Convex function exercise [on hold]
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
asked 2 days ago
user310148
22
put on hold as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin yesterday
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 |
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
asked 2 days ago
user310148
22
put on hold as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin yesterday
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.
May be n dimensional Jensen's inequality may fit the bill?
– Satish Ramanathan
2 days ago
1
This needs rewriting. It is strangely confusing.
– zhw.
2 days ago
add a comment |
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
asked 2 days ago
user310148
22
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 2 days ago
user310148
22
asked 2 days ago
user310148
22
asked 2 days ago
user310148
22
asked 2 days ago
user310148
22
asked 2 days ago
user310148
22
22
put on hold as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin yesterday
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.
put on hold as off-topic by amWhy, Paul Frost, Eevee Trainer, Saad, Lord_Farin yesterday
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.
May be n dimensional Jensen's inequality may fit the bill?
– Satish Ramanathan
2 days ago
1
This needs rewriting. It is strangely confusing.
– zhw.
2 days ago
add a comment |
May be n dimensional Jensen's inequality may fit the bill?
– Satish Ramanathan
2 days ago
1
This needs rewriting. It is strangely confusing.
– zhw.
2 days ago
May be n dimensional Jensen's inequality may fit the bill?
– Satish Ramanathan
2 days ago
May be n dimensional Jensen's inequality may fit the bill?
– Satish Ramanathan
2 days ago
1
1
This needs rewriting. It is strangely confusing.
– zhw.
2 days ago
This needs rewriting. It is strangely confusing.
– zhw.
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
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 2 days ago
hamam_Abdallah
37.9k21634
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
hamam_Abdallah
37.9k21634
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
add a comment |
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 2 days ago
hamam_Abdallah
37.9k21634
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
add a comment |
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 2 days ago
hamam_Abdallah
37.9k21634
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 2 days ago
hamam_Abdallah
37.9k21634
answered 2 days ago
hamam_Abdallah
37.9k21634
answered 2 days ago
hamam_Abdallah
37.9k21634
answered 2 days ago
hamam_Abdallah
37.9k21634
37.9k21634
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
add a comment |
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
Hey thanks yea, sorry i forgot to say i got that part already sorry
– user310148
2 days ago
add a comment |
May be n dimensional Jensen's inequality may fit the bill?
– Satish Ramanathan
2 days ago
1
This needs rewriting. It is strangely confusing.
– zhw.
2 days ago