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Title: Hölder’s Inequality
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Series: Functional Analysis
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YouTube-Title: Functional Analysis 19 | Hölder’s Inequality
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Subtitle on GitHub: fa19_sub_eng.srt
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Timestamps
00:00 Introduction
01:42 Hölder’s Inequality
02:19 Young’s Inequality
05:11 Proof
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Subtitle in English
1 00:00:00,550 –> 00:00:02,289 Hello and welcome back to
2 00:00:02,299 –> 00:00:03,589 functional analysis
3 00:00:03,710 –> 00:00:04,929 and as always, I want to
4 00:00:04,940 –> 00:00:06,320 thank all the nice people
5 00:00:06,329 –> 00:00:07,590 that support this channel
6 00:00:07,599 –> 00:00:09,039 on Steady or PayPal.
7 00:00:09,729 –> 00:00:11,689 In today’s part 19, we
8 00:00:11,699 –> 00:00:13,420 will go back and prove Hölder’s
9 00:00:13,430 –> 00:00:14,369 inequality.
10 00:00:15,250 –> 00:00:16,530 There are a lot of different
11 00:00:16,540 –> 00:00:17,930 inequalities named after
12 00:00:17,940 –> 00:00:19,350 Hölder and indeed, they are
13 00:00:19,360 –> 00:00:20,309 all related,
14 00:00:20,319 –> 00:00:21,850 but here we look at the simplest
15 00:00:21,860 –> 00:00:23,639 one. It’s the one that
16 00:00:23,649 –> 00:00:24,979 holds in F^n
17 00:00:24,989 –> 00:00:26,739 and for p greater than one.
18 00:00:27,450 –> 00:00:29,190 For each such number p, there
19 00:00:29,200 –> 00:00:30,870 is a Hölder conjugate which
20 00:00:30,879 –> 00:00:32,119 we call p prime.
21 00:00:32,909 –> 00:00:34,080 The definition is what you
22 00:00:34,090 –> 00:00:35,479 should really memorize.
23 00:00:35,490 –> 00:00:37,020 The reciprocals should add
24 00:00:37,029 –> 00:00:37,610 to one.
25 00:00:38,180 –> 00:00:39,729 This is an important relation,
26 00:00:39,740 –> 00:00:41,130 because every time we use
27 00:00:41,139 –> 00:00:42,750 p prime, this is what we
28 00:00:42,759 –> 00:00:44,529 mean. Instead of p
29 00:00:44,540 –> 00:00:46,099 prime, a lot of people just
30 00:00:46,110 –> 00:00:46,830 use q.
31 00:00:47,630 –> 00:00:48,979 However, here I want to use
32 00:00:48,990 –> 00:00:50,790 q as another variable to
33 00:00:50,799 –> 00:00:52,590 tell you again what the p-
34 00:00:52,599 –> 00:00:53,380 norm was.
35 00:00:54,020 –> 00:00:55,240 Or today, I would better
36 00:00:55,250 –> 00:00:57,049 say the q-norm of a vector
37 00:00:57,060 –> 00:00:57,450 x.
38 00:00:58,250 –> 00:01:00,080 It’s defined by the sum over
39 00:01:00,090 –> 00:01:01,569 the entries in the absolute
40 00:01:01,580 –> 00:01:03,049 value to the power q
41 00:01:03,709 –> 00:01:05,370 and then we take the q-th root
42 00:01:05,379 –> 00:01:06,290 of the whole thing
43 00:01:07,069 –> 00:01:08,470 and this is what we do for
44 00:01:08,480 –> 00:01:09,980 all q between one and
45 00:01:09,989 –> 00:01:11,769 infinity including one.
46 00:01:12,470 –> 00:01:14,269 Now Hölder inequality will
47 00:01:14,279 –> 00:01:15,849 connect the p-norm the p
48 00:01:15,860 –> 00:01:17,550 prime-norm and the 1-
49 00:01:17,559 –> 00:01:19,489 norm. Indeed, it’s one
50 00:01:19,500 –> 00:01:20,839 nice short formula,
51 00:01:20,849 –> 00:01:22,160 you always should remember.
