00:00 Introduction

00:30 One-dimensional example

01:21 Zero-dimensional example

02:10 l^p-space

1 00:00:00,449 –> 00:00:02,269 Hello and welcome back to

2 00:00:02,279 –> 00:00:03,490 functional analysis.

3 00:00:03,559 –> 00:00:04,670 And as always, I want to

4 00:00:04,679 –> 00:00:06,099 thank all the nice people

5 00:00:06,110 –> 00:00:07,269 that support this channel

6 00:00:07,280 –> 00:00:08,779 on Steady or paypal.

7 00:00:09,569 –> 00:00:11,140 Today we have part seven

8 00:00:11,149 –> 00:00:12,829 where we discuss examples

9 00:00:12,840 –> 00:00:14,060 of Banach spaces.

10 00:00:14,739 –> 00:00:16,040 Please recall the picture

11 00:00:16,049 –> 00:00:17,299 from the last time where

12 00:00:17,309 –> 00:00:18,450 we learned that the Banach

13 00:00:18,489 –> 00:00:19,969 space is on the one hand,

14 00:00:19,979 –> 00:00:21,340 a real or complex vector

15 00:00:21,350 –> 00:00:21,920 space.

16 00:00:21,930 –> 00:00:23,170 And on the other hand, a

17 00:00:23,180 –> 00:00:24,920 complete metric space where

18 00:00:24,930 –> 00:00:26,659 the connection is given by

19 00:00:26,670 –> 00:00:27,280 the norm.

20 00:00:28,159 –> 00:00:29,899 Of course, now we can find

21 00:00:29,909 –> 00:00:31,420 a lot of examples.

22 00:00:32,209 –> 00:00:33,419 Let’s start with the simplest

23 00:00:33,430 –> 00:00:35,139 one, you can just consider

24 00:00:35,150 –> 00:00:36,380 the real number line.

25 00:00:37,380 –> 00:00:38,860 Therefore, we already know

26 00:00:38,889 –> 00:00:40,840 it’s a one dimensional real

27 00:00:40,849 –> 00:00:41,720 vector space.

28 00:00:42,500 –> 00:00:43,919 And we also know that we

29 00:00:43,930 –> 00:00:45,459 can calculate lengths

30 00:00:45,540 –> 00:00:47,180 if we consider the absolute

31 00:00:47,189 –> 00:00:48,799 value in the real numbers.

32 00:00:49,630 –> 00:00:50,819 Indeed with our

33 00:00:50,830 –> 00:00:52,400 definition, this is a

34 00:00:52,409 –> 00:00:54,400 norm and then

35 00:00:54,409 –> 00:00:56,180 you should know the associated

36 00:00:56,189 –> 00:00:57,959 metric is given by this

37 00:00:57,970 –> 00:00:58,599 formula.

38 00:00:59,509 –> 00:01:00,909 And that’s just the normal

39 00:01:00,919 –> 00:01:02,349 metric we have in the real

40 00:01:02,360 –> 00:01:02,930 numbers.

41 00:01:03,720 –> 00:01:05,319 And now from real analysis,

42 00:01:05,330 –> 00:01:06,809 you should know that all

43 00:01:06,819 –> 00:01:08,699 the Cauchy sequences indeed

44 00:01:08,709 –> 00:01:09,980 converge in R.

45 00:01:10,830 –> 00:01:12,760 In other words, are together

46 00:01:12,769 –> 00:01:14,180 with this norm is a Banach

47 00:01:14,370 –> 00:01:14,989 space.

48 00:01:15,980 –> 00:01:17,519 So this is our first.

49 00:01:17,529 –> 00:01:19,129 And as you can see a very

50 00:01:19,139 –> 00:01:20,209 simple example,

51 00:01:20,839 –> 00:01:22,290 however, we can consider

52 00:01:22,300 –> 00:01:23,910 even a simple example.

53 00:01:24,760 –> 00:01:26,059 And you might already guess

54 00:01:26,069 –> 00:01:27,410 instead of a one dimensional

55 00:01:27,419 –> 00:01:28,980 vector space, we choose a

56 00:01:28,989 –> 00:01:30,699 zero dimensional vector space.

57 00:01:31,599 –> 00:01:32,900 This means that the zero

58 00:01:32,910 –> 00:01:34,769 vector is the only vector

59 00:01:34,779 –> 00:01:35,430 in the space.

