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Title: Example Operator Norm
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Series: Functional Analysis
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YouTube-Title: Functional Analysis 14 | Example Operator Norm
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Bright video: https://youtu.be/YMm-UZwmuF0
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Dark video: https://youtu.be/JgasOQng_jU
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Ad-free video: Watch Vimeo video
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Quiz: Test your knowledge
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Dark-PDF: Download PDF version of the dark video
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Thumbnail (bright): Download PNG
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Thumbnail (dark): Download PNG
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Subtitle on GitHub: fa14_sub_eng.srt
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Timestamps
00:00 Introduction
00:15 Example
01:27 First estimate
03:30 Second estimate
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Subtitle in English
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 normed
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 codomain, 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 the supremum 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 Lebesgue 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 T_g is
56 00:01:30,949 –> 00:01:32,349 given by a supremum.
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 supremum 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 supremum norm.
104 00:02:41,589 –> 00:02:42,660 w Which means that we have
105 00:02:42,669 –> 00:02:43,639 here a 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 supremum 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 T_g?
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 supremum
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 supremum
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 supremum norm of h 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 T_g 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 the 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 g(t)
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 T_g
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 normed 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.
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Quiz Content
Q1: Let $X = (C([0,1]), |\cdot |_\infty)$, $Y = (\mathbb{F}, |\cdot |)$. For $g \in X$, we define $T_g: X \rightarrow Y$ as $T_g(f) = \int_0^1 g(t) f(t) , dt$. What is not correct in general?
A1: $T_g$ is a linear operator
A2: $| T_g | \leq \int_0^1 |g(t)| , dt$
A3: $| T_g | = \int_0^1 g(t) , dt$
A4: $| T_g f | \leq | T_g | \cdot | f |_\infty$
A5: $| T_g | = \sup \Big{ | T_g f | ~ \Big| ~ f \in X \text{ with } | f |_\infty = 1 \Big} $
Q2: Let $X = (C([0,1]), |\cdot |_\infty)$, $Y = (\mathbb{F}, |\cdot |)$. For $g \in X$, we define $T_g: X \rightarrow Y$ as $T_g(f) = \int_0^1 g(t) f(t) , dt$. Is $T_g$ a bounded linear operator?
A1: Yes!
A2: No, it’s not linear.
A3: No, it’s unbounded.
A4: One needs more information.
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Last update: 2024-10
