Dtjytyk

Review your own paper in Mathematics

What to do when eye contact makes your coworker uncomfortable?

What's the name of the logical fallacy where a debater extends a statement far beyond the original statement to make it true?

Why should universal income be universal?

Why Shazam when there is already Superman?

I found an audio circuit and I built it just fine, but I find it a bit too quiet. How do I amplify the output so that it is a bit louder?

What is the highest possible scrabble score for placing a single tile

PTIJ: Why is Haman obsessed with Bose?

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

Why do ¬, ∀ and ∃ have the same precedence?

Is this toilet slogan correct usage of the English language?

What does "Scientists rise up against statistical significance" mean? (Comment in Nature)

What kind of floor tile is this?

What fields between the rationals and the reals allow a good notion of 2D distance?

Can you use Vicious Mockery to win an argument or gain favours?

When were female captains banned from Starfleet?

Non-trope happy ending?

How do I fix the group tension caused by my character stealing and possibly killing without provocation?

awk assign to multiple variables at once

Why can't the Brexit deadlock in the UK parliament be solved with a plurality vote?

What is the English pronunciation of "pain au chocolat"?

Has any country ever had 2 former presidents in jail simultaneously?

How does electrical safety system work on ISS?

Why is so much work done on numerical verification of the Riemann Hypothesis?

How to show $(f(x_k))$ converges to the minimum when $f(x)$ is strongly convex? [closed]

$F(x) = f(x) + g(x) + h(x)$, where h(x) is strongly convex , is also strongly convexIf the sequence ${y_k}$ is bounded and $sum |x_k|$ converges, then $sum x_k y_k$ converges.Gluing two strongly convex functionshow that if a subsequence of a cauchy sequence converges, then the whole sequence convergesshow that $X_n=X_k$ in a convergent monotone sequenceConfusion over the definition of strongly convexProving that a strongly convex function is coercivelevel set strongly convex and smooth functionsIf $f(y)=limsup_k|y-x_k|^2$ then $f$ is a strictly convex functionHow to show two different definitions of $alpha$-strongly convex are equivalent?

1

0

$begingroup$

Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a $C^2$ strongly convex function on $mathbb{R}^n$, i.e., $z^{top}nabla^2f(x)z geq |z|_2^2$ for all $z in mathbb{R}^n$. Let $(x_k)$ be a vector sequence such that the real sequence $(f(x_k))$ is strictly decreasing, i.e, $f(x_k) > f(x_{k+1})$ for each $k$. Let $(x_{k_j})$ be a convergent subsequence of $(x_k)$, i.e., $(x_{k_j}) rightarrow x_*$, where $nabla f(x_*)=0$.

Show that $(f(x_k))$ converges to $f(x_*)$.

real-analysis convergence optimization convex-optimization

share|cite|improve this question

asked Mar 13 at 6:23

SepideSepide

4938

$endgroup$

closed as off-topic by Saad, José Carlos Santos, Gibbs, Song, mau Mar 13 at 15:06

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." – Saad, José Carlos Santos, Gibbs, Song, mau

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

1

0

$begingroup$

Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a $C^2$ strongly convex function on $mathbb{R}^n$, i.e., $z^{top}nabla^2f(x)z geq |z|_2^2$ for all $z in mathbb{R}^n$. Let $(x_k)$ be a vector sequence such that the real sequence $(f(x_k))$ is strictly decreasing, i.e, $f(x_k) > f(x_{k+1})$ for each $k$. Let $(x_{k_j})$ be a convergent subsequence of $(x_k)$, i.e., $(x_{k_j}) rightarrow x_*$, where $nabla f(x_*)=0$.

Show that $(f(x_k))$ converges to $f(x_*)$.

real-analysis convergence optimization convex-optimization

share|cite|improve this question

asked Mar 13 at 6:23

SepideSepide

4938

$endgroup$

closed as off-topic by Saad, José Carlos Santos, Gibbs, Song, mau Mar 13 at 15:06

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." – Saad, José Carlos Santos, Gibbs, Song, mau

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

$begingroup$

Let $f: mathbb{R}^n rightarrow mathbb{R}$ be a $C^2$ strongly convex function on $mathbb{R}^n$, i.e., $z^{top}nabla^2f(x)z geq |z|_2^2$ for all $z in mathbb{R}^n$. Let $(x_k)$ be a vector sequence such that the real sequence $(f(x_k))$ is strictly decreasing, i.e, $f(x_k) > f(x_{k+1})$ for each $k$. Let $(x_{k_j})$ be a convergent subsequence of $(x_k)$, i.e., $(x_{k_j}) rightarrow x_*$, where $nabla f(x_*)=0$.

Show that $(f(x_k))$ converges to $f(x_*)$.

real-analysis convergence optimization convex-optimization

share|cite|improve this question

asked Mar 13 at 6:23

SepideSepide

4938

$endgroup$

share|cite|improve this question

asked Mar 13 at 6:23

SepideSepide

4938

share|cite|improve this question

asked Mar 13 at 6:23

SepideSepide

4938

asked Mar 13 at 6:23

SepideSepide

4938

asked Mar 13 at 6:23

SepideSepide

4938