00:00 Introduction

00:15 Example

01:27 First estimate

03:30 Second estimate

1 00:00:00,370 –> 00:00:02,269 Hello and welcome back to

2 00:00:02,279 –> 00:00:03,869 functional analysis.

3 00:00:03,880 –> 00:00:05,329 And many, many thanks to

4 00:00:05,340 –> 00:00:06,690 all the nice people that

5 00:00:06,699 –> 00:00:08,619 support me on Steady or paypal

6 00:00:09,329 –> 00:00:09,729 today.

7 00:00:09,739 –> 00:00:11,619 In part 14, I want to show

8 00:00:11,630 –> 00:00:13,119 you one example how to

9 00:00:13,130 –> 00:00:14,720 calculate the operator norm.

10 00:00:14,840 –> 00:00:16,729 In this example, our norm

11 00:00:16,739 –> 00:00:18,590 space X should be given by

12 00:00:18,600 –> 00:00:19,930 the continuous functions.

13 00:00:20,420 –> 00:00:21,590 The domain of the functions

14 00:00:21,600 –> 00:00:23,079 should be the unit interval

15 00:00:23,120 –> 00:00:25,010 and the co domain, the complex

16 00:00:25,020 –> 00:00:26,159 or real numbers.

17 00:00:26,520 –> 00:00:27,719 And the common norm for the

18 00:00:27,729 –> 00:00:29,600 space is this a premium norm.

19 00:00:30,489 –> 00:00:32,250 And the second space Y should

20 00:00:32,259 –> 00:00:33,740 be just the complex or real

21 00:00:33,750 –> 00:00:35,139 numbers together with the

22 00:00:35,150 –> 00:00:36,020 absolute value.

23 00:00:36,500 –> 00:00:38,060 Hence today, we will consider

24 00:00:38,069 –> 00:00:39,580 a map T that sends a

25 00:00:39,590 –> 00:00:41,400 continuous function to just

26 00:00:41,409 –> 00:00:41,880 a number.

27 00:00:42,750 –> 00:00:44,009 In order to define this

28 00:00:44,020 –> 00:00:45,869 operator, I want to fix a

29 00:00:45,880 –> 00:00:47,419 continuous function G

30 00:00:48,110 –> 00:00:49,380 to make our life a little

31 00:00:49,389 –> 00:00:50,200 bit easier.

32 00:00:50,209 –> 00:00:51,580 The function G shouldn’t

33 00:00:51,590 –> 00:00:53,029 have any zeros.

34 00:00:53,189 –> 00:00:54,939 Now, for each such G, we

35 00:00:54,950 –> 00:00:56,939 define an operator T and

36 00:00:56,950 –> 00:00:58,340 therefore I put an index

37 00:00:58,349 –> 00:00:59,139 G to it.

38 00:01:00,119 –> 00:01:01,549 Now the operator sends a

39 00:01:01,560 –> 00:01:03,150 function F to a number.

40 00:01:03,159 –> 00:01:04,709 By the following definition,

41 00:01:05,669 –> 00:01:06,930 it’s the integral of the

42 00:01:06,940 –> 00:01:08,930 function G multiplied

43 00:01:08,940 –> 00:01:10,290 with the function F

44 00:01:11,120 –> 00:01:12,669 now because we are only dealing

45 00:01:12,680 –> 00:01:14,180 with continuous functions.

46 00:01:14,190 –> 00:01:15,889 You can read this as an Riemann

47 00:01:15,900 –> 00:01:17,849 integral or an leak integral.

48 00:01:17,860 –> 00:01:19,459 It does not matter the

49 00:01:19,470 –> 00:01:21,150 question for us is, is

50 00:01:21,160 –> 00:01:22,739 this a bounded operator

51 00:01:23,690 –> 00:01:25,019 or a similar question?

52 00:01:25,029 –> 00:01:26,500 What is the operator norm

53 00:01:27,019 –> 00:01:27,580 for this?

54 00:01:27,589 –> 00:01:29,279 Please recall that the operator

55 00:01:29,290 –> 00:01:30,940 norm of TG is

56 00:01:30,949 –> 00:01:32,349 given by a suprema

57 00:01:32,930 –> 00:01:34,419 more concretely we have the

58 00:01:34,430 –> 00:01:35,889 norm of the output

59 00:01:35,900 –> 00:01:37,660 divided by the norm of the

60 00:01:37,669 –> 00:01:38,139 input.

61 00:01:38,779 –> 00:01:40,559 And then we go over all functions

62 00:01:40,569 –> 00:01:42,419 of X with the exception of

63 00:01:42,430 –> 00:01:43,379 the zero vector.

