Jensen's type inequality for inner product
Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
$$ z := tx + (1-t)y , quad t in (0,1) , $$
then we have the following inequality
$$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.
The question is when we only have
$$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.
A naive approach gives
$$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$
If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?
inequality inner-product-space jensen-inequality
asked Nov 29 '18 at 22:01
mortalmortal
397414
add a comment |
Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
$$ z := tx + (1-t)y , quad t in (0,1) , $$
then we have the following inequality
$$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.
The question is when we only have
$$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.
A naive approach gives
$$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$
If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?
inequality inner-product-space jensen-inequality
asked Nov 29 '18 at 22:01
mortalmortal
397414
add a comment |
Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
$$ z := tx + (1-t)y , quad t in (0,1) , $$
then we have the following inequality
$$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.
The question is when we only have
$$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.
A naive approach gives
$$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$
If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?
inequality inner-product-space jensen-inequality
asked Nov 29 '18 at 22:01
mortalmortal
397414
Let $mathcal{H}$ be a real Hilbert space and $x,y,z in mathcal{H}$. If $x$ and $y$ are a convex combination of $z$, that is
$$ z := tx + (1-t)y , quad t in (0,1) , $$
then we have the following inequality
$$ leftlVert z rightrVert ^{2} leq t leftlVert x rightrVert ^{2} + (1-t)leftlVert y rightrVert ^{2} tag{1}label{1} . $$
Notice that eqref{1} is better than the Cauchy - Schwarz inequality in the sense that both coefficients are strictly less than $1$.
The question is when we only have
$$ leftlangle z , v rightrangle = leftlangle tx + (1-t)y , v rightrangle , quad t in (0,1) , tag{2}label{2} $$
and $v in mathcal{H}$ is arbitrary given, can we still derive the inequality eqref{1}.
A naive approach gives
$$ leftlVert z rightrVert ^{2} - leftlVert z-v rightrVert ^{2} = t leftlVert x rightrVert ^{2} + (1-t) leftlVert y rightrVert ^{2} - t (1-t) leftlVert x-y rightrVert ^{2} - leftlVert tx + (1-t)y-v rightrVert ^{2} $$
If we can not get eqref{1}, is there any alternative one in which I can avoid or at least weaken the term $- leftlVert z-v rightrVert ^{2}$, that is, to have $-c leftlVert z-v rightrVert ^{2}$ for some $c in (0,1)$?
inequality inner-product-space jensen-inequality
inequality inner-product-space jensen-inequality
asked Nov 29 '18 at 22:01
mortalmortal
397414
asked Nov 29 '18 at 22:01
mortalmortal
397414
asked Nov 29 '18 at 22:01
mortalmortal
397414
asked Nov 29 '18 at 22:01
mortalmortal
397414
asked Nov 29 '18 at 22:01
mortalmortal
397414
397414
add a comment |
add a comment |
1 Answer
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It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
1
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
1
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
|
show 1 more comment
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It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
1
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
1
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
|
show 1 more comment
It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
1
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
1
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
|
show 1 more comment
It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
It fails in $mathcal{H}=mathbb{R}^3$ with the usual inner product. Let's consider the following particular case: $y=(0,0,0)$, $x=(2,0,0)$, $z=(lambda,0,0)$ with $lambda^2>2$, $v=(0,1,0)$ and $t=1/2$. Then $langle z,v rangle =0 =langle tx+(1-t)y,vrangle $ but $||z||^2=lambda^2>2=t||x||^2$.
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
answered Nov 29 '18 at 22:19
Dante GrevinoDante Grevino
94319
94319
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
1
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
1
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
|
show 1 more comment
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
1
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
1
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
thanks for that. but what if $leftlangle z , v rightrangle neq 0$? since I don't think that it would be my case..
– mortal
Nov 29 '18 at 22:29
1
1
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
I think there is no chance because, in general, your hypothesis is very weak. For example, modify the example above by $z=(lambda,1,0)$ with $lambda^2>3$, $x=(2,2,0)$.
– Dante Grevino
Nov 29 '18 at 22:35
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
if then could you please have a look on the case with $$ z = a - t cdot c , x = b - t cdot d , z = - t cdot d , v = a - b . $$ That is the case I'm looking for. I hope it would help.
– mortal
Nov 29 '18 at 22:39
1
1
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
No problem, but you should post that question as a separate question and giving the precise hypothesis.
– Dante Grevino
Nov 29 '18 at 22:43
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
thank you! I'll do it.
– mortal
Nov 29 '18 at 22:45
|
show 1 more comment
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