60 00:01:36,279 –> 00:01:37,760 And of course, there’s only

61 00:01:37,769 –> 00:01:39,699 one norm we can define

62 00:01:39,709 –> 00:01:41,569 here because the zero

63 00:01:41,580 –> 00:01:43,410 vector has to have length

64 00:01:43,419 –> 00:01:43,930 zero.

65 00:01:44,959 –> 00:01:46,300 Hence everything we need

66 00:01:46,309 –> 00:01:47,569 here is fulfilled.

67 00:01:47,739 –> 00:01:48,889 We have a Banach space.

68 00:01:50,069 –> 00:01:51,430 Of course, this is a very

69 00:01:51,440 –> 00:01:53,160 boring one and not very

70 00:01:53,169 –> 00:01:53,779 helpful.

71 00:01:54,589 –> 00:01:56,080 Indeed, often it’s a

72 00:01:56,089 –> 00:01:58,019 pathological one which means

73 00:01:58,029 –> 00:01:59,730 that some theorem we prove

74 00:01:59,739 –> 00:02:01,680 later for Banach spaces will

75 00:02:01,690 –> 00:02:03,260 not hold for this particular

76 00:02:03,269 –> 00:02:03,739 one here.

77 00:02:04,660 –> 00:02:06,050 Please keep that in mind

78 00:02:06,059 –> 00:02:07,489 if you want to use the zero

79 00:02:07,500 –> 00:02:09,449 Banach space as your example.

80 00:02:10,229 –> 00:02:11,820 However, now let’s continue

81 00:02:11,830 –> 00:02:13,300 with one of the most important

82 00:02:13,309 –> 00:02:14,059 examples.

83 00:02:14,919 –> 00:02:16,889 This is the LP space written

84 00:02:16,899 –> 00:02:18,190 as LP

85 00:02:18,380 –> 00:02:19,330 NF

86 00:02:20,089 –> 00:02:21,729 and denotes the natural numbers

87 00:02:21,740 –> 00:02:23,130 and F is our field.

88 00:02:23,139 –> 00:02:25,070 So the real or complex numbers

89 00:02:25,759 –> 00:02:27,339 and P is a real number

90 00:02:27,350 –> 00:02:28,960 greater or equal than one,

91 00:02:29,039 –> 00:02:30,740 but it could also be infinity.

92 00:02:31,550 –> 00:02:33,339 However, infinity is a special

93 00:02:33,350 –> 00:02:34,699 case we consider later.

94 00:02:34,820 –> 00:02:36,600 So here I exclude it.

95 00:02:37,690 –> 00:02:39,020 Now, as I said, this

96 00:02:39,029 –> 00:02:40,949 LPN F is defined

97 00:02:40,960 –> 00:02:42,509 as all the sequences in

98 00:02:42,520 –> 00:02:44,470 F that fulfill one condition.

99 00:02:45,229 –> 00:02:46,690 So you immediately see the

100 00:02:46,699 –> 00:02:48,009 notation here makes sense.

101 00:02:48,020 –> 00:02:49,610 This is the domain and this

102 00:02:49,619 –> 00:02:51,009 is the co domain for the

103 00:02:51,020 –> 00:02:52,729 map given by the sequence.

104 00:02:53,500 –> 00:02:54,899 And Now this condition is

105 00:02:54,910 –> 00:02:56,559 given by the series where

106 00:02:56,570 –> 00:02:58,039 you add up all the sequence

107 00:02:58,050 –> 00:02:59,960 members and we do it in the

108 00:02:59,970 –> 00:03:01,720 absolute value and to the

109 00:03:01,729 –> 00:03:02,589 power P.

110 00:03:03,740 –> 00:03:05,080 And if this series now

111 00:03:05,089 –> 00:03:06,960 converges, which means it’s

112 00:03:06,970 –> 00:03:08,399 less than infinity, then

113 00:03:08,410 –> 00:03:09,990 the sequence is an LP.

114 00:03:11,009 –> 00:03:12,449 Often from the context, it’s

115 00:03:12,460 –> 00:03:13,630 clear if we are dealing with

116 00:03:13,639 –> 00:03:15,229 real or complex numbers and

117 00:03:15,240 –> 00:03:16,789 what the index set of the

118 00:03:16,800 –> 00:03:17,820 sequences is.

