00:00 Introduction

00:20 Definition - continuity

09:56 Orthogonal complement is closed

1 00:00:00,330 –> 00:00:02,269 Hello and welcome back to

2 00:00:02,279 –> 00:00:03,569 functional analysis

3 00:00:04,239 –> 00:00:05,719 and many, many thanks to

4 00:00:05,730 –> 00:00:07,179 all my great supporters on

5 00:00:07,190 –> 00:00:08,390 Steady and PayPal.

6 00:00:08,539 –> 00:00:10,029 You really give me the motivation

7 00:00:10,039 –> 00:00:11,829 to put all this work in making

8 00:00:11,840 –> 00:00:12,810 these videos.

9 00:00:12,819 –> 00:00:14,770 So thanks a lot! In

10 00:00:14,779 –> 00:00:15,800 functional analysis

11 00:00:15,810 –> 00:00:17,159 we’ve already reached part

12 00:00:17,170 –> 00:00:17,629 12

13 00:00:17,719 –> 00:00:19,329 and today we will talk about

14 00:00:19,340 –> 00:00:20,479 continuity

15 00:00:21,500 –> 00:00:22,549 to be more precise.

16 00:00:22,559 –> 00:00:24,309 We will talk about continuity

17 00:00:24,319 –> 00:00:25,709 for metric spaces.

18 00:00:26,510 –> 00:00:28,069 For this, we just need two

19 00:00:28,079 –> 00:00:29,209 metric spaces.

20 00:00:29,219 –> 00:00:30,809 So one set X with a

21 00:00:30,819 –> 00:00:32,790 metric d_X and a set

22 00:00:32,799 –> 00:00:34,250 Y with a metric d_Y

23 00:00:35,340 –> 00:00:37,080 and now we consider a map

24 00:00:37,090 –> 00:00:38,840 from X to Y.

25 00:00:39,880 –> 00:00:41,779 And now such a map is called

26 00:00:41,790 –> 00:00:42,810 continuous,

27 00:00:43,630 –> 00:00:45,389 If the preimages of

28 00:00:45,400 –> 00:00:47,299 open sets are always

29 00:00:47,310 –> 00:00:48,819 open. More

30 00:00:48,830 –> 00:00:50,080 concretely, we would write

31 00:00:50,090 –> 00:00:51,709 the preimage of a set

32 00:00:51,720 –> 00:00:53,020 B in Y.

33 00:00:54,380 –> 00:00:56,130 So now this usually denotes

34 00:00:56,139 –> 00:00:57,580 the preimage of B, but I

35 00:00:57,590 –> 00:00:58,509 want to make that really

36 00:00:58,520 –> 00:00:59,049 clear.

37 00:00:59,060 –> 00:01:00,529 So I want to use brackets

38 00:01:00,540 –> 00:01:00,840 here.

39 00:01:01,819 –> 00:01:03,520 I think that’s helpful, because

40 00:01:03,529 –> 00:01:05,190 every time you see the brackets,

41 00:01:05,199 –> 00:01:06,489 you know this is not the

42 00:01:06,500 –> 00:01:07,710 inverse map, it’s the

43 00:01:07,740 –> 00:01:08,430 preimage.

44 00:01:09,629 –> 00:01:09,919 OK.

45 00:01:09,930 –> 00:01:11,559 Now we say this set is

46 00:01:11,569 –> 00:01:13,080 open in

47 00:01:13,089 –> 00:01:14,760 X for

48 00:01:14,769 –> 00:01:16,739 any open set B in

49 00:01:16,750 –> 00:01:18,540 Y. OK.

50 00:01:18,550 –> 00:01:20,379 So this is the formal definition

51 00:01:20,519 –> 00:01:21,639 and now I can give you a

52 00:01:21,650 –> 00:01:23,300 picture to remember it.

53 00:01:24,110 –> 00:01:25,519 So we have our metric space

54 00:01:25,529 –> 00:01:26,849 X on the left-hand side,

55 00:01:26,860 –> 00:01:27,900 on the right-hand side, we

56 00:01:27,910 –> 00:01:29,379 have our metric space Y

57 00:01:29,440 –> 00:01:31,069 and in between, we have the

58 00:01:31,080 –> 00:01:31,820 map f.