52 00:01:22,989 –> 00:01:24,510 I will now use a possibly
53 00:01:24,519 –> 00:01:26,110 strange notation to help
54 00:01:26,120 –> 00:01:27,699 you to remember the important
55 00:01:27,709 –> 00:01:29,610 inequality. Putting two
56 00:01:29,620 –> 00:01:31,250 vectors together just
57 00:01:31,260 –> 00:01:33,110 denotes a new vector which
58 00:01:33,120 –> 00:01:34,949 has the product in the components.
59 00:01:35,239 –> 00:01:36,930 So x_2, y_2 in the
60 00:01:36,940 –> 00:01:38,559 second component and
61 00:01:38,569 –> 00:01:39,849 x_n times y_n
62 00:01:39,860 –> 00:01:41,580 in the last component.
63 00:01:41,940 –> 00:01:43,760 Then for all vectors x and
64 00:01:43,769 –> 00:01:45,650 y, we have that the 1-norm
65 00:01:45,660 –> 00:01:47,440 of the vector xy
66 00:01:47,940 –> 00:01:49,430 is less or equal than the
67 00:01:49,440 –> 00:01:51,309 p-norm of x times the p
68 00:01:51,319 –> 00:01:52,720 prime-norm of y
69 00:01:53,379 –> 00:01:54,830 and that’s what we call
70 00:01:54,839 –> 00:01:56,029 Hölder’s inequality.
71 00:01:56,580 –> 00:01:57,720 Often you see it written
72 00:01:57,730 –> 00:01:59,199 with the sums, but I think
73 00:01:59,209 –> 00:02:00,910 it’s better to write it immediately
74 00:02:00,919 –> 00:02:02,599 in this way. On the one
75 00:02:02,610 –> 00:02:04,269 hand, it’s easier to remember
76 00:02:04,360 –> 00:02:05,680 and on the other hand, we
77 00:02:05,690 –> 00:02:06,919 will generalize the whole
78 00:02:06,930 –> 00:02:07,860 thing later on.
79 00:02:08,470 –> 00:02:09,899 Indeed, there we will have
80 00:02:09,910 –> 00:02:11,199 the whole inequality for
81 00:02:11,210 –> 00:02:13,160 functions defined on an abstract
82 00:02:13,169 –> 00:02:13,929 measure space.
83 00:02:14,429 –> 00:02:15,479 However, that’s not what
84 00:02:15,490 –> 00:02:16,770 we do today. Today
85 00:02:16,779 –> 00:02:18,149 we prove it for vectors
86 00:02:18,160 –> 00:02:18,869 in F^n.
87 00:02:19,649 –> 00:02:20,880 In order to prove Hölder’s
88 00:02:20,889 –> 00:02:22,610 inequality, we first need
89 00:02:22,619 –> 00:02:24,130 another inequality which
90 00:02:24,139 –> 00:02:25,490 is known as Young’s
91 00:02:25,500 –> 00:02:26,440 inequality.
92 00:02:27,110 –> 00:02:28,649 In fact, this one is very
93 00:02:28,660 –> 00:02:29,250 simple.
94 00:02:29,259 –> 00:02:30,589 We just look at positive
95 00:02:30,600 –> 00:02:32,550 numbers a, b and
96 00:02:32,559 –> 00:02:33,850 then we conclude that the
97 00:02:33,860 –> 00:02:35,529 product of ab is
98 00:02:35,539 –> 00:02:37,210 always less or equal than
99 00:02:37,220 –> 00:02:38,309 the following sum.
100 00:02:38,470 –> 00:02:39,630 So what you should see here
101 00:02:39,639 –> 00:02:41,160 is that p greater than one
102 00:02:41,169 –> 00:02:42,850 goes in and the Hölder
103 00:02:42,860 –> 00:02:43,610 conjugate.
104 00:02:44,259 –> 00:02:45,940 So as before you can choose
105 00:02:45,949 –> 00:02:47,669 p greater one as you want,
106 00:02:47,690 –> 00:02:49,470 but then p prime is fixed.
107 00:02:50,000 –> 00:02:51,610 For example, for p equals
108 00:02:51,619 –> 00:02:53,570 to two, we have p prime equals
109 00:02:53,580 –> 00:02:54,110 to two.
