quantifier.

122 00:03:21,869 –> 00:03:23,660 So here we have, there exists

123 00:03:23,669 –> 00:03:25,619 a compact set such that for

124 00:03:25,630 –> 00:03:27,029 all M and all

125 00:03:27,080 –> 00:03:28,740 C there exists a

126 00:03:28,750 –> 00:03:29,949 test function phi

127 00:03:30,649 –> 00:03:32,320 and the negation of the implication

128 00:03:32,330 –> 00:03:33,789 is just that the support

129 00:03:33,800 –> 00:03:35,110 is still in K.

130 00:03:35,339 –> 00:03:37,029 But the inequality goes the

131 00:03:37,039 –> 00:03:38,789 other way around this

132 00:03:38,800 –> 00:03:40,179 means that we have a greater

133 00:03:40,190 –> 00:03:41,139 sign here.

134 00:03:42,000 –> 00:03:43,619 So this is our assumption

135 00:03:43,630 –> 00:03:44,779 and where we want to get

136 00:03:44,789 –> 00:03:46,649 to is that T is

137 00:03:46,660 –> 00:03:47,699 not continuous.

138 00:03:48,460 –> 00:03:49,839 Now, we call this whole thing

139 00:03:49,850 –> 00:03:51,130 means there is a compact

140 00:03:51,139 –> 00:03:52,720 set but then no matter

141 00:03:52,729 –> 00:03:54,440 which M and C we choose,

142 00:03:54,449 –> 00:03:56,149 we always find a corresponding

143 00:03:56,160 –> 00:03:56,600 Phi.

144 00:03:57,399 –> 00:03:58,669 Therefore, let’s go through

145 00:03:58,679 –> 00:04:00,160 all the natural numbers K

146 00:04:00,169 –> 00:04:01,979 for C and M and take the

147 00:04:01,990 –> 00:04:03,380 corresponding phi k.

148 00:04:04,240 –> 00:04:05,820 This means that we have T

149 00:04:05,830 –> 00:04:07,729 phi K greater than K

150 00:04:07,740 –> 00:04:09,610 times the sum where we have

151 00:04:09,619 –> 00:04:10,050 here.

152 00:04:10,059 –> 00:04:11,320 The maximum order for the

153 00:04:11,330 –> 00:04:12,619 derivatives as k

154 00:04:13,399 –> 00:04:15,100 please don’t forget the absolute

155 00:04:15,110 –> 00:04:16,519 value around a multi

156 00:04:16,649 –> 00:04:18,250 index just means that you

157 00:04:18,260 –> 00:04:20,040 add up all the components.

158 00:04:20,868 –> 00:04:22,588 So no matter how you combine

159 00:04:22,598 –> 00:04:24,299 all the partial derivatives,

160 00:04:24,419 –> 00:04:25,848 the maximum order you get

161 00:04:25,859 –> 00:04:27,118 out is always key.

162 00:04:28,089 –> 00:04:29,609 Also important to note is

163 00:04:29,619 –> 00:04:31,140 that we have the zero multi

164 00:04:31,230 –> 00:04:33,130 index as well in the sum,

165 00:04:33,200 –> 00:04:34,779 which means we have the supremum

166 00:04:34,899 –> 00:04:36,420 norm of Phi K here.

167 00:04:37,339 –> 00:04:38,559 Now, the key idea of the

168 00:04:38,570 –> 00:04:39,910 proof is that we define a

169 00:04:39,920 –> 00:04:41,829 new test function Phi

170 00:04:41,839 –> 00:04:43,640 K by scaling the

171 00:04:43,649 –> 00:04:44,359 Phi case.

172 00:04:45,299 –> 00:04:46,720 And of course, the correct

173 00:04:46,730 –> 00:04:48,619 scaling factor is exactly

174 00:04:48,630 –> 00:04:50,500 one over the absolute value

175 00:04:50,510 –> 00:04:51,839 of T of Phi k

176 00:04:52,529 –> 00:04:54,279 because we know it’s increasing

177 00:04:54,290 –> 00:04:56,179 with K and it increases

178 00:04:56,190 –> 00:04:58,179 faster than the supremum

179 00:04:58,190 –> 00:04:59,230 norm of Phi K.