64 00:01:44,269 –> 00:01:45,589 So what we can always do

65 00:01:45,599 –> 00:01:47,500 is apply the scaling as we

66 00:01:47,510 –> 00:01:49,330 did it last time we can

67 00:01:49,339 –> 00:01:50,839 do that such that we only

68 00:01:50,849 –> 00:01:51,839 divide by one.

69 00:01:51,849 –> 00:01:53,669 So we consider only functions

70 00:01:53,680 –> 00:01:55,300 with supreme norm

71 00:01:55,309 –> 00:01:55,839 one.

72 00:01:56,379 –> 00:01:57,360 Of course, you don’t have

73 00:01:57,370 –> 00:01:58,540 to do that, but it makes

74 00:01:58,550 –> 00:02:00,110 our life easier because we

75 00:02:00,120 –> 00:02:01,459 don’t have to write down

76 00:02:01,470 –> 00:02:02,440 a fraction here.

77 00:02:03,000 –> 00:02:04,019 In the next step, we just

78 00:02:04,029 –> 00:02:05,089 fill in what we know from

79 00:02:05,099 –> 00:02:06,430 the image we have here.

80 00:02:06,440 –> 00:02:07,760 The absolute value of this

81 00:02:07,769 –> 00:02:08,460 integral.

82 00:02:08,960 –> 00:02:10,309 This integral is of course

83 00:02:10,320 –> 00:02:11,830 what we need to calculate.

84 00:02:11,880 –> 00:02:13,509 But maybe a better first

85 00:02:13,520 –> 00:02:14,630 step would be to find an

86 00:02:14,639 –> 00:02:15,389 estimate.

87 00:02:15,830 –> 00:02:17,229 Now, I want to use the knowledge

88 00:02:17,240 –> 00:02:18,389 that pulling the absolute

89 00:02:18,399 –> 00:02:20,270 value into the integral

90 00:02:20,279 –> 00:02:21,770 only makes it bigger or it

91 00:02:21,779 –> 00:02:22,509 stays the same.

92 00:02:23,279 –> 00:02:24,429 And then we just use the

93 00:02:24,440 –> 00:02:26,110 multiplicativity of the absolute

94 00:02:26,119 –> 00:02:27,779 value, which means we can

95 00:02:27,789 –> 00:02:29,399 pull out the multiplication

96 00:02:29,410 –> 00:02:29,860 sign.

97 00:02:30,710 –> 00:02:31,830 Then in the next step, we

98 00:02:31,839 –> 00:02:32,970 see that we have here the

99 00:02:32,979 –> 00:02:34,839 absolute value of F at some

100 00:02:34,850 –> 00:02:35,770 point T.

101 00:02:36,259 –> 00:02:37,869 However, this can’t be bigger

102 00:02:37,880 –> 00:02:39,529 than the largest value, which

103 00:02:39,539 –> 00:02:41,089 is given by the supreme norm,

104 00:02:41,589 –> 00:02:42,660 which means that we have

105 00:02:42,669 –> 00:02:43,639 here constant.

106 00:02:43,649 –> 00:02:44,970 And we already fixed that

107 00:02:44,979 –> 00:02:45,720 to one.

108 00:02:46,839 –> 00:02:48,389 In summary, the supreme can’t

109 00:02:48,399 –> 00:02:49,949 change anything anymore.

110 00:02:50,039 –> 00:02:51,630 So we know the operator norm

111 00:02:51,639 –> 00:02:53,110 can’t be bigger than the

112 00:02:53,119 –> 00:02:54,509 integral of G.

113 00:02:55,179 –> 00:02:56,639 And we already know this

114 00:02:56,649 –> 00:02:58,279 is a finite number because

115 00:02:58,289 –> 00:03:00,080 we have a positive continuous

116 00:03:00,089 –> 00:03:01,800 function inside the integral.

117 00:03:02,250 –> 00:03:04,139 So indeed, this is a bounded

118 00:03:04,149 –> 00:03:04,880 operator.

119 00:03:05,630 –> 00:03:07,009 The last question that remains

120 00:03:07,020 –> 00:03:08,589 is what exactly is the

121 00:03:08,600 –> 00:03:10,050 operator norm of TG?

122 00:03:11,089 –> 00:03:12,610 Do we have in fact an equality

123 00:03:12,619 –> 00:03:14,250 sign here or is the

124 00:03:14,259 –> 00:03:15,729 operator norm really less

125 00:03:15,740 –> 00:03:16,869 than this integral?

126 00:03:17,649 –> 00:03:19,119 In order to answer this question,

127 00:03:19,130 –> 00:03:20,520 it’s always good to look

128 00:03:20,529 –> 00:03:22,250 at some examples because

129 00:03:22,259 –> 00:03:23,610 then you get a lower bound

130 00:03:23,619 –> 00:03:24,690 for the whole supreme.