59 00:01:33,000 –> 00:01:34,389 Now, for any open set

60 00:01:34,400 –> 00:01:35,779 B in Y,

61 00:01:36,760 –> 00:01:38,040 we find the corresponding

62 00:01:38,050 –> 00:01:39,540 preimage on the left-hand

63 00:01:39,550 –> 00:01:39,940 side.

64 00:01:40,720 –> 00:01:42,480 Now, the property continuous

65 00:01:42,489 –> 00:01:44,000 means that, if you start with

66 00:01:44,010 –> 00:01:45,419 an open set here, you’ll

67 00:01:45,430 –> 00:01:47,019 also get an open set here.

68 00:01:47,550 –> 00:01:48,919 By thinking about boundary

69 00:01:48,930 –> 00:01:49,339 points,

70 00:01:49,349 –> 00:01:50,980 this means that if you don’t

71 00:01:50,989 –> 00:01:52,199 have any boundary points

72 00:01:52,209 –> 00:01:53,940 in this set, you also don’t

73 00:01:53,949 –> 00:01:55,250 have any boundary points

74 00:01:55,260 –> 00:01:56,059 in this set.

75 00:01:56,480 –> 00:01:57,709 In fact, you can also read

76 00:01:57,720 –> 00:01:59,519 that from left to right

77 00:01:59,529 –> 00:02:01,209 by looking here at an input

78 00:02:01,970 –> 00:02:03,309 and you change the input

79 00:02:03,319 –> 00:02:05,099 a little bit by a small distance.

80 00:02:05,110 –> 00:02:06,230 You know, this is always

81 00:02:06,239 –> 00:02:07,699 possible, because you are

82 00:02:07,709 –> 00:02:08,589 in an open set

83 00:02:09,268 –> 00:02:10,458 and then you know, on the

84 00:02:10,467 –> 00:02:12,378 right-hand side, the output

85 00:02:12,389 –> 00:02:13,968 is also only changed

86 00:02:13,979 –> 00:02:15,289 inside this open set.

87 00:02:16,210 –> 00:02:17,669 Of course, this notion seems

88 00:02:17,679 –> 00:02:18,770 a little bit abstract.

89 00:02:18,779 –> 00:02:19,979 Therefore, now we look at

90 00:02:19,990 –> 00:02:21,770 something we call sequentially

91 00:02:21,779 –> 00:02:22,550 continuous.

92 00:02:23,360 –> 00:02:25,000 As the name tells you this

93 00:02:25,009 –> 00:02:26,250 has something to do with

94 00:02:26,259 –> 00:02:27,300 sequences.

95 00:02:27,979 –> 00:02:29,259 In some sense, I can repeat

96 00:02:29,270 –> 00:02:30,820 what I told you before. We

97 00:02:30,830 –> 00:02:32,479 choose any output on the

98 00:02:32,490 –> 00:02:33,380 left-hand side

99 00:02:33,960 –> 00:02:35,639 and let’s call it x tilde

100 00:02:36,779 –> 00:02:38,619 and also let’s look at any

101 00:02:38,630 –> 00:02:40,479 sequence x_n that

102 00:02:40,490 –> 00:02:42,039 converges to x tilde

103 00:02:42,770 –> 00:02:44,350 and then it should hold that

104 00:02:44,360 –> 00:02:46,059 the new sequence f of

105 00:02:46,070 –> 00:02:47,869 x_n is convergent

106 00:02:47,880 –> 00:02:49,339 to f of x tilde.

107 00:02:50,380 –> 00:02:51,720 In other words, every

108 00:02:51,729 –> 00:02:52,940 convergent sequence on the

109 00:02:52,949 –> 00:02:54,300 left-hand side gets

110 00:02:54,309 –> 00:02:56,080 translated into a convergent

111 00:02:56,089 –> 00:02:56,520 sequence

112 00:02:56,529 –> 00:02:57,699 on the right-hand side.

113 00:02:58,449 –> 00:02:59,860 you might already guess for

114 00:02:59,869 –> 00:03:01,039 us, it might be easier to

115 00:03:01,050 –> 00:03:02,820 work with the notion sequentially

116 00:03:02,830 –> 00:03:04,679 continuous, because we just

117 00:03:04,690 –> 00:03:05,960 have to deal with sequences

118 00:03:05,970 –> 00:03:06,279 there.

