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inf sup problem need help

Prove that a sequence converges to a finite limit iff lim inf equals lim supProve $inf_xin D [f(x)g(x)]lesup_xin Df(x)cdotinf_xin D g(x).$Find all differentiable functions $f:[0;2] to BbbR$ such that $int_0^2xf(x)dx=f(0)+f(2)$Prove this $sup f_n$ has infinite integralProve supremum has infinite integrationShow that inf $f(x)$ is achieved. Find $inf f(x)$.Rolle's Theorem in reverseWhat are the conditions on $f$ for $sup f$ and $f(sup)$ be interchangeable?Showing that the pointwise limit of continuous functions equals its supremum somewhere on compact domainThe definition and norm of $(mathbbR^*)^2$

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

4

2

$begingroup$

Let $BbbM $ be the set of decreasing smooth functions in $[0,1]$
for which $f(1)=0.$

Find $$inf_f in BbbMsup_x in [0;1] fracx*f(x)int_0^1f(t)dt
$$

Let now $F(x)=x*f(x)$.
I only find out that $ F(0)=0 $ and $F(1)=0$. Only idea is to proof somehow that F is concave and use Rolle's theorem. Can you give some hints?

real-analysis inequality supremum-and-infimum

share|cite|improve this question

edited Jul 22 at 10:01

Динар Ахметзянов

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
  $endgroup$
  – José Carlos Santos
  Jul 22 at 7:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The only thing I found is that this infimum is at most $1/e$. This can be shown considering $f(x)=(1-x)^n$ and sending $n to infty$.
  $endgroup$
  – Crostul
  Jul 22 at 8:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What operation is meant by $*$? Multiplication or convolution?
  $endgroup$
  – daw
  Jul 22 at 9:34
  

  
  
  
  

add a comment | 

4

2

$begingroup$

Let $BbbM $ be the set of decreasing smooth functions in $[0,1]$
for which $f(1)=0.$

Find $$inf_f in BbbMsup_x in [0;1] fracx*f(x)int_0^1f(t)dt
$$

Let now $F(x)=x*f(x)$.
I only find out that $ F(0)=0 $ and $F(1)=0$. Only idea is to proof somehow that F is concave and use Rolle's theorem. Can you give some hints?

real-analysis inequality supremum-and-infimum

share|cite|improve this question

edited Jul 22 at 10:01

Динар Ахметзянов

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
  $endgroup$
  – José Carlos Santos
  Jul 22 at 7:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The only thing I found is that this infimum is at most $1/e$. This can be shown considering $f(x)=(1-x)^n$ and sending $n to infty$.
  $endgroup$
  – Crostul
  Jul 22 at 8:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What operation is meant by $*$? Multiplication or convolution?
  $endgroup$
  – daw
  Jul 22 at 9:34
  

  
  
  
  

add a comment | 

$begingroup$

Let $BbbM $ be the set of decreasing smooth functions in $[0,1]$
for which $f(1)=0.$

Find $$inf_f in BbbMsup_x in [0;1] fracx*f(x)int_0^1f(t)dt
$$

Let now $F(x)=x*f(x)$.
I only find out that $ F(0)=0 $ and $F(1)=0$. Only idea is to proof somehow that F is concave and use Rolle's theorem. Can you give some hints?

real-analysis inequality supremum-and-infimum

share|cite|improve this question

edited Jul 22 at 10:01

Динар Ахметзянов

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

$endgroup$

share|cite|improve this question

edited Jul 22 at 10:01

Динар Ахметзянов

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

share|cite|improve this question

edited Jul 22 at 10:01

Динар Ахметзянов

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

asked Jul 22 at 7:20

Динар АхметзяновДинар Ахметзянов

2344 bronze badges

1 Answer
1

active

oldest

votes

4

$begingroup$

The inf-sup is zero. Clearly, the quantity in question is non-negative. Let me show that the infimum is indeed zero..

Let $s<1$, define $f(x) = x^-s-1$. Then $int_0^1 f(x) = frac11-s-1 = frac ss-1$. The function $x f(x)$ has derivative $(1-s)x^-s-1$. Hence the product is minimal at $x_s=(1-s)^frac1s$.

Then
$$
fracsup_x xf(x) int f dt = ( (1-s)^frac1-ss-(1-s)^frac1s) cdot fracs-1s
= frac1s ((1-s)^frac1s-(1