Dtjytyk

Should I be concerned about student access to a test bank?

Have any astronauts/cosmonauts died in space?

DisplayForm problem with pi in FractionBox

Do native speakers use "ultima" and "proxima" frequently in spoken English?

When should a starting writer get his own webpage?

Norwegian Refugee travel document

Does convergence of polynomials imply that of its coefficients?

Why does Surtur say that Thor is Asgard's doom?

How can a new country break out from a developed country without war?

How are passwords stolen from companies if they only store hashes?

Print a physical multiplication table

Recursively updating the MLE as new observations stream in

Hot air balloons as primitive bombers

Nested Dynamic SOQL Query

PTIJ: At the Passover Seder, is one allowed to speak more than once during Maggid?

Is there any common country to visit for uk and schengen visa?

"Marked down as someone wanting to sell shares." What does that mean?

Imaginary part of expression too difficult to calculate

Asserting that Atheism and Theism are both faith based positions

The English Debate

Why I don't get the wanted width of tcbox?

Symbolism of 18 Journeyers

What is the reasoning behind standardization (dividing by standard deviation)?

Is VPN a layer 3 concept?

Why is it necessary to show subsequence convergence in the extreme value theorem?

Proof of Extreme Value Theorem. Modus operandi. (S. Abbott pp 115 t4.4.3)Convergent subsequence proofAn issue with Weierstrass theorem's proof (extreme value theorem)show that if a subsequence of a cauchy sequence converges, then the whole sequence convergesExtreme value theorem using continuous image of compact is compact + Heine BorelLet $F_n(x) = int_a^bf_n(t) dt, xin [a,b]$, show that there exists a subsequence $F_{n_k}$ which converges uniformly$L^1$ convergence and pointwise convergence of subsequence (counterexample)The proofs for every convergence subsequenceQuestion about a proof of the extreme value theorem.Bolzano-Weierstrass implies the monotone convergence theorem.

1

$begingroup$

I’m probably making this more complicated than it needs to be, but I’m trying to figure out why it is necessary to prove the convergence of a subsequence in proving the extreme value theorem (EVT). I typically see the EVT proved as follows:

Given an interval $f:[a,b]rightarrowmathbb{R}$:

• 1) prove that the sequence $f(x_n)$ converges to the supremum of the set $S={f:[a,b]}$.


• 2) prove that a subseqeunce $f(x_{n_k})$ converges to $f(x_{0})$, and since a subsequence of a convergent sequence must converge to the same limit, then that means that $f(x_{0})=sup S$.

Why is the second bullet point necessary, if we already proved that $f$ attains the supremum?

real-analysis calculus continuity real-numbers supremum-and-infimum

share|cite|improve this question

edited Mar 12 at 7:03

guimption

asked Mar 12 at 3:46

guimptionguimption

1309

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The first bullet point doesn't say that $f$ attains the supremum, it only says that there is a sequence $x_n$ so that $f(x_n)$ approaches the supremum. You still have to prove that there is a point $x_0 in [a,b]$ with $f(x_0) = sup f$.
  $endgroup$
  – Jair Taylor
  Mar 12 at 3:50
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In general, if your interval is not closed, for example, the sequence $x_n$ may not converge in your interval. $x_n = 1-1/n$ for $f(x) = x$ on the interval $(0,1)$. You have to somehow 'trap' a limit point of your sequence so it doesn't escape.
  $endgroup$
  – Jair Taylor
  Mar 12 at 3:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or, even if the interval is closed $x_n$ might still not converge - e.g. if $f$ has more than one maximizing point and $x_n$ alternates between which it gets close to. Thus you'll have to pick a subsequence that converges to one $x$ in particular.
  $endgroup$
  – Jair Taylor
  Mar 12 at 4:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For 1), yes, the issue is that even though $f(x_n)$ converges to the supremum that doesn't tell you that $x_n$ converges.
  $endgroup$
  – Jair Taylor
  Mar 12 at 7:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the second question, yes, all three are required. But I don't quite think of it as "proving the supremum and the function value are equal" - it is an existence proof. You can't prove $sup f = f(x)$ until you have defined what $x$ is. I would put it as "showing that $f$ attains its supremum at a particular input $x$."
  $endgroup$
  – Jair Taylor
  Mar 12 at 7:40
  