119 00:03:07,039 –> 00:03:08,339 Now, the good thing is that

120 00:03:08,350 –> 00:03:10,179 for metric spaces, both

121 00:03:10,190 –> 00:03:11,740 terms mean indeed the same

122 00:03:11,750 –> 00:03:12,130 thing.

123 00:03:13,020 –> 00:03:14,740 In other words, one can prove

124 00:03:14,750 –> 00:03:16,580 that they are in fact equivalent.

125 00:03:17,369 –> 00:03:18,550 For that reason, we will

126 00:03:18,559 –> 00:03:20,369 simply use the word continuous,

127 00:03:20,479 –> 00:03:21,929 but often mean this

128 00:03:21,940 –> 00:03:23,759 characterization with sequences.

129 00:03:24,660 –> 00:03:25,889 However, I wanted to show

130 00:03:25,899 –> 00:03:27,649 you both definitions, because

131 00:03:27,660 –> 00:03:29,190 if you don’t work in metric

132 00:03:29,199 –> 00:03:29,889 spaces,

133 00:03:29,899 –> 00:03:31,669 but in general topological

134 00:03:31,679 –> 00:03:33,309 spaces, in general

135 00:03:33,320 –> 00:03:34,720 both notions are not

136 00:03:34,729 –> 00:03:35,429 equivalent.

137 00:03:36,309 –> 00:03:36,759 OK.

138 00:03:36,770 –> 00:03:38,160 So I write down the fact

139 00:03:38,169 –> 00:03:40,160 and then we can look at examples.

140 00:03:40,759 –> 00:03:42,199 OK, maybe you already know

141 00:03:42,210 –> 00:03:43,440 a lot of examples.

142 00:03:43,520 –> 00:03:45,080 So let’s look at some simple

143 00:03:45,089 –> 00:03:45,619 ones.

144 00:03:46,410 –> 00:03:48,190 The first one would be choosing

145 00:03:48,199 –> 00:03:49,630 (X, d_x) as the

146 00:03:49,639 –> 00:03:51,110 discrete metric space.

147 00:03:51,820 –> 00:03:53,649 So where X is any non

148 00:03:53,720 –> 00:03:54,429 empty set.

149 00:03:55,160 –> 00:03:56,339 Now, on the right-hand side,

150 00:03:56,350 –> 00:03:57,419 it does not matter which

151 00:03:57,429 –> 00:03:58,580 metric space we choose.

152 00:03:58,589 –> 00:04:00,490 So let’s say it’s any metric

153 00:04:00,500 –> 00:04:00,940 space.

154 00:04:02,089 –> 00:04:03,330 Now let’s look at a map

155 00:04:03,339 –> 00:04:05,259 f from X to Y

156 00:04:06,339 –> 00:04:07,720 and then we can conclude

157 00:04:07,729 –> 00:04:09,520 that this one has to be

158 00:04:09,529 –> 00:04:10,320 continuous.

159 00:04:11,020 –> 00:04:12,669 This is simply the case, because

160 00:04:12,679 –> 00:04:14,039 all subsets in a

161 00:04:14,050 –> 00:04:16,040 discrete metric space are

162 00:04:16,048 –> 00:04:16,709 open sets.

163 00:04:17,410 –> 00:04:18,839 This means that checking

164 00:04:18,850 –> 00:04:20,608 the definition of continuity

165 00:04:20,690 –> 00:04:22,089 is immediately given here.

166 00:04:22,890 –> 00:04:24,540 So keep that in mind, whenever

167 00:04:24,549 –> 00:04:25,790 you see a discrete metric

168 00:04:25,799 –> 00:04:27,299 space on the left-hand side.

169 00:04:27,940 –> 00:04:29,500 Now another general example

170 00:04:29,510 –> 00:04:31,260 like this would be to consider

171 00:04:31,269 –> 00:04:32,440 a constant function.