119 00:03:17,830 –> 00:03:19,520 And therefore one just writes

120 00:03:19,529 –> 00:03:20,860 LP in this case.

121 00:03:22,059 –> 00:03:23,029 Now, the first thing you

122 00:03:23,039 –> 00:03:24,440 should note is that if we

123 00:03:24,449 –> 00:03:25,869 ignore this condition and

124 00:03:25,880 –> 00:03:27,649 just consider the sequences,

125 00:03:27,759 –> 00:03:29,559 then they form a vector space

126 00:03:30,240 –> 00:03:32,009 because adding and scaling

127 00:03:32,020 –> 00:03:33,729 is just given in a component

128 00:03:33,740 –> 00:03:34,529 wise sense.

129 00:03:35,449 –> 00:03:36,750 Therefore, here we just have

130 00:03:36,759 –> 00:03:38,410 to show that LP is a

131 00:03:38,419 –> 00:03:40,279 subspace of this bigger vector

132 00:03:40,289 –> 00:03:40,779 space.

133 00:03:41,990 –> 00:03:43,729 However, what we also want

134 00:03:43,740 –> 00:03:45,229 is a norm for this vector

135 00:03:45,240 –> 00:03:47,100 space and

136 00:03:47,110 –> 00:03:48,710 usually one just uses a

137 00:03:48,720 –> 00:03:50,369 P as an index.

138 00:03:51,250 –> 00:03:52,550 Now if we give the whole

139 00:03:52,559 –> 00:03:54,199 sequence, the name X,

140 00:03:54,500 –> 00:03:56,350 then we can calculate

141 00:03:56,360 –> 00:03:58,190 the P norm of X

142 00:03:58,199 –> 00:04:00,130 by using this number because

143 00:04:00,139 –> 00:04:01,419 we know it’s finite.

144 00:04:01,929 –> 00:04:03,130 However, you might already

145 00:04:03,139 –> 00:04:04,460 see that the second property

146 00:04:04,470 –> 00:04:05,729 of a norm is not fulfilled.

147 00:04:05,740 –> 00:04:07,559 In this case, we also need

148 00:04:07,570 –> 00:04:09,110 the PEF root outside.

149 00:04:10,080 –> 00:04:11,910 If you set P equals to two,

150 00:04:12,020 –> 00:04:13,380 you will recognize something

151 00:04:13,389 –> 00:04:15,039 similar to the Euclidean

152 00:04:15,050 –> 00:04:16,160 norm in RD.

153 00:04:17,048 –> 00:04:18,630 Indeed, with similar arguments,

154 00:04:18,640 –> 00:04:20,260 we can show that this now

155 00:04:20,269 –> 00:04:21,459 gives us a norm.

156 00:04:22,350 –> 00:04:23,739 I emphasize that this is

157 00:04:23,750 –> 00:04:25,200 not trivial to show that.

158 00:04:25,260 –> 00:04:26,850 And indeed, I skip that here,

159 00:04:26,859 –> 00:04:28,059 but we will do it later

160 00:04:29,200 –> 00:04:29,470 here.

161 00:04:29,480 –> 00:04:30,579 In this video, I want to

162 00:04:30,589 –> 00:04:32,549 focus how we get the completeness

163 00:04:32,559 –> 00:04:33,339 of the space.

164 00:04:34,170 –> 00:04:35,589 Hence, our claim is

165 00:04:35,600 –> 00:04:37,109 LP together with this

166 00:04:37,119 –> 00:04:38,899 norm is a Banach space.

167 00:04:39,700 –> 00:04:40,980 The first part of the proof

168 00:04:40,989 –> 00:04:42,910 has to be that LP is an F

169 00:04:42,920 –> 00:04:44,700 vector space and that this

170 00:04:44,709 –> 00:04:46,589 P norm is indeed a norm

171 00:04:46,600 –> 00:04:47,079 on it.

172 00:04:47,559 –> 00:04:49,239 I already told you we don’t

173 00:04:49,250 –> 00:04:49,839 do it here.

174 00:04:49,850 –> 00:04:51,549 We assume it, we will prove

175 00:04:51,559 –> 00:04:52,089 it later.

