Convex function exercise [closed]
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having a bit of trouble with this exercise;
Let f be a function, convex on $mathbb{I}$, by writing $sum_{k=1}^{n}x_klambda_k $ for $ n ge 3 $ under the form $x_nlambda_n+(1-lambda_n)y_n$ with $y_n$ expressed with $x_k, kin [|1,n|] $ and $ lambda_q, qin[|1,n|]$ and $lambda_iin]0,1[$and $sum_{n}^{i=1}lambda_i = 1$ .
Prove that :
$sum_{k=1}^{n}lambda_kf(x_k)ge f(sum_{k=1}^{n}x_klambda_k)$.
The previous question was to demonstrate that ;
$zin ]x,y[ <=> existslambda, z=xlambda + (1-lambda)y$
convex-analysis
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closed as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin Dec 26 '18 at 9:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
having a bit of trouble with this exercise;
Let f be a function, convex on $mathbb{I}$, by writing $sum_{k=1}^{n}x_klambda_k $ for $ n ge 3 $ under the form $x_nlambda_n+(1-lambda_n)y_n$ with $y_n$ expressed with $x_k, kin [|1,n|] $ and $ lambda_q, qin[|1,n|]$ and $lambda_iin]0,1[$and $sum_{n}^{i=1}lambda_i = 1$ .
Prove that :
$sum_{k=1}^{n}lambda_kf(x_k)ge f(sum_{k=1}^{n}x_klambda_k)$.
The previous question was to demonstrate that ;
$zin ]x,y[ <=> existslambda, z=xlambda + (1-lambda)y$
convex-analysis
$endgroup$
closed as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin Dec 26 '18 at 9:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
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May be n dimensional Jensen's inequality may fit the bill?
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– Satish Ramanathan
Dec 25 '18 at 17:26
1
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This needs rewriting. It is strangely confusing.
$endgroup$
– zhw.
Dec 25 '18 at 17:59
add a comment |
$begingroup$
having a bit of trouble with this exercise;
Let f be a function, convex on $mathbb{I}$, by writing $sum_{k=1}^{n}x_klambda_k $ for $ n ge 3 $ under the form $x_nlambda_n+(1-lambda_n)y_n$ with $y_n$ expressed with $x_k, kin [|1,n|] $ and $ lambda_q, qin[|1,n|]$ and $lambda_iin]0,1[$and $sum_{n}^{i=1}lambda_i = 1$ .
Prove that :
$sum_{k=1}^{n}lambda_kf(x_k)ge f(sum_{k=1}^{n}x_klambda_k)$.
The previous question was to demonstrate that ;
$zin ]x,y[ <=> existslambda, z=xlambda + (1-lambda)y$
convex-analysis
$endgroup$
having a bit of trouble with this exercise;
Let f be a function, convex on $mathbb{I}$, by writing $sum_{k=1}^{n}x_klambda_k $ for $ n ge 3 $ under the form $x_nlambda_n+(1-lambda_n)y_n$ with $y_n$ expressed with $x_k, kin [|1,n|] $ and $ lambda_q, qin[|1,n|]$ and $lambda_iin]0,1[$and $sum_{n}^{i=1}lambda_i = 1$ .
Prove that :
$sum_{k=1}^{n}lambda_kf(x_k)ge f(sum_{k=1}^{n}x_klambda_k)$.
The previous question was to demonstrate that ;
$zin ]x,y[ <=> existslambda, z=xlambda + (1-lambda)y$
convex-analysis
convex-analysis
asked Dec 25 '18 at 16:54
user310148user310148
43
43
closed as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin Dec 26 '18 at 9:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin Dec 26 '18 at 9:33
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
May be n dimensional Jensen's inequality may fit the bill?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:26
1
$begingroup$
This needs rewriting. It is strangely confusing.
$endgroup$
– zhw.
Dec 25 '18 at 17:59
add a comment |
$begingroup$
May be n dimensional Jensen's inequality may fit the bill?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:26
1
$begingroup$
This needs rewriting. It is strangely confusing.
$endgroup$
– zhw.
Dec 25 '18 at 17:59
$begingroup$
May be n dimensional Jensen's inequality may fit the bill?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:26
$begingroup$
May be n dimensional Jensen's inequality may fit the bill?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:26
1
1
$begingroup$
This needs rewriting. It is strangely confusing.
$endgroup$
– zhw.
Dec 25 '18 at 17:59
$begingroup$
This needs rewriting. It is strangely confusing.
$endgroup$
– zhw.
Dec 25 '18 at 17:59
add a comment |
1 Answer
1
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oldest
votes
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hint
$$sum_{i=1}^nlambda_ix_i=$$
$$lambda_nx_n+(1-lambda_n)sum_{i=1}^{n-1}frac{lambda_i x_i}{1-lambda_n}=$$
$$lambda_nx_n+(1-lambda_n)y_n$$
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$begingroup$
Hey thanks yea, sorry i forgot to say i got that part already sorry
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– user310148
Dec 25 '18 at 18:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hint
$$sum_{i=1}^nlambda_ix_i=$$
$$lambda_nx_n+(1-lambda_n)sum_{i=1}^{n-1}frac{lambda_i x_i}{1-lambda_n}=$$
$$lambda_nx_n+(1-lambda_n)y_n$$
$endgroup$
$begingroup$
Hey thanks yea, sorry i forgot to say i got that part already sorry
$endgroup$
– user310148
Dec 25 '18 at 18:50
add a comment |
$begingroup$
hint
$$sum_{i=1}^nlambda_ix_i=$$
$$lambda_nx_n+(1-lambda_n)sum_{i=1}^{n-1}frac{lambda_i x_i}{1-lambda_n}=$$
$$lambda_nx_n+(1-lambda_n)y_n$$
$endgroup$
$begingroup$
Hey thanks yea, sorry i forgot to say i got that part already sorry
$endgroup$
– user310148
Dec 25 '18 at 18:50
add a comment |
$begingroup$
hint
$$sum_{i=1}^nlambda_ix_i=$$
$$lambda_nx_n+(1-lambda_n)sum_{i=1}^{n-1}frac{lambda_i x_i}{1-lambda_n}=$$
$$lambda_nx_n+(1-lambda_n)y_n$$
$endgroup$
hint
$$sum_{i=1}^nlambda_ix_i=$$
$$lambda_nx_n+(1-lambda_n)sum_{i=1}^{n-1}frac{lambda_i x_i}{1-lambda_n}=$$
$$lambda_nx_n+(1-lambda_n)y_n$$
answered Dec 25 '18 at 18:45
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
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Hey thanks yea, sorry i forgot to say i got that part already sorry
$endgroup$
– user310148
Dec 25 '18 at 18:50
add a comment |
$begingroup$
Hey thanks yea, sorry i forgot to say i got that part already sorry
$endgroup$
– user310148
Dec 25 '18 at 18:50
$begingroup$
Hey thanks yea, sorry i forgot to say i got that part already sorry
$endgroup$
– user310148
Dec 25 '18 at 18:50
$begingroup$
Hey thanks yea, sorry i forgot to say i got that part already sorry
$endgroup$
– user310148
Dec 25 '18 at 18:50
add a comment |
$begingroup$
May be n dimensional Jensen's inequality may fit the bill?
$endgroup$
– Satish Ramanathan
Dec 25 '18 at 17:26
1
$begingroup$
This needs rewriting. It is strangely confusing.
$endgroup$
– zhw.
Dec 25 '18 at 17:59