172 00:04:32,570 –> 00:04:34,429 For this, we can choose X

173 00:04:34,440 –> 00:04:35,950 and Y as general metric

174 00:04:35,959 –> 00:04:37,640 spaces and then

175 00:04:37,649 –> 00:04:39,070 fix a y_0

176 00:04:39,079 –> 00:04:39,720 in Y.

177 00:04:40,600 –> 00:04:42,179 In this case, then we

178 00:04:42,190 –> 00:04:43,899 know that the map

179 00:04:43,910 –> 00:04:45,540 f that sends all the

180 00:04:45,549 –> 00:04:47,260 x to y_0

181 00:04:47,420 –> 00:04:49,059 is always continuous.

182 00:04:50,029 –> 00:04:51,250 Indeed, this is what you

183 00:04:51,260 –> 00:04:52,549 see immediately by looking

184 00:04:52,559 –> 00:04:54,220 at the definition of sequentially

185 00:04:54,230 –> 00:04:55,010 continuous.

186 00:04:55,760 –> 00:04:56,910 Simply by looking at this

187 00:04:56,920 –> 00:04:58,309 part, where you know this

188 00:04:58,320 –> 00:04:59,670 is always fulfilled.

189 00:05:00,390 –> 00:05:01,670 The left-hand side is always

190 00:05:01,679 –> 00:05:02,410 y_0.

191 00:05:02,420 –> 00:05:03,790 The right-hand side is always

192 00:05:03,799 –> 00:05:04,429 y_0.

193 00:05:04,440 –> 00:05:06,209 So we stay at the limit point

194 00:05:06,220 –> 00:05:08,089 for all n. OK,

195 00:05:08,100 –> 00:05:09,269 with this in mind, let’s

196 00:05:09,279 –> 00:05:11,040 go to two other examples

197 00:05:11,049 –> 00:05:12,559 I find very important.

198 00:05:13,380 –> 00:05:14,820 For the first example, I

199 00:05:14,829 –> 00:05:16,679 want to look at a normed space.

200 00:05:17,309 –> 00:05:18,549 Of course, we always have

201 00:05:18,559 –> 00:05:19,950 the induced metric space

202 00:05:19,959 –> 00:05:20,440 in mind.

203 00:05:21,299 –> 00:05:23,170 And Y should be just R

204 00:05:23,230 –> 00:05:24,700 with the standard metric

205 00:05:24,709 –> 00:05:26,239 given by the absolute value

206 00:05:26,920 –> 00:05:28,339 and now the map I want to

207 00:05:28,350 –> 00:05:30,059 consider is f given

208 00:05:30,070 –> 00:05:31,700 by X to R,

209 00:05:32,269 –> 00:05:33,760 where an element x is

210 00:05:33,769 –> 00:05:35,299 sent to the norm of

211 00:05:35,309 –> 00:05:35,839 x

212 00:05:36,339 –> 00:05:38,140 and now our claim is; that

213 00:05:38,149 –> 00:05:39,799 is always continuous.

214 00:05:40,920 –> 00:05:41,260 OK.

215 00:05:41,269 –> 00:05:42,459 So for this, let’s write

216 00:05:42,470 –> 00:05:43,339 down a proof.

217 00:05:43,350 –> 00:05:44,980 So we check that the map

218 00:05:44,989 –> 00:05:46,320 is indeed sequentially

219 00:05:46,329 –> 00:05:47,059 continuous.

220 00:05:47,959 –> 00:05:49,559 Hence let x_n be a

221 00:05:49,570 –> 00:05:51,290 sequence in X with

222 00:05:51,299 –> 00:05:53,019 limit point x tilde

223 00:05:53,880 –> 00:05:55,380 and then we look what happens

224 00:05:55,390 –> 00:05:56,980 when we apply the map f.

225 00:05:57,700 –> 00:05:59,619 Of course, f of x_n is just

226 00:05:59,630 –> 00:06:01,579 by definition the norm of

227 00:06:01,589 –> 00:06:02,100 x_n.

228 00:06:03,130 –> 00:06:04,410 Now we have to bring in the

229 00:06:04,420 –> 00:06:06,109 limit point, because that’s

230 00:06:06,119 –> 00:06:07,489 the only thing we know about

231 00:06:07,500 –> 00:06:09,070 the sequence, it has a limit.