131 00:03:25,380 –> 00:03:27,089 And if you do that, you might

132 00:03:27,100 –> 00:03:28,979 find that you also can use

133 00:03:28,990 –> 00:03:30,850 the function G as an input.

134 00:03:31,240 –> 00:03:32,669 However, then we can’t use

135 00:03:32,679 –> 00:03:34,479 this line here because

136 00:03:34,490 –> 00:03:36,119 here we need that the supreme

137 00:03:36,130 –> 00:03:37,479 norm of the function we put

138 00:03:37,490 –> 00:03:38,940 in is exactly one.

139 00:03:39,539 –> 00:03:40,360 But of course, you know,

140 00:03:40,369 –> 00:03:41,750 we could change the function

141 00:03:41,759 –> 00:03:43,300 G in such a way just by

142 00:03:43,309 –> 00:03:44,529 dividing by the absolute

143 00:03:44,539 –> 00:03:45,399 value of G.

144 00:03:46,080 –> 00:03:47,330 So let’s call this function

145 00:03:47,339 –> 00:03:49,179 H and I also want to

146 00:03:49,190 –> 00:03:50,729 add a complex conjugation

147 00:03:50,740 –> 00:03:52,350 here which will be helpful

148 00:03:52,360 –> 00:03:52,740 later.

149 00:03:53,580 –> 00:03:55,070 Now it’s easy to see the

150 00:03:55,080 –> 00:03:56,789 supreme norm of age is

151 00:03:56,800 –> 00:03:57,789 exactly one.

152 00:03:58,600 –> 00:03:59,899 Now putting this function

153 00:03:59,910 –> 00:04:01,770 into TG and looking

154 00:04:01,779 –> 00:04:03,020 at the absolute value.

155 00:04:03,710 –> 00:04:05,179 Then we know this has to

156 00:04:05,190 –> 00:04:06,679 be less or equal than the

157 00:04:06,690 –> 00:04:07,940 operator norm itself.

158 00:04:08,419 –> 00:04:09,729 On the other hand, we know

159 00:04:09,740 –> 00:04:10,940 this is the absolute value

160 00:04:10,949 –> 00:04:12,500 of the integral where we

161 00:04:12,509 –> 00:04:14,320 have to function G times

162 00:04:14,330 –> 00:04:14,779 H.

163 00:04:15,220 –> 00:04:16,820 And now you see why we added

164 00:04:16,829 –> 00:04:18,700 the complex conjugation before

165 00:04:18,890 –> 00:04:20,720 because now we have the absolute

166 00:04:20,730 –> 00:04:22,679 value squared in the numerator,

167 00:04:23,230 –> 00:04:24,320 we can omit the absolute

168 00:04:24,329 –> 00:04:26,239 value outside because everything

169 00:04:26,250 –> 00:04:27,200 is positive here.

170 00:04:27,239 –> 00:04:28,799 And we also can just

171 00:04:28,809 –> 00:04:30,709 cancel out one absolute value

172 00:04:30,720 –> 00:04:31,440 of GT.

173 00:04:32,220 –> 00:04:33,820 And then what remains is

174 00:04:33,829 –> 00:04:35,440 exactly the same integral

175 00:04:35,450 –> 00:04:36,200 as before.

176 00:04:36,980 –> 00:04:38,329 And now we can put both

177 00:04:38,339 –> 00:04:39,839 inequalities together and

178 00:04:39,850 –> 00:04:41,459 find out that the operator

179 00:04:41,470 –> 00:04:43,339 norm of the operator TG

180 00:04:43,410 –> 00:04:44,700 is exactly this

181 00:04:44,709 –> 00:04:45,480 integral.

182 00:04:46,059 –> 00:04:46,359 OK.

183 00:04:46,369 –> 00:04:48,269 So this was one quick example

184 00:04:48,279 –> 00:04:49,700 of calculating the operator

185 00:04:49,709 –> 00:04:51,510 norm of an operator between

186 00:04:51,519 –> 00:04:52,709 two norm spaces.

187 00:04:53,399 –> 00:04:54,559 In the next video, I want

188 00:04:54,570 –> 00:04:56,459 to start talking about operators

189 00:04:56,470 –> 00:04:57,890 between Hilbert spaces.

190 00:04:58,540 –> 00:04:59,880 So thanks for listening.

191 00:04:59,890 –> 00:05:01,209 Thanks for supporting me

192 00:05:01,220 –> 00:05:02,600 and see you next time.

193 00:05:02,649 –> 00:05:03,279 Bye.