176 00:04:52,619 –> 00:04:53,769 That’s just because we need

177 00:04:53,779 –> 00:04:55,239 a lot of technical details

178 00:04:55,250 –> 00:04:55,959 for this.

179 00:04:55,989 –> 00:04:57,779 And I think it will distract

180 00:04:57,790 –> 00:04:59,429 us from the important completeness

181 00:04:59,440 –> 00:05:00,679 part of the proof here.

182 00:05:01,339 –> 00:05:02,559 Now, in order to show the

183 00:05:02,570 –> 00:05:04,070 completeness, we have to

184 00:05:04,079 –> 00:05:05,720 choose an arbitrary

185 00:05:05,950 –> 00:05:06,839 Cauchy sequence.

186 00:05:07,660 –> 00:05:09,079 So this might be hard to

187 00:05:09,089 –> 00:05:09,600 understand.

188 00:05:09,609 –> 00:05:11,600 Now the queen X is

189 00:05:11,609 –> 00:05:13,410 a sequence in F and

190 00:05:13,420 –> 00:05:14,920 now we have to consider

191 00:05:14,929 –> 00:05:16,380 sequences of

192 00:05:16,390 –> 00:05:17,200 sequences.

193 00:05:18,059 –> 00:05:19,230 Therefore, I choose here

194 00:05:19,239 –> 00:05:21,149 an upper index K to

195 00:05:21,160 –> 00:05:23,149 denote the different sequences

196 00:05:23,160 –> 00:05:25,029 we have in our Cauchy sequence.

197 00:05:25,750 –> 00:05:27,579 And now we have to show that

198 00:05:27,589 –> 00:05:29,160 all the sequences here in

199 00:05:29,170 –> 00:05:31,149 a row converge to another

200 00:05:31,160 –> 00:05:32,410 sequence, a limit.

201 00:05:33,529 –> 00:05:33,989 OK?

202 00:05:34,000 –> 00:05:35,519 To get the right idea, let’s

203 00:05:35,529 –> 00:05:37,470 write down some of the sequences.

204 00:05:38,410 –> 00:05:39,970 So here you see our Cauchy

205 00:05:39,980 –> 00:05:41,470 sequence which goes from

206 00:05:41,480 –> 00:05:42,320 top to bottom.

207 00:05:43,230 –> 00:05:44,859 Now, for example, our X one

208 00:05:44,869 –> 00:05:46,290 is a sequence in LP.

209 00:05:46,420 –> 00:05:48,079 So it consists of real or

210 00:05:48,089 –> 00:05:49,630 complex numbers in a row.

211 00:05:50,290 –> 00:05:51,350 And the same we can write

212 00:05:51,359 –> 00:05:53,209 down for X two X three and

213 00:05:53,220 –> 00:05:53,709 so on.

214 00:05:54,410 –> 00:05:55,500 Now, if you look at this

215 00:05:55,510 –> 00:05:57,230 picture, you see a lot of

216 00:05:57,239 –> 00:05:59,109 numbers in F in two different

217 00:05:59,119 –> 00:06:00,799 directions from left to

218 00:06:00,809 –> 00:06:02,500 right and from top to bottom.

219 00:06:03,269 –> 00:06:04,970 And please keep in mind in

220 00:06:04,980 –> 00:06:06,929 the end, we want the limit

221 00:06:06,940 –> 00:06:08,010 here on the left.

222 00:06:09,000 –> 00:06:10,459 In order to find this, we

223 00:06:10,470 –> 00:06:11,799 can just look at the different

224 00:06:11,809 –> 00:06:13,640 sequences we have here from

225 00:06:13,649 –> 00:06:14,549 top to bottom.

226 00:06:15,670 –> 00:06:16,989 So maybe you just take the

227 00:06:17,000 –> 00:06:18,529 fourth column here and then

228 00:06:18,540 –> 00:06:19,890 we look at the difference

229 00:06:19,899 –> 00:06:21,369 between two members of the

230 00:06:21,380 –> 00:06:22,010 sequence.

231 00:06:22,890 –> 00:06:24,470 So maybe we use K

232 00:06:24,600 –> 00:06:26,559 and L for the corresponding

233 00:06:26,570 –> 00:06:27,170 index.