  
  

  
  
  
  

 | 
show 13 more comments

1

$begingroup$

I’m probably making this more complicated than it needs to be, but I’m trying to figure out why it is necessary to prove the convergence of a subsequence in proving the extreme value theorem (EVT). I typically see the EVT proved as follows:

Given an interval $f:[a,b]rightarrowmathbb{R}$:

• 1) prove that the sequence $f(x_n)$ converges to the supremum of the set $S={f:[a,b]}$.


• 2) prove that a subseqeunce $f(x_{n_k})$ converges to $f(x_{0})$, and since a subsequence of a convergent sequence must converge to the same limit, then that means that $f(x_{0})=sup S$.

Why is the second bullet point necessary, if we already proved that $f$ attains the supremum?

real-analysis calculus continuity real-numbers supremum-and-infimum

share|cite|improve this question

edited Mar 12 at 7:03

guimption

asked Mar 12 at 3:46

guimptionguimption

1309

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The first bullet point doesn't say that $f$ attains the supremum, it only says that there is a sequence $x_n$ so that $f(x_n)$ approaches the supremum. You still have to prove that there is a point $x_0 in [a,b]$ with $f(x_0) = sup f$.
  $endgroup$
  – Jair Taylor
  Mar 12 at 3:50
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  In general, if your interval is not closed, for example, the sequence $x_n$ may not converge in your interval. $x_n = 1-1/n$ for $f(x) = x$ on the interval $(0,1)$. You have to somehow 'trap' a limit point of your sequence so it doesn't escape.
  $endgroup$
  – Jair Taylor
  Mar 12 at 3:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or, even if the interval is closed $x_n$ might still not converge - e.g. if $f$ has more than one maximizing point and $x_n$ alternates between which it gets close to. Thus you'll have to pick a subsequence that converges to one $x$ in particular.
  $endgroup$
  – Jair Taylor
  Mar 12 at 4:53
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For 1), yes, the issue is that even though $f(x_n)$ converges to the supremum that doesn't tell you that $x_n$ converges.
  $endgroup$
  – Jair Taylor
  Mar 12 at 7:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For the second question, yes, all three are required. But I don't quite think of it as "proving the supremum and the function value are equal" - it is an existence proof. You can't prove $sup f = f(x)$ until you have defined what $x$ is. I would put it as "showing that $f$ attains its supremum at a particular input $x$."
  $endgroup$
  – Jair Taylor
  Mar 12 at 7:40
  
  
  

  
  
  
  

 | 
show 13 more comments

$begingroup$

I’m probably making this more complicated than it needs to be, but I’m trying to figure out why it is necessary to prove the convergence of a subsequence in proving the extreme value theorem (EVT). I typically see the EVT proved as follows:

Given an interval $f:[a,b]rightarrowmathbb{R}$:

• 1) prove that the sequence $f(x_n)$ converges to the supremum of the set $S={f:[a,b]}$.


• 2) prove that a subseqeunce $f(x_{n_k})$ converges to $f(x_{0})$, and since a subsequence of a convergent sequence must converge to the same limit, then that means that $f(x_{0})=sup S$.

Why is the second bullet point necessary, if we already proved that $f$ attains the supremum?

real-analysis calculus continuity real-numbers supremum-and-infimum

share|cite|improve this question

edited Mar 12 at 7:03

guimption

asked Mar 12 at 3:46

guimptionguimption

1309

$endgroup$

share|cite|improve this question

edited Mar 12 at 7:03

guimption

asked Mar 12 at 3:46

guimptionguimption

1309

share|cite|improve this question

edited Mar 12 at 7:03

guimption

asked Mar 12 at 3:46

guimptionguimption

1309

asked Mar 12 at 3:46

guimptionguimption

1309

asked Mar 12 at 3:46

guimptionguimption

1309