110 00:02:54,600 –> 00:02:55,910 However, in this case, you
111 00:02:55,919 –> 00:02:57,619 already know that the inequality
112 00:02:57,630 –> 00:02:58,509 here is correct.
113 00:02:59,300 –> 00:03:00,479 So please check that,
114 00:03:00,490 –> 00:03:01,759 but for all other cases,
115 00:03:01,770 –> 00:03:03,380 we have to write down a proof.
116 00:03:04,190 –> 00:03:05,570 What we can use here is that
117 00:03:05,580 –> 00:03:07,210 the common exponential function
118 00:03:07,220 –> 00:03:08,929 is a so called convex
119 00:03:08,940 –> 00:03:09,369 function.
120 00:03:10,059 –> 00:03:11,190 This means that when you
121 00:03:11,199 –> 00:03:12,279 look at the graph,
122 00:03:12,970 –> 00:03:14,389 then you can choose any two
123 00:03:14,399 –> 00:03:14,779 points
124 00:03:14,789 –> 00:03:15,630 as you want.
125 00:03:15,710 –> 00:03:16,929 The direct connection would
126 00:03:16,940 –> 00:03:18,630 be always above the graph.
127 00:03:19,110 –> 00:03:20,940 This property is called convex,
128 00:03:20,949 –> 00:03:22,509 because the red line is the
129 00:03:22,520 –> 00:03:24,100 convex combination of the
130 00:03:24,110 –> 00:03:24,839 two points.
131 00:03:25,440 –> 00:03:26,479 Therefore, we can easily
132 00:03:26,490 –> 00:03:28,029 put that into a formula when
133 00:03:28,039 –> 00:03:29,449 we call the function just
134 00:03:29,460 –> 00:03:29,910 f.
135 00:03:30,360 –> 00:03:31,990 Now, by denoting the two points
136 00:03:32,000 –> 00:03:33,669 by x and y, we can form a
137 00:03:33,679 –> 00:03:35,410 convex combination of x and
138 00:03:35,419 –> 00:03:37,369 y and put that into the function
139 00:03:37,380 –> 00:03:39,300 f. Which means we get out
140 00:03:39,309 –> 00:03:40,220 the blue line here.
141 00:03:40,589 –> 00:03:42,339 Please recall a convex
142 00:03:42,350 –> 00:03:43,940 combination is just a special
143 00:03:43,949 –> 00:03:45,179 linear combination, where
144 00:03:45,190 –> 00:03:46,940 we only have one lambda which
145 00:03:46,949 –> 00:03:48,029 comes from the interval
146 00:03:48,039 –> 00:03:49,199 0 to 1.
147 00:03:49,350 –> 00:03:50,660 Now with the red line, we
148 00:03:50,669 –> 00:03:52,289 also get a convex combination
149 00:03:52,300 –> 00:03:53,789 with the images. Which means
150 00:03:53,800 –> 00:03:55,029 we have here instead of x
151 00:03:55,039 –> 00:03:56,899 and y, just f(x)
152 00:03:56,940 –> 00:03:57,899 and f(y).
153 00:03:58,309 –> 00:03:59,770 Therefore being always above
154 00:03:59,779 –> 00:04:01,199 the blue graph means we have
155 00:04:01,210 –> 00:04:02,669 here our inequality.
156 00:04:03,410 –> 00:04:04,949 Which we now want to use
157 00:04:04,960 –> 00:04:06,229 for some special numbers.
158 00:04:06,960 –> 00:04:08,339 Lambda should be one over
159 00:04:08,350 –> 00:04:10,110 p and then one minus
160 00:04:10,119 –> 00:04:11,509 lambda should be one over
161 00:04:11,520 –> 00:04:12,240 p prime.
162 00:04:12,729 –> 00:04:13,789 Maybe that’s not so clear
163 00:04:13,800 –> 00:04:15,360 but for x I want to put in
164 00:04:15,369 –> 00:04:16,720 the natural logarithm of
165 00:04:16,730 –> 00:04:18,108 a to the power p
166 00:04:18,459 –> 00:04:19,640 and the similar thing for
167 00:04:19,649 –> 00:04:21,559 y but now with b and p
168 00:04:21,570 –> 00:04:23,540 prime. Of course, this
169 00:04:23,549 –> 00:04:25,059 all fits together, because
170 00:04:25,070 –> 00:04:26,329 by applying the logarithm
171 00:04:26,339 –> 00:04:27,779 rules, we can bring this
172 00:04:27,790 –> 00:04:29,559 power in front, where it cancels
173 00:04:29,570 –> 00:04:30,779 out. Hence,
174 00:04:30,790 –> 00:04:31,779 on the left-hand side, we
175 00:04:31,790 –> 00:04:33,029 have the two logarithms in
176 00:04:33,040 –> 00:04:33,940 the function f.