234 00:06:28,040 –> 00:06:29,489 Now we want to put in the

235 00:06:29,500 –> 00:06:31,230 information we already have,

236 00:06:31,540 –> 00:06:32,910 which means we consider here

237 00:06:32,920 –> 00:06:34,079 the absolute value to the

238 00:06:34,089 –> 00:06:35,010 power of P.

239 00:06:35,700 –> 00:06:37,420 And then we know this is

240 00:06:37,429 –> 00:06:39,220 less or equal than

241 00:06:39,230 –> 00:06:40,570 summing up all other

242 00:06:40,579 –> 00:06:41,480 possibilities.

243 00:06:42,450 –> 00:06:43,809 That just means that we have

244 00:06:43,820 –> 00:06:45,529 here all the other

245 00:06:45,540 –> 00:06:46,549 columns as well.

246 00:06:47,690 –> 00:06:49,519 Of course, we do that because

247 00:06:49,529 –> 00:06:51,269 here we have our P

248 00:06:51,279 –> 00:06:52,579 norm to the power P

249 00:06:53,420 –> 00:06:55,119 and there we find our original

250 00:06:55,130 –> 00:06:56,600 sequences again where we

251 00:06:56,609 –> 00:06:58,320 know that they form a Cauchy

252 00:06:58,329 –> 00:06:58,920 sequence.

253 00:06:59,779 –> 00:07:01,200 Don’t worry, I’ll remind

254 00:07:01,209 –> 00:07:02,690 you what it means to be a

255 00:07:02,820 –> 00:07:03,750 Cauchy sequence here.

256 00:07:04,420 –> 00:07:06,209 For all epsilon greater zero,

257 00:07:06,220 –> 00:07:07,950 we find an index and

258 00:07:07,959 –> 00:07:09,209 maybe we call it capital

259 00:07:09,220 –> 00:07:11,070 K such that

260 00:07:11,079 –> 00:07:12,149 for all indices

261 00:07:12,160 –> 00:07:13,750 afterwards K and

262 00:07:13,760 –> 00:07:15,679 L we have that

263 00:07:15,690 –> 00:07:16,799 the distance between two

264 00:07:16,809 –> 00:07:18,679 members here is less than

265 00:07:18,690 –> 00:07:19,350 epsilon.

266 00:07:20,700 –> 00:07:22,390 Now, if you look at the inequality

267 00:07:22,399 –> 00:07:24,220 above, you see that this

268 00:07:24,230 –> 00:07:26,200 thing here is always greater

269 00:07:26,209 –> 00:07:27,279 than just looking at the

270 00:07:27,290 –> 00:07:29,119 absolute value in one column.

271 00:07:30,000 –> 00:07:31,279 Let’s put that in here.

272 00:07:31,290 –> 00:07:32,959 And then you can see this

273 00:07:32,970 –> 00:07:34,799 sequence here of normal

274 00:07:34,809 –> 00:07:36,070 numbers in F.

275 00:07:36,079 –> 00:07:37,670 So this column here is

276 00:07:37,679 –> 00:07:39,500 also a Cauchy sequence in

277 00:07:39,510 –> 00:07:39,940 F.

278 00:07:40,859 –> 00:07:42,390 The value is just the normal

279 00:07:42,399 –> 00:07:43,540 norm in F.

280 00:07:43,549 –> 00:07:44,559 So the real numbers or the

281 00:07:44,570 –> 00:07:45,399 complex numbers.

282 00:07:46,140 –> 00:07:47,390 And we already discussed

283 00:07:47,399 –> 00:07:48,880 this in the first example,

284 00:07:48,929 –> 00:07:50,399 this is a Banach space.

285 00:07:51,470 –> 00:07:53,119 In other words, this

286 00:07:53,130 –> 00:07:54,540 sequence has a limit

287 00:07:55,429 –> 00:07:56,609 and a suitable name should

288 00:07:56,619 –> 00:07:58,589 be x tilde with an index

289 00:07:58,600 –> 00:07:59,010 four.

290 00:08:00,010 –> 00:08:01,910 And this limit in F is what

291 00:08:01,920 –> 00:08:03,829 we find if we go from top

292 00:08:03,839 –> 00:08:05,410 to bottom in the fourth

293 00:08:05,420 –> 00:08:07,109 column, of

294 00:08:07,119 –> 00:08:08,450 course, the four was just

295 00:08:08,459 –> 00:08:09,250 an example.