232 00:06:09,980 –> 00:06:11,570 So what I do is subtract

233 00:06:11,579 –> 00:06:13,119 the limit and add it again.

234 00:06:13,160 –> 00:06:14,779 So we add a zero here

235 00:06:14,790 –> 00:06:16,600 inside the norm. We

236 00:06:16,609 –> 00:06:18,489 do that, because then we

237 00:06:18,500 –> 00:06:19,730 can apply the triangle

238 00:06:19,739 –> 00:06:20,529 inequality.

239 00:06:20,540 –> 00:06:22,299 So I write x_n minus

240 00:06:22,309 –> 00:06:23,839 x tilde in the norm

241 00:06:23,850 –> 00:06:25,529 plus the norm of x tilde

242 00:06:26,630 –> 00:06:28,140 and then we know, the first

243 00:06:28,149 –> 00:06:29,529 part is just the distance

244 00:06:29,540 –> 00:06:31,239 between x_n and x

245 00:06:31,250 –> 00:06:31,700 tilde

246 00:06:31,910 –> 00:06:33,779 and the second part is just

247 00:06:33,790 –> 00:06:34,970 f of x tilde.

248 00:06:35,820 –> 00:06:37,549 By assumption we now know

249 00:06:37,589 –> 00:06:38,730 that the distance between

250 00:06:38,739 –> 00:06:40,640 x_n and x tilde goes to

251 00:06:40,649 –> 00:06:42,260 zero when n goes to

252 00:06:42,269 –> 00:06:42,959 infinity.

253 00:06:43,799 –> 00:06:45,220 Hence doing the limit on both

254 00:06:45,230 –> 00:06:45,679 sides,

255 00:06:45,690 –> 00:06:47,140 we get that the limit of

256 00:06:47,149 –> 00:06:48,910 f(x_n) is less or

257 00:06:48,920 –> 00:06:50,559 equal than f(x tilde)

258 00:06:51,640 –> 00:06:51,959 OK.

259 00:06:51,970 –> 00:06:53,339 So you see that’s already

260 00:06:53,350 –> 00:06:54,450 half of the work we have

261 00:06:54,459 –> 00:06:54,920 to do.

262 00:06:55,040 –> 00:06:56,380 So now let’s go from the

263 00:06:56,390 –> 00:06:57,779 other side and show the other

264 00:06:57,790 –> 00:06:58,619 inequality.

265 00:06:59,149 –> 00:07:00,209 Therefore, we start with

266 00:07:00,220 –> 00:07:01,989 f of x tilde, which is just

267 00:07:02,000 –> 00:07:03,579 the norm of x tilde.

268 00:07:04,380 –> 00:07:05,609 Doing a similar thing than

269 00:07:05,619 –> 00:07:07,160 before we just put in

270 00:07:07,170 –> 00:07:08,239 x_n.

271 00:07:09,140 –> 00:07:10,959 So subtracting and adding

272 00:07:10,970 –> 00:07:12,660 again. Of course,

273 00:07:12,670 –> 00:07:14,179 now we doing the triangle

274 00:07:14,190 –> 00:07:16,179 inequality gives us x tilde

275 00:07:16,190 –> 00:07:17,690 minus x_n in the

276 00:07:17,700 –> 00:07:19,450 norm plus the norm of

277 00:07:19,459 –> 00:07:21,239 x_n, which is a

278 00:07:21,250 –> 00:07:22,820 similar result as before,

279 00:07:22,829 –> 00:07:24,660 because we have x tilde,

280 00:07:24,670 –> 00:07:26,149 x_n in the metric

281 00:07:26,230 –> 00:07:27,769 plus f of

282 00:07:27,779 –> 00:07:28,250 x_n.

283 00:07:29,010 –> 00:07:29,989 Now, I don’t have to tell

284 00:07:30,000 –> 00:07:31,309 you, applying the limit on

285 00:07:31,320 –> 00:07:32,109 both sides,

286 00:07:32,119 –> 00:07:33,829 we get the inequality the

287 00:07:33,839 –> 00:07:34,589 other way around

288 00:07:35,429 –> 00:07:36,720 and with that, our proof

289 00:07:36,730 –> 00:07:38,130 is finished, because we have

290 00:07:38,140 –> 00:07:39,799 shown that the sequence f

291 00:07:39,809 –> 00:07:41,790 of x_n converges

292 00:07:41,799 –> 00:07:43,279 with respect to the metric

293 00:07:43,290 –> 00:07:45,170 in R to f of x tilde.