180 00:05:00,059 –> 00:05:02,049 Therefore, by this definition,

181 00:05:02,059 –> 00:05:03,339 the psi K go

182 00:05:03,350 –> 00:05:05,269 uniformly to the zero function

183 00:05:05,279 –> 00:05:06,989 when K tends to infinity.

184 00:05:07,850 –> 00:05:09,339 In fact, the whole argument

185 00:05:09,350 –> 00:05:11,250 works for every multi index

186 00:05:11,329 –> 00:05:13,200 which means we have the uniform

187 00:05:13,209 –> 00:05:14,630 convergence for every

188 00:05:14,640 –> 00:05:15,359 derivative

189 00:05:15,910 –> 00:05:17,890 combining this with the supports

190 00:05:17,899 –> 00:05:19,480 being all in K.

191 00:05:19,640 –> 00:05:21,029 We have indeed the D

192 00:05:21,040 –> 00:05:21,929 convergence.

193 00:05:22,170 –> 00:05:23,829 However, now you might already

194 00:05:23,839 –> 00:05:25,660 see the images under

195 00:05:25,670 –> 00:05:27,279 T don’t converge to

196 00:05:27,290 –> 00:05:29,220 zero to see this.

197 00:05:29,230 –> 00:05:30,359 Just look at the absolute

198 00:05:30,369 –> 00:05:32,010 value of T of Phi

199 00:05:32,019 –> 00:05:32,640 K

200 00:05:33,809 –> 00:05:35,269 which is by the linearity

201 00:05:35,279 –> 00:05:36,910 just one over the absolute

202 00:05:36,920 –> 00:05:38,510 value of T of Phi

203 00:05:38,519 –> 00:05:40,160 K times the absolute

204 00:05:40,170 –> 00:05:41,750 value of T of Phi

205 00:05:41,760 –> 00:05:43,549 K, which is of

206 00:05:43,559 –> 00:05:45,140 course always one.

207 00:05:46,269 –> 00:05:47,829 In other words, it does not

208 00:05:47,839 –> 00:05:49,029 converge to zero.

209 00:05:49,910 –> 00:05:51,500 Now this simply means the

210 00:05:51,510 –> 00:05:53,359 linear map T is not

211 00:05:53,369 –> 00:05:55,279 continuous and that

212 00:05:55,290 –> 00:05:56,350 is what we wanted to show

213 00:05:56,359 –> 00:05:57,200 from the beginning.

214 00:05:57,209 –> 00:05:58,690 So our proof is finished

215 00:05:59,450 –> 00:06:01,019 with this technical proposition.

216 00:06:01,029 –> 00:06:02,679 Out of the way we can go

217 00:06:02,690 –> 00:06:04,500 to the definition of a regular

218 00:06:04,510 –> 00:06:05,339 distribution.

219 00:06:05,809 –> 00:06:07,209 For this, we first need a

220 00:06:07,220 –> 00:06:09,000 notion of a locally

221 00:06:09,010 –> 00:06:10,339 integrable function.

222 00:06:10,760 –> 00:06:12,239 Let’s call the function F

223 00:06:12,250 –> 00:06:13,640 and it can have values in

224 00:06:13,649 –> 00:06:14,769 R or in C

225 00:06:15,480 –> 00:06:17,040 and we call it locally

226 00:06:17,049 –> 00:06:18,869 integral if the function

227 00:06:18,880 –> 00:06:20,459 is integrable, when we

228 00:06:20,470 –> 00:06:22,239 restrict it to any compact

229 00:06:22,250 –> 00:06:22,670 set.