296 00:08:09,260 –> 00:08:11,029 You can do all the things

297 00:08:11,040 –> 00:08:12,350 here with another column

298 00:08:12,359 –> 00:08:12,940 as well.

299 00:08:12,950 –> 00:08:14,510 So we can introduce an

300 00:08:14,519 –> 00:08:16,100 index M for example,

301 00:08:16,820 –> 00:08:18,079 and then I can change it

302 00:08:18,089 –> 00:08:19,500 here everywhere.

303 00:08:20,209 –> 00:08:21,899 Now, this means for all the

304 00:08:21,910 –> 00:08:23,320 columns here, we find the

305 00:08:23,329 –> 00:08:25,100 corresponding limit X till

306 00:08:25,109 –> 00:08:26,739 the M we have here at the

307 00:08:26,750 –> 00:08:27,200 bottom.

308 00:08:28,100 –> 00:08:29,850 So indeed, what we get here

309 00:08:29,859 –> 00:08:31,540 on the right is a new

310 00:08:31,549 –> 00:08:33,159 sequence with numbers in

311 00:08:33,169 –> 00:08:33,549 F.

312 00:08:34,780 –> 00:08:36,140 Of course, we can and we

313 00:08:36,150 –> 00:08:37,429 should give it the name X

314 00:08:37,469 –> 00:08:38,969 tilde, but we don’t know

315 00:08:38,979 –> 00:08:40,530 yet if it’s the limit here

316 00:08:40,539 –> 00:08:41,359 on the left.

317 00:08:42,299 –> 00:08:43,669 This is what we still have

318 00:08:43,679 –> 00:08:44,408 to show now.

319 00:08:45,369 –> 00:08:45,760 OK.

320 00:08:45,770 –> 00:08:47,659 So the plan is to show that

321 00:08:47,669 –> 00:08:49,659 XK minus X tilde

322 00:08:49,669 –> 00:08:51,619 in the P norm goes to zero.

323 00:08:52,260 –> 00:08:53,599 Or in other words, for an

324 00:08:53,609 –> 00:08:55,489 arbitrary epsilon, this should

325 00:08:55,500 –> 00:08:56,929 be smaller than epsilon.

326 00:08:57,770 –> 00:08:58,119 OK.

327 00:08:58,130 –> 00:08:59,299 So then let’s write down

328 00:08:59,309 –> 00:09:00,229 the P norm again.

329 00:09:01,099 –> 00:09:02,729 So we have this series and

330 00:09:02,739 –> 00:09:04,309 we need the P food again.

331 00:09:04,450 –> 00:09:05,590 But it’s easier to write

332 00:09:05,599 –> 00:09:06,479 the power of P.

333 00:09:06,489 –> 00:09:07,390 On the other side.

334 00:09:08,250 –> 00:09:09,559 This just means that we used

335 00:09:09,570 –> 00:09:11,469 to p-th root just in the end.

336 00:09:12,250 –> 00:09:13,400 The first idea now would

337 00:09:13,409 –> 00:09:15,090 be to push the limit process

338 00:09:15,099 –> 00:09:15,940 here outside.

339 00:09:15,950 –> 00:09:17,909 So we write the limit of

340 00:09:17,919 –> 00:09:19,219 N to infinity.

341 00:09:20,140 –> 00:09:21,530 So you see it’s completely

342 00:09:21,539 –> 00:09:22,159 the same.

343 00:09:22,169 –> 00:09:23,710 But now we can discuss this

344 00:09:23,719 –> 00:09:25,359 finite sum here on the right.

345 00:09:26,130 –> 00:09:27,580 However, we know that this

346 00:09:27,590 –> 00:09:29,039 X till the N is

347 00:09:29,049 –> 00:09:30,239 also a limit.

348 00:09:30,710 –> 00:09:32,280 Hence, we can push that limit

349 00:09:32,289 –> 00:09:34,260 out as well because the absolute

350 00:09:34,270 –> 00:09:35,989 value is a continuous function

351 00:09:36,000 –> 00:09:37,119 and we just have a finite

352 00:09:37,130 –> 00:09:37,679 sum here.