294 00:07:45,899 –> 00:07:47,649 So we conclude the norm is

295 00:07:47,660 –> 00:07:49,309 always a continuous map.

296 00:07:49,910 –> 00:07:50,410 OK.

297 00:07:50,420 –> 00:07:51,700 So you might guess with the

298 00:07:51,709 –> 00:07:53,329 next example, we go to an

299 00:07:53,339 –> 00:07:54,410 inner product space.

300 00:07:54,950 –> 00:07:56,579 So let’s choose X as an inner

301 00:07:56,589 –> 00:07:58,130 product space, where we already

302 00:07:58,140 –> 00:07:59,910 know it’s also a metric

303 00:07:59,920 –> 00:08:00,470 space

304 00:08:01,049 –> 00:08:02,369 and Y should just be the

305 00:08:02,380 –> 00:08:04,089 complex numbers with the

306 00:08:04,100 –> 00:08:05,709 standard metric given by

307 00:08:05,720 –> 00:08:06,649 the absolute value.

308 00:08:07,029 –> 00:08:08,239 Now, since an inner product

309 00:08:08,250 –> 00:08:09,700 has two entries, I want to

310 00:08:09,709 –> 00:08:11,209 fix the first one with a

311 00:08:11,220 –> 00:08:12,670 given x_zero.

312 00:08:13,260 –> 00:08:14,730 Now the map we consider is

313 00:08:14,739 –> 00:08:16,570 f defined from X to

314 00:08:16,579 –> 00:08:17,079 Y.

315 00:08:17,089 –> 00:08:18,799 So C and the

316 00:08:18,809 –> 00:08:20,630 vector X is sent to the inner

317 00:08:20,640 –> 00:08:22,549 product x_0 with

318 00:08:22,559 –> 00:08:23,079 x

319 00:08:23,670 –> 00:08:25,279 and the claim is again, this

320 00:08:25,290 –> 00:08:27,200 map is a continuous one.

321 00:08:28,089 –> 00:08:29,679 As before, I really want

322 00:08:29,690 –> 00:08:30,910 to show you a proof of this

323 00:08:30,920 –> 00:08:31,510 statement.

324 00:08:32,020 –> 00:08:33,109 Of course, the procedure

325 00:08:33,119 –> 00:08:34,400 should be the same as before.

326 00:08:34,409 –> 00:08:35,919 We consider any point x

327 00:08:35,950 –> 00:08:37,659 tilde and a sequence that

328 00:08:37,669 –> 00:08:39,099 is convergent to this point.

329 00:08:39,659 –> 00:08:41,030 However, now, maybe we can

330 00:08:41,039 –> 00:08:42,929 do it in one step when considering

331 00:08:42,940 –> 00:08:44,739 the absolute value of f of

332 00:08:44,750 –> 00:08:46,729 x_n minus f of x

333 00:08:46,739 –> 00:08:47,150 tilde.

334 00:08:47,869 –> 00:08:49,010 In other words, this is the

335 00:08:49,020 –> 00:08:50,469 inner product x_0 with

336 00:08:50,479 –> 00:08:52,340 x_n and x_0 with

337 00:08:52,349 –> 00:08:54,280 x tilde. Using

338 00:08:54,289 –> 00:08:55,609 the linearity, we can put

339 00:08:55,619 –> 00:08:56,219 that together.

340 00:08:56,229 –> 00:08:57,510 So we have just one inner

341 00:08:57,520 –> 00:08:58,119 product.

342 00:08:58,849 –> 00:09:00,270 Now, every time you see something

343 00:09:00,280 –> 00:09:01,630 like that, you know, you

344 00:09:01,640 –> 00:09:03,270 can apply Cauchy-Schwarz.