177 00:04:34,320 –> 00:04:36,089 Now using the next logarithm
178 00:04:36,100 –> 00:04:36,619 rule
179 00:04:36,630 –> 00:04:38,239 and the fact that the exponential
180 00:04:38,250 –> 00:04:39,820 function is the inverse function
181 00:04:39,829 –> 00:04:41,179 of the logarithm, we have
182 00:04:41,190 –> 00:04:42,859 just ab on the left-hand
183 00:04:42,869 –> 00:04:43,290 side
184 00:04:43,750 –> 00:04:44,980 and now you see the idea
185 00:04:44,989 –> 00:04:46,390 of the whole proof. We want
186 00:04:46,399 –> 00:04:48,140 to show this one and we’re
187 00:04:48,149 –> 00:04:49,380 already finished with the
188 00:04:49,390 –> 00:04:50,179 left part.
189 00:04:50,920 –> 00:04:51,940 Regarding the right-hand
190 00:04:51,950 –> 00:04:53,220 side, we just put in all
191 00:04:53,230 –> 00:04:54,609 the numbers we already know.
192 00:04:55,290 –> 00:04:56,869 Now this looks more complicated
193 00:04:56,880 –> 00:04:58,230 than it really is, because
194 00:04:58,239 –> 00:04:59,500 we have again the inverse
195 00:04:59,510 –> 00:05:00,670 function of the logarithm
196 00:05:00,679 –> 00:05:01,029 here.
197 00:05:01,579 –> 00:05:03,170 Hence this all vanishes
198 00:05:03,179 –> 00:05:04,309 and what remains is what
199 00:05:04,320 –> 00:05:05,309 we wanted to show.
200 00:05:05,799 –> 00:05:07,320 So Young’s inequality is
201 00:05:07,329 –> 00:05:08,829 correct and we can use it
202 00:05:08,839 –> 00:05:10,750 to prove Hölder’s inequality.
203 00:05:11,260 –> 00:05:12,410 Starting with the proof,
204 00:05:12,420 –> 00:05:13,480 let’s first consider the
205 00:05:13,489 –> 00:05:14,420 simplest case.
206 00:05:15,079 –> 00:05:16,320 What I mean by that is that
207 00:05:16,329 –> 00:05:17,549 we look what happens when
208 00:05:17,559 –> 00:05:19,239 x is the zero vector or
209 00:05:19,250 –> 00:05:20,579 y is the zero vector,
210 00:05:21,179 –> 00:05:22,459 but of course, this is not
211 00:05:22,470 –> 00:05:23,899 a problem for us, because
212 00:05:23,910 –> 00:05:25,269 by the definition of all
213 00:05:25,279 –> 00:05:26,799 the norms here, we have that
214 00:05:26,809 –> 00:05:28,250 the left-hand side is zero
215 00:05:28,260 –> 00:05:29,459 and the right-hand side is
216 00:05:29,470 –> 00:05:29,850 zero.
217 00:05:29,859 –> 00:05:31,820 So the inequality is fulfilled.
218 00:05:32,480 –> 00:05:33,739 Therefore, for the second
219 00:05:33,750 –> 00:05:35,529 case, we can divide and bring
220 00:05:35,540 –> 00:05:36,660 everything to the left-hand
221 00:05:36,670 –> 00:05:37,070 side.
222 00:05:37,700 –> 00:05:38,880 Of course, now we want to
223 00:05:38,890 –> 00:05:40,609 bring this inside the norm
224 00:05:40,619 –> 00:05:42,290 and divide x by its p-
225 00:05:42,299 –> 00:05:44,250 norm and y by its p prime-
226 00:05:44,260 –> 00:05:44,640 norm.