353 00:09:38,510 –> 00:09:39,830 Now, if you look back, you

354 00:09:39,840 –> 00:09:41,460 will see that this thing

355 00:09:41,469 –> 00:09:42,969 here is what we can keep

356 00:09:42,979 –> 00:09:44,729 as small as we want.

357 00:09:45,599 –> 00:09:46,900 We use again that we have

358 00:09:46,909 –> 00:09:48,320 here the property of the

359 00:09:48,530 –> 00:09:49,349 Cauchy sequence.

360 00:09:50,400 –> 00:09:51,659 Therefore, for our given

361 00:09:51,669 –> 00:09:53,400 epsilon, we choose our capital

362 00:09:53,409 –> 00:09:55,140 K such that we have this

363 00:09:56,039 –> 00:09:56,799 at this point.

364 00:09:56,809 –> 00:09:58,080 In the proof, we can’t know

365 00:09:58,090 –> 00:09:59,640 how small this really should

366 00:09:59,650 –> 00:10:00,099 be.

367 00:10:00,109 –> 00:10:01,559 Therefore, to be careful,

368 00:10:01,570 –> 00:10:03,229 maybe we introduce an epsilon

369 00:10:03,239 –> 00:10:05,080 prime and define it later

370 00:10:05,090 –> 00:10:05,469 on.

371 00:10:06,140 –> 00:10:07,599 Now, what we know from above

372 00:10:07,609 –> 00:10:09,599 is that this whole thing

373 00:10:09,840 –> 00:10:11,609 is less than epsilon prime

374 00:10:11,619 –> 00:10:12,559 to the power P

375 00:10:13,520 –> 00:10:15,330 as long as K and L

376 00:10:15,340 –> 00:10:16,760 are bigger than the capital

377 00:10:16,770 –> 00:10:17,219 K.

378 00:10:17,229 –> 00:10:18,539 It’s the same reasoning as

379 00:10:18,549 –> 00:10:18,900 here.

380 00:10:20,309 –> 00:10:20,679 OK.

381 00:10:20,690 –> 00:10:21,979 So this is good information

382 00:10:21,989 –> 00:10:23,799 for us because it tells you

383 00:10:23,809 –> 00:10:24,789 we have here the nonne

384 00:10:25,270 –> 00:10:26,880 numbers that are bounded

385 00:10:26,890 –> 00:10:28,679 from above by this constant,

386 00:10:29,210 –> 00:10:30,640 of course, then comes some

387 00:10:30,650 –> 00:10:31,710 limit process.

388 00:10:31,750 –> 00:10:33,309 But we know in the worst

389 00:10:33,320 –> 00:10:34,869 case, this limit is

390 00:10:34,880 –> 00:10:36,599 exactly this upper bound.

391 00:10:37,359 –> 00:10:39,320 So in summary, we know for

392 00:10:39,330 –> 00:10:40,859 all K greater or capital

393 00:10:40,869 –> 00:10:42,739 K, this whole thing

394 00:10:42,750 –> 00:10:44,609 is less or equal

395 00:10:44,619 –> 00:10:45,780 than epsilon prime to the

396 00:10:45,789 –> 00:10:46,580 power P.

397 00:10:47,440 –> 00:10:48,869 And now is the point to take

398 00:10:48,880 –> 00:10:50,380 the pea food because we get

399 00:10:50,390 –> 00:10:52,119 rid of these two peas.

400 00:10:53,169 –> 00:10:54,710 And now to get to the correct

401 00:10:54,719 –> 00:10:56,349 result, we need epsilon prime

402 00:10:56,359 –> 00:10:58,229 to be less than epsilon itself,

403 00:10:58,559 –> 00:11:00,109 which means we can define

404 00:11:00,119 –> 00:11:01,909 epsilon prime, for example,

405 00:11:02,030 –> 00:11:03,390 as epsilon half.