230 00:06:23,190 –> 00:06:24,179 So you can put it in the

231 00:06:24,190 –> 00:06:26,119 way that the integral over

232 00:06:26,130 –> 00:06:28,019 compact set K of the

233 00:06:28,029 –> 00:06:29,920 function F with the absolute

234 00:06:29,929 –> 00:06:31,269 value makes

235 00:06:31,279 –> 00:06:32,540 sense and is

236 00:06:32,549 –> 00:06:33,299 finite.

237 00:06:34,170 –> 00:06:35,260 When you know how to deal

238 00:06:35,269 –> 00:06:36,619 with Lebesgue integrals, it

239 00:06:36,630 –> 00:06:38,019 will make your life much

240 00:06:38,029 –> 00:06:39,579 easier here because you

241 00:06:39,589 –> 00:06:41,380 know, you need a measurable

242 00:06:41,390 –> 00:06:42,859 function F and the

243 00:06:42,869 –> 00:06:44,420 integral should be finite.

244 00:06:44,980 –> 00:06:46,500 However, it’s also possible

245 00:06:46,510 –> 00:06:47,980 to think of a normal Riemann

246 00:06:47,989 –> 00:06:49,750 integral here and do everything

247 00:06:49,760 –> 00:06:50,950 with the Riemann integral.

248 00:06:50,980 –> 00:06:52,399 But then of course, we don’t

249 00:06:52,410 –> 00:06:54,130 get out the full general

250 00:06:54,140 –> 00:06:54,820 result.

251 00:06:55,369 –> 00:06:57,100 Therefore, I would say knowing

252 00:06:57,109 –> 00:06:58,890 some measure theory and the

253 00:06:58,910 –> 00:07:00,500 Lebesgue integral is indeed

254 00:07:00,510 –> 00:07:01,959 helpful here in the theory

255 00:07:01,970 –> 00:07:02,980 of distributions.

256 00:07:03,670 –> 00:07:05,279 Now, for such locally integrable

257 00:07:05,600 –> 00:07:07,239 functions, we use a common

258 00:07:07,250 –> 00:07:08,890 notation, namely we

259 00:07:08,899 –> 00:07:10,850 write a curved L

260 00:07:10,859 –> 00:07:12,500 with one and lo

261 00:07:13,950 –> 00:07:15,850 and often we put in the domain

262 00:07:15,859 –> 00:07:17,529 which is here

263 00:07:17,540 –> 00:07:19,339 and please

264 00:07:19,350 –> 00:07:21,019 note of course, all

265 00:07:21,440 –> 00:07:22,859 integrable functions are also

266 00:07:22,869 –> 00:07:24,200 locally integrable.

267 00:07:24,630 –> 00:07:26,140 We just have more functions

268 00:07:26,149 –> 00:07:26,579 here.

269 00:07:27,130 –> 00:07:28,619 For example, if you consider

270 00:07:28,630 –> 00:07:30,380 the function F which sends

271 00:07:30,390 –> 00:07:31,820 R to R

272 00:07:32,540 –> 00:07:34,519 and X to X squared, then

273 00:07:34,529 –> 00:07:35,779 this function is not

274 00:07:35,790 –> 00:07:37,779 integrable because the integral

275 00:07:37,790 –> 00:07:39,170 over the whole domain are

276 00:07:39,179 –> 00:07:40,279 would be infinity.

277 00:07:40,920 –> 00:07:42,529 However, it’s locally

278 00:07:42,540 –> 00:07:44,019 integrable because it’s a

279 00:07:44,029 –> 00:07:45,369 continuous function.

280 00:07:45,459 –> 00:07:47,190 And when we integrate over

281 00:07:47,200 –> 00:07:49,059 a bounded set here, we don’t

282 00:07:49,070 –> 00:07:50,299 have any problem, we get

283 00:07:50,309 –> 00:07:51,839 out a finite integral.

284 00:07:52,570 –> 00:07:54,230 Nevertheless, please note

285 00:07:54,239 –> 00:07:56,149 that we don’t need the continuity

286 00:07:56,160 –> 00:07:57,540 for a locally integrable

287 00:07:57,549 –> 00:07:58,119 function.

