Using Jensen inequality to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$
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My goal is to have a chain of inequalities so that $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$
$|sum_{i=1}^{n}a_ix_i|lesum_{i=1}^{n}|a_ix_i|=sum_{i=1}^{n}sqrt{(a_ix_i)^2}lesqrt{sum_{i=1}^{n}(a_ix_i)^2}lesqrt{max_{i=1,ldots,n}a_i^2sum_{i=1}^{n}x_i^2}=max_{i=1,ldots,n}|a_i|sqrt{sum_{i=1}^{n}x_i^2}$
Is this correct?
inequality
asked Nov 15 at 15:00
user610431
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up vote
0
down vote
favorite
My goal is to have a chain of inequalities so that $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$
$|sum_{i=1}^{n}a_ix_i|lesum_{i=1}^{n}|a_ix_i|=sum_{i=1}^{n}sqrt{(a_ix_i)^2}lesqrt{sum_{i=1}^{n}(a_ix_i)^2}lesqrt{max_{i=1,ldots,n}a_i^2sum_{i=1}^{n}x_i^2}=max_{i=1,ldots,n}|a_i|sqrt{sum_{i=1}^{n}x_i^2}$
Is this correct?
inequality
asked Nov 15 at 15:00
user610431
627
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My goal is to have a chain of inequalities so that $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$
$|sum_{i=1}^{n}a_ix_i|lesum_{i=1}^{n}|a_ix_i|=sum_{i=1}^{n}sqrt{(a_ix_i)^2}lesqrt{sum_{i=1}^{n}(a_ix_i)^2}lesqrt{max_{i=1,ldots,n}a_i^2sum_{i=1}^{n}x_i^2}=max_{i=1,ldots,n}|a_i|sqrt{sum_{i=1}^{n}x_i^2}$
Is this correct?
inequality
asked Nov 15 at 15:00
user610431
627
My goal is to have a chain of inequalities so that $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$
$|sum_{i=1}^{n}a_ix_i|lesum_{i=1}^{n}|a_ix_i|=sum_{i=1}^{n}sqrt{(a_ix_i)^2}lesqrt{sum_{i=1}^{n}(a_ix_i)^2}lesqrt{max_{i=1,ldots,n}a_i^2sum_{i=1}^{n}x_i^2}=max_{i=1,ldots,n}|a_i|sqrt{sum_{i=1}^{n}x_i^2}$
Is this correct?
inequality
inequality
asked Nov 15 at 15:00
user610431
627
asked Nov 15 at 15:00
user610431
627
asked Nov 15 at 15:00
user610431
627
asked Nov 15 at 15:00
user610431
627
asked Nov 15 at 15:00
user610431
627
627
add a comment |
add a comment |
1 Answer
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Jensen inequality means for a convex function, $f$, we have
$$f(E(X)) le E(f(X))$$
square root is a concave function.
$$frac{sum_{i=1}^nsqrt{(a_ix_i)^2}}{n} le sqrt{frac{sum_{i=1}^n(a_ix_i)^2}{n}}$$
Hence we have
$$sum_{i=1}^nsqrt{(a_ix_i)^2}le sqrt{n}sqrt{sum_{i=1}^n(a_ix_i)^2}$$
To see that $sqrt{n}$ can't be omitted, let $a_i=1, n=2, x_1=3, x_2=4$.
Without $sqrt{n}$, we have $$3+4 le sqrt{3^2+4^2}$$ which is not true.
Edit:
By Cauchy-Schwarz,
$$|langle a, x rangle | le |a|_2 |x|_2$$
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Jensen inequality means for a convex function, $f$, we have
$$f(E(X)) le E(f(X))$$
square root is a concave function.
$$frac{sum_{i=1}^nsqrt{(a_ix_i)^2}}{n} le sqrt{frac{sum_{i=1}^n(a_ix_i)^2}{n}}$$
Hence we have
$$sum_{i=1}^nsqrt{(a_ix_i)^2}le sqrt{n}sqrt{sum_{i=1}^n(a_ix_i)^2}$$
To see that $sqrt{n}$ can't be omitted, let $a_i=1, n=2, x_1=3, x_2=4$.