227 00:05:45,019 –> 00:05:46,380 However, since I use this
228 00:05:46,390 –> 00:05:47,760 strange notation here, I
229 00:05:47,769 –> 00:05:49,339 bring in now the whole sum
230 00:05:49,350 –> 00:05:51,299 of the 1-norm. Here we have
231 00:05:51,309 –> 00:05:52,649 it and we can pull in the
232 00:05:52,660 –> 00:05:54,250 norm into the sum and then
233 00:05:54,260 –> 00:05:55,299 into the absolute value
234 00:05:55,989 –> 00:05:56,929 and there you should see
235 00:05:56,940 –> 00:05:58,679 now, that we have two positive
236 00:05:58,690 –> 00:06:00,149 numbers, which we could call
237 00:06:00,160 –> 00:06:01,989 a and b and then apply
238 00:06:02,000 –> 00:06:03,899 Young’s inequality inside
239 00:06:03,910 –> 00:06:04,459 the sum.
240 00:06:04,980 –> 00:06:06,380 So at this point, we get
241 00:06:06,390 –> 00:06:07,690 the inequality sign in.
242 00:06:08,399 –> 00:06:08,700 OK.
243 00:06:08,709 –> 00:06:10,010 The first term is one over
244 00:06:10,019 –> 00:06:11,940 p times x_j
245 00:06:11,950 –> 00:06:13,739 to the power p divided by
246 00:06:13,750 –> 00:06:15,450 the p-norm to the power p.
247 00:06:15,950 –> 00:06:17,410 The second term looks similar
248 00:06:17,420 –> 00:06:18,570 and I already distributed
249 00:06:18,579 –> 00:06:20,089 the sum over both parts.
250 00:06:20,739 –> 00:06:22,329 So here we just have p prime
251 00:06:22,339 –> 00:06:23,799 and j instead of x.
252 00:06:24,220 –> 00:06:25,769 However, that’s not so important.
253 00:06:25,779 –> 00:06:26,869 The important thing you should
254 00:06:26,880 –> 00:06:28,279 see is what we have in
255 00:06:28,290 –> 00:06:29,600 the numerator together with
256 00:06:29,609 –> 00:06:31,109 the sum, is the same as the
257 00:06:31,119 –> 00:06:32,809 denominator and the
258 00:06:32,820 –> 00:06:34,119 same in the second part.
259 00:06:34,549 –> 00:06:35,760 Hence, the only things that
260 00:06:35,769 –> 00:06:37,640 remain is one over p here
261 00:06:37,649 –> 00:06:39,160 and one over p prime here
262 00:06:39,630 –> 00:06:40,880 and by the definition of
263 00:06:40,890 –> 00:06:42,470 the Hölder conjugate this
264 00:06:42,480 –> 00:06:43,480 is simply one
265 00:06:44,000 –> 00:06:45,540 and with this, we have proven
266 00:06:45,549 –> 00:06:47,200 Hölder’s inequality, because
267 00:06:47,209 –> 00:06:48,630 you can bring this one on
268 00:06:48,640 –> 00:06:49,760 the right-hand side again,
269 00:06:49,769 –> 00:06:50,440 if you want.
270 00:06:51,130 –> 00:06:52,459 So you see the proof was
271 00:06:52,470 –> 00:06:53,869 not so hard, but we will
272 00:06:53,880 –> 00:06:55,630 need Hölder’s inequality to
273 00:06:55,640 –> 00:06:57,309 prove another inequality
274 00:06:57,739 –> 00:06:58,839 and this one will be the
275 00:06:58,850 –> 00:07:00,149 so called Minkowski
276 00:07:00,160 –> 00:07:01,010 inequality.
277 00:07:01,320 –> 00:07:02,790 Essentially, it’s just the
278 00:07:02,940 –> 00:07:04,869 triangle inequality for our l^p
279 00:07:04,880 –> 00:07:05,470 space.
280 00:07:06,010 –> 00:07:07,220 Therefore, I hope I see you
281 00:07:07,230 –> 00:07:08,070 in the next video.
282 00:07:08,079 –> 00:07:09,579 Thanks for listening and
283 00:07:09,589 –> 00:07:10,410 see you then.
284 00:07:10,420 –> 00:07:11,109 Bye.
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Last update: 2024-10