288 00:07:58,299 –> 00:08:00,070 We have much more functions

289 00:08:00,079 –> 00:08:01,440 here than just the continuous

290 00:08:01,450 –> 00:08:01,869 ones.

291 00:08:02,670 –> 00:08:04,040 Now, the nice thing we want

292 00:08:04,049 –> 00:08:05,619 to do now with such a function

293 00:08:05,630 –> 00:08:07,489 is to define a distribution

294 00:08:08,200 –> 00:08:09,869 for such a function F I want

295 00:08:09,880 –> 00:08:11,820 to use the notation TF

296 00:08:11,829 –> 00:08:13,700 for the corresponding distribution.

297 00:08:14,540 –> 00:08:16,320 And it is defined for test

298 00:08:16,329 –> 00:08:17,279 function Phi.

299 00:08:17,540 –> 00:08:18,559 And you might already know

300 00:08:18,570 –> 00:08:20,109 this because we did it for

301 00:08:20,119 –> 00:08:21,559 continuous functions in the

302 00:08:21,570 –> 00:08:23,200 last video by the

303 00:08:23,209 –> 00:08:24,079 integral

304 00:08:24,799 –> 00:08:25,890 FX phi

305 00:08:25,899 –> 00:08:27,390 XDX.

306 00:08:28,250 –> 00:08:29,420 This is a well defined

307 00:08:29,429 –> 00:08:31,140 distribution because the

308 00:08:31,149 –> 00:08:32,739 integral is actually

309 00:08:32,750 –> 00:08:34,510 just an integral over a compact

310 00:08:34,520 –> 00:08:36,500 set, namely the support of

311 00:08:36,510 –> 00:08:38,390 Phi, it’s also

312 00:08:38,400 –> 00:08:40,250 linear and it fulfills our

313 00:08:40,260 –> 00:08:41,679 estimate from the beginning.

314 00:08:42,380 –> 00:08:43,690 This one we really should

315 00:08:43,700 –> 00:08:44,409 write down.

316 00:08:44,419 –> 00:08:45,690 So we have the absolute value

317 00:08:45,700 –> 00:08:47,510 of T f phi, we can

318 00:08:47,520 –> 00:08:49,070 pull in the absolute value

319 00:08:49,080 –> 00:08:51,070 into the integral and we

320 00:08:51,080 –> 00:08:53,030 can also easily split

321 00:08:53,039 –> 00:08:53,469 it up.

322 00:08:54,719 –> 00:08:56,419 Now instead of RN, we can

323 00:08:56,429 –> 00:08:58,210 just write the support of

324 00:08:58,219 –> 00:09:00,049 Phi simply because

325 00:09:00,059 –> 00:09:01,469 outside of function, Phi

326 00:09:01,479 –> 00:09:03,140 is zero, which has no

327 00:09:03,150 –> 00:09:04,780 contribution to the integral.

328 00:09:05,260 –> 00:09:07,169 Also we know we can estimate

329 00:09:07,179 –> 00:09:08,809 this one with the suprema

330 00:09:08,969 –> 00:09:09,859 norm of Phi,

331 00:09:10,690 –> 00:09:12,400 hence, we have the integral

332 00:09:12,409 –> 00:09:13,969 of F in the absolute

333 00:09:13,979 –> 00:09:14,549 value

334 00:09:15,580 –> 00:09:17,260 times the sup norm of

335 00:09:17,270 –> 00:09:17,710 Phi.

336 00:09:18,609 –> 00:09:20,109 Of course, we can integrate

337 00:09:20,119 –> 00:09:22,070 here over any compact set

338 00:09:22,080 –> 00:09:24,049 K as long as it is a

339 00:09:24,059 –> 00:09:25,789 superset of the support of

340 00:09:25,799 –> 00:09:26,260 Phi.

341 00:09:26,960 –> 00:09:28,500 Now please remember what

342 00:09:28,510 –> 00:09:29,979 the estimate is we want to

343 00:09:29,989 –> 00:09:30,419 show.