Without $sqrt{n}$, we have $$3+4 le sqrt{3^2+4^2}$$ which is not true.
Edit:
By Cauchy-Schwarz,
$$|langle a, x rangle | le |a|_2 |x|_2$$
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
add a comment |
up vote
0
down vote
Jensen inequality means for a convex function, $f$, we have
$$f(E(X)) le E(f(X))$$
square root is a concave function.
$$frac{sum_{i=1}^nsqrt{(a_ix_i)^2}}{n} le sqrt{frac{sum_{i=1}^n(a_ix_i)^2}{n}}$$
Hence we have
$$sum_{i=1}^nsqrt{(a_ix_i)^2}le sqrt{n}sqrt{sum_{i=1}^n(a_ix_i)^2}$$
To see that $sqrt{n}$ can't be omitted, let $a_i=1, n=2, x_1=3, x_2=4$.
Without $sqrt{n}$, we have $$3+4 le sqrt{3^2+4^2}$$ which is not true.
Edit:
By Cauchy-Schwarz,
$$|langle a, x rangle | le |a|_2 |x|_2$$
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
add a comment |
up vote
0
down vote
up vote
0
down vote
Jensen inequality means for a convex function, $f$, we have
$$f(E(X)) le E(f(X))$$
square root is a concave function.
$$frac{sum_{i=1}^nsqrt{(a_ix_i)^2}}{n} le sqrt{frac{sum_{i=1}^n(a_ix_i)^2}{n}}$$
Hence we have
$$sum_{i=1}^nsqrt{(a_ix_i)^2}le sqrt{n}sqrt{sum_{i=1}^n(a_ix_i)^2}$$
To see that $sqrt{n}$ can't be omitted, let $a_i=1, n=2, x_1=3, x_2=4$.
Without $sqrt{n}$, we have $$3+4 le sqrt{3^2+4^2}$$ which is not true.
Edit:
By Cauchy-Schwarz,
$$|langle a, x rangle | le |a|_2 |x|_2$$
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
Jensen inequality means for a convex function, $f$, we have
$$f(E(X)) le E(f(X))$$
square root is a concave function.
$$frac{sum_{i=1}^nsqrt{(a_ix_i)^2}}{n} le sqrt{frac{sum_{i=1}^n(a_ix_i)^2}{n}}$$
Hence we have
$$sum_{i=1}^nsqrt{(a_ix_i)^2}le sqrt{n}sqrt{sum_{i=1}^n(a_ix_i)^2}$$
To see that $sqrt{n}$ can't be omitted, let $a_i=1, n=2, x_1=3, x_2=4$.
Without $sqrt{n}$, we have $$3+4 le sqrt{3^2+4^2}$$ which is not true.
Edit:
By Cauchy-Schwarz,
$$|langle a, x rangle | le |a|_2 |x|_2$$
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
edited Nov 15 at 15:23
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
answered Nov 15 at 15:12
Siong Thye Goh
93.1k1462114
93.1k1462114
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
add a comment |
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
Hmm okay, I need it for part b) here: math.stackexchange.com/questions/2999496/… to show $|sum_{i=1}^{n}a_ix_i|le ldots le csqrt{sum_{i=1}^{n}x_i^2}$ but I don't know what to do. Maybe it's C-S inequality or something completely different I don't know
– user610431
Nov 15 at 15:18
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
You mean $c$ can't depend on $n$? most of your working is correct.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
CS is a good idea too.
– Siong Thye Goh
Nov 15 at 15:20
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
The $c$ should be some norm of $A$ I guess, so maybe $c=|A|_infty$ or $c=|A|_2$, something like this but I can't do it
– user610431
Nov 15 at 15:21
add a comment |
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