406 00:11:04,440 –> 00:11:05,599 And now in the end, you can

407 00:11:05,609 –> 00:11:07,150 put everything together for

408 00:11:07,159 –> 00:11:08,840 an arbitrarily chosen epsilon

409 00:11:08,849 –> 00:11:10,169 greater zero, you find a

410 00:11:10,179 –> 00:11:12,049 capital K such that for

411 00:11:12,059 –> 00:11:13,950 all indices after the capital

412 00:11:13,960 –> 00:11:15,380 K, you find the

413 00:11:15,390 –> 00:11:16,940 inequality here, the

414 00:11:16,950 –> 00:11:18,489 distance between XK

415 00:11:18,500 –> 00:11:20,359 minus X tilde is less

416 00:11:20,369 –> 00:11:22,349 than epsilon and there we

417 00:11:22,359 –> 00:11:23,830 have it X tilde is

418 00:11:23,840 –> 00:11:25,469 indeed the limit of

419 00:11:25,479 –> 00:11:27,190 XK with respect to the

420 00:11:27,200 –> 00:11:27,619 p-norm.

421 00:11:28,909 –> 00:11:30,299 The only thing missing is

422 00:11:30,309 –> 00:11:32,090 the explanation why X tilde

423 00:11:32,099 –> 00:11:33,539 is also an element in

424 00:11:33,549 –> 00:11:34,130 LP.

425 00:11:34,809 –> 00:11:36,289 However, this now follows

426 00:11:36,299 –> 00:11:38,090 from all we know we know

427 00:11:38,099 –> 00:11:39,650 here that the difference

428 00:11:39,659 –> 00:11:40,729 is an LP.

429 00:11:41,000 –> 00:11:42,669 And we also know that XK

430 00:11:42,679 –> 00:11:43,570 is an LP.

431 00:11:44,400 –> 00:11:45,669 Therefore, if you write X

432 00:11:45,679 –> 00:11:47,049 tilde in this way, which

433 00:11:47,059 –> 00:11:48,969 means you subtract XK and

434 00:11:48,979 –> 00:11:49,760 add it again.

435 00:11:49,859 –> 00:11:51,729 Then you see the linear combination,

436 00:11:52,710 –> 00:11:54,609 which means here’s one part

437 00:11:54,619 –> 00:11:56,539 in LP and here’s one

438 00:11:56,549 –> 00:11:57,530 part in LP.

439 00:11:58,690 –> 00:12:00,159 Hence also the sum is an

440 00:12:00,169 –> 00:12:02,070 LP because we are dealing

441 00:12:02,080 –> 00:12:03,179 with a vector space.

442 00:12:04,109 –> 00:12:05,239 That’s the part of the proof

443 00:12:05,250 –> 00:12:06,409 at the beginning where we

444 00:12:06,419 –> 00:12:07,539 skip the details.

445 00:12:07,549 –> 00:12:08,780 But here we can use it.

446 00:12:09,570 –> 00:12:11,049 Of course, we will discuss

447 00:12:11,059 –> 00:12:12,690 this even further in upcoming

448 00:12:12,700 –> 00:12:13,489 videos.

449 00:12:13,500 –> 00:12:14,979 But at this point, I really

450 00:12:14,989 –> 00:12:16,830 wanted to show you an important

451 00:12:16,849 –> 00:12:18,549 and non trivial example of

452 00:12:18,559 –> 00:12:19,349 a Banach space.

453 00:12:20,200 –> 00:12:20,599 OK?

454 00:12:20,609 –> 00:12:21,580 I hope I see you in the next

455 00:12:21,590 –> 00:12:23,369 videos where I first explain

456 00:12:23,380 –> 00:12:24,869 more concepts in functional

457 00:12:24,880 –> 00:12:25,580 analysis.

458 00:12:25,590 –> 00:12:27,099 And afterwards we go into

459 00:12:27,109 –> 00:12:28,940 more details for examples.

460 00:12:29,919 –> 00:12:31,440 So thanks for listening and

461 00:12:31,450 –> 00:12:32,210 see you later.

462 00:12:32,219 –> 00:12:32,859 Bye.

Q1: Is $\mathbb{R}$ given with $| x | = 2 |x|$ a Banach space?

A1: Yes!

A2: No!

A3: One needs more information.

Q2: Is $\mathbb{R}^2$ given with $\left| \binom{x_1}{x_2} \right| = \max(x_1, x_2)$ a Banach space?

A1: No!

A2: Yes!

A3: One needs more information.

Q3: Is $\mathbb{R}^2$ given with $\left| \binom{x_1}{x_2} \right| = |x_1| + |x_2|$ a Banach space?

A1: Yes!

A2: No!

A3: One needs more information.