344 00:09:30,570 –> 00:09:32,140 So I put it here in a

345 00:09:32,150 –> 00:09:33,539 box as a reminder.

346 00:09:34,809 –> 00:09:36,349 And there you see the

347 00:09:36,359 –> 00:09:38,169 integral here is the constant

348 00:09:38,179 –> 00:09:38,950 C here.

349 00:09:39,190 –> 00:09:41,039 And the sum is just the sum

350 00:09:41,049 –> 00:09:42,349 where M is zero.

351 00:09:42,429 –> 00:09:44,030 So we just have the suprema

352 00:09:44,070 –> 00:09:45,909 number Phi in

353 00:09:45,919 –> 00:09:47,619 other words, we have indeed

354 00:09:47,630 –> 00:09:48,539 a distribution.

355 00:09:49,250 –> 00:09:50,849 So you see this is similar

356 00:09:50,859 –> 00:09:52,270 to what we had in the last

357 00:09:52,280 –> 00:09:53,729 video where we considered

358 00:09:53,739 –> 00:09:55,349 continuous functions.

359 00:09:55,460 –> 00:09:56,830 The distribution here is

360 00:09:56,840 –> 00:09:58,469 essentially given by a

361 00:09:58,479 –> 00:09:59,770 normal function F.

362 00:10:00,559 –> 00:10:01,960 And therefore, these

363 00:10:01,969 –> 00:10:03,619 distributions are just called

364 00:10:03,630 –> 00:10:05,179 regular distributions.

365 00:10:06,020 –> 00:10:07,630 To put this into a definition,

366 00:10:07,640 –> 00:10:09,159 you would say a distribution

367 00:10:09,169 –> 00:10:11,070 T is called regular

368 00:10:11,789 –> 00:10:13,150 if there is a locally

369 00:10:13,159 –> 00:10:14,909 integrable function F

370 00:10:14,919 –> 00:10:16,840 such that T is the

371 00:10:16,849 –> 00:10:18,450 same as TF

372 00:10:19,010 –> 00:10:20,729 so the linear map T can

373 00:10:20,739 –> 00:10:22,479 actually be written as an

374 00:10:22,489 –> 00:10:24,080 integral in this sense.

375 00:10:24,960 –> 00:10:26,669 So you could say these are

376 00:10:26,679 –> 00:10:28,200 the distributions that behave

377 00:10:28,210 –> 00:10:30,099 like normal functions as

378 00:10:30,109 –> 00:10:31,039 we already know it.

379 00:10:31,460 –> 00:10:33,090 But we also know there are

380 00:10:33,099 –> 00:10:34,710 much more distributions than

381 00:10:34,719 –> 00:10:35,020 that.

382 00:10:35,909 –> 00:10:37,760 Indeed, the delta distribution

383 00:10:37,770 –> 00:10:38,979 is not regular.

384 00:10:39,150 –> 00:10:40,419 And we can show this in the

385 00:10:40,429 –> 00:10:41,210 next video.

386 00:10:41,890 –> 00:10:43,289 Therefore, I hope I see you

387 00:10:43,299 –> 00:10:44,909 there and have a nice day.

388 00:10:44,989 –> 00:10:45,669 Bye.

Q1: What is the definition of a locally integrable function $f: \mathbb{R}^n \rightarrow \mathbb{R}$?

A1: $\int_K | f(x) | , dx < \infty$ for all compact sets $K \subseteq \mathbb{R}^n$.

A2: $\int_K f(x) , dx = 0$ for all compact sets $K \subseteq \mathbb{R}^n$.

A3: $\int_K f(x) , dx > 0$ for all compact sets $K \subseteq \mathbb{R}^n$.

Q2: For a locally integrable function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, we define $T_f$. What is not correct?

A1: $T_f$ is always a distribution.

A2: $T_f$ is called a regular distribution.

A3: $T_f(\varphi) = \int_{\mathbb{R}^n} f(x) \varphi(x) , dx$

A4: $T_f(0) = 1$

Q3: Is $T = 0$ a regular distribution?

A1: Yes!

A2: No!