345 00:09:04,000 –> 00:09:05,159 This means that we have the

346 00:09:05,169 –> 00:09:07,099 inequality I will abbreviate

347 00:09:07,109 –> 00:09:08,229 with C.S

348 00:09:08,940 –> 00:09:10,159 and it tells us we have the

349 00:09:10,169 –> 00:09:11,760 norm of x_0 times the

350 00:09:11,770 –> 00:09:13,460 norm of x_n minus

351 00:09:13,510 –> 00:09:14,260 x tilde.

352 00:09:14,750 –> 00:09:16,179 Since this last part is the

353 00:09:16,190 –> 00:09:17,890 distance between x_n and x

354 00:09:17,900 –> 00:09:19,820 tilde, we know this goes

355 00:09:19,830 –> 00:09:21,320 to zero when n goes to

356 00:09:21,330 –> 00:09:21,979 infinity.

357 00:09:22,710 –> 00:09:24,090 Moreover, the whole right-

358 00:09:24,099 –> 00:09:25,650 hand side goes to zero

359 00:09:25,659 –> 00:09:27,239 when n goes to infinity.

360 00:09:27,940 –> 00:09:29,260 In conclusion, also the left-

361 00:09:29,270 –> 00:09:31,030 hand side goes to zero, because

362 00:09:31,039 –> 00:09:32,359 it was non-negative from

363 00:09:32,369 –> 00:09:32,940 the beginning

364 00:09:33,830 –> 00:09:35,479 and then we know the sequence

365 00:09:35,489 –> 00:09:37,440 f(x_n) converges to f

366 00:09:37,450 –> 00:09:38,150 of x tilde.

367 00:09:39,090 –> 00:09:39,469 OK.

368 00:09:39,479 –> 00:09:40,809 So we have shown that the

369 00:09:40,820 –> 00:09:42,450 inner product is a continuous

370 00:09:42,460 –> 00:09:44,070 map in the second argument

371 00:09:44,700 –> 00:09:46,070 and by symmetry or using

372 00:09:46,080 –> 00:09:47,359 the same argument as here,

373 00:09:47,369 –> 00:09:48,719 we can also show that it

374 00:09:48,729 –> 00:09:50,239 is a continuous map in the

375 00:09:50,250 –> 00:09:51,280 first argument.

376 00:09:51,719 –> 00:09:52,859 This is important to know

377 00:09:52,869 –> 00:09:54,039 for a lot of proofs we will

378 00:09:54,049 –> 00:09:54,960 do in the future

379 00:09:55,059 –> 00:09:56,109 and I’ll show you one

380 00:09:56,119 –> 00:09:58,000 now. Indeed,

381 00:09:58,010 –> 00:09:59,130 this is the cliffhanger from

382 00:09:59,140 –> 00:09:59,960 the last video.

383 00:09:59,969 –> 00:10:01,039 So we have an inner product

384 00:10:01,049 –> 00:10:02,840 space and a subset U

385 00:10:03,520 –> 00:10:05,090 and then U perp, the orthogonal

386 00:10:05,099 –> 00:10:06,630 complement of U,

387 00:10:06,640 –> 00:10:08,559 is always a closed subset.

388 00:10:09,270 –> 00:10:10,549 Proving this is not hard

389 00:10:10,559 –> 00:10:12,280 at all, because now we know

390 00:10:12,289 –> 00:10:13,500 all the things we need.

391 00:10:13,890 –> 00:10:15,559 Now, please recall part four

392 00:10:15,570 –> 00:10:17,000 of the series where I

393 00:10:17,010 –> 00:10:18,789 described closed sets with

394 00:10:18,799 –> 00:10:19,919 the help of sequences.

395 00:10:20,559 –> 00:10:21,700 So what we have to do is

396 00:10:21,710 –> 00:10:23,059 choose a sequence in the

397 00:10:23,070 –> 00:10:24,760 set itself where we know

398 00:10:24,770 –> 00:10:25,900 it’s convergent,

399 00:10:25,909 –> 00:10:27,640 but first just in the space

400 00:10:27,650 –> 00:10:29,570 X. Now when we can

401 00:10:29,580 –> 00:10:31,169 show that the limit x tilde

402 00:10:31,179 –> 00:10:33,130 is also in U perp, then

403 00:10:33,140 –> 00:10:34,770 we know U perp is closed.

404 00:10:35,429 –> 00:10:35,799 OK.

405 00:10:35,809 –> 00:10:37,239 So now we just have to use

406 00:10:37,250 –> 00:10:38,700 the definition of U perp,

407 00:10:39,299 –> 00:10:41,020 which means that inner product

408 00:10:41,030 –> 00:10:42,940 of x_n with u is

409 00:10:42,950 –> 00:10:44,590 always zero. No matter

410 00:10:44,599 –> 00:10:45,840 which u we choose.

411 00:10:46,330 –> 00:10:47,739 Since this equation holds

412 00:10:47,750 –> 00:10:49,400 for all n, we can look at

413 00:10:49,409 –> 00:10:51,039 the limit and know this is

414 00:10:51,049 –> 00:10:51,940 also zero

415 00:10:52,719 –> 00:10:54,500 and now we know we can pull

416 00:10:54,510 –> 00:10:56,059 in the limit here, because

417 00:10:56,070 –> 00:10:57,900 this is just the continuity

418 00:10:57,909 –> 00:10:59,859 of the map g we stated before.

419 00:11:00,739 –> 00:11:02,210 In other words, we just have

420 00:11:02,219 –> 00:11:04,030 x tilde with u is equal to

421 00:11:04,039 –> 00:11:04,530 zero

422 00:11:05,000 –> 00:11:06,390 and since this holds for

423 00:11:06,400 –> 00:11:08,340 all u in U, you see

424 00:11:08,349 –> 00:11:09,739 this is just the definition

425 00:11:09,750 –> 00:11:11,669 of U perp. x tilde is

426 00:11:11,679 –> 00:11:12,830 also in U perp

427 00:11:13,530 –> 00:11:14,669 and this is exactly what

428 00:11:14,679 –> 00:11:16,469 we wanted to show the orthogonal

429 00:11:16,479 –> 00:11:17,950 complement is always a

430 00:11:17,960 –> 00:11:18,820 closed set

431 00:11:19,539 –> 00:11:21,349 and you have seen, this immediately

432 00:11:21,359 –> 00:11:22,919 comes from the continuity

433 00:11:22,929 –> 00:11:23,989 of the inner product.

434 00:11:24,799 –> 00:11:25,200 OK.

435 00:11:25,210 –> 00:11:26,159 I think that’s good enough

436 00:11:26,169 –> 00:11:26,700 for today.

437 00:11:26,710 –> 00:11:28,260 Thank you very much for listening

438 00:11:28,270 –> 00:11:29,210 and please check out the

439 00:11:29,219 –> 00:11:31,179 PDFs if you want. Of

440 00:11:31,190 –> 00:11:32,130 course, I hope I see you

441 00:11:32,140 –> 00:11:33,659 next time when we go further

442 00:11:33,669 –> 00:11:34,950 into the topic of functional

443 00:11:34,960 –> 00:11:35,710 analysis.

444 00:11:36,260 –> 00:11:37,849 So have a nice day and see

445 00:11:37,859 –> 00:11:38,340 you then.

446 00:11:38,349 –> 00:11:38,940 Bye.

Q1: Let $(X, d)$ be a metric space. What is not correct?

A1: $d(x,z) \leq d(x,y) + d(y,z)$ (triangle inequality)

A2: $d(x,z) \geq | d(x,y) - d(y,z) |$ (reverse triangle inequality)

A3: $d(x,z) = 0$ if and only if $x = z$.

A4: $d(x,z)+d(z,x) = d(x,x)$ for all $x,z \in X$.

A5: $d(x,z) \geq 0 $ for all $x,z \in X$.

Q2: Let $(X, d)$ be a metric space and $z \in X$. Is the following map $x \mapsto d(x, z)$ continuous?

A1: Yes, by the reverse triangle inequality.

A2: No, never!

A3: Only in special cases.

Q3: Let $(X, \langle \cdot, \cdot \rangle)$ be a Hilbert space. Which claim is not correct in general?

A1: $x \mapsto \langle x,x \rangle$ is continuous.

A2: ${0}^\perp$ is an open set.

A3: ${0}^\perp$ is a closed set.

A4: $U^\perp$ is closed for any open set $U \subseteq X$.

A5: $U^\perp$ is open for any closed set $U \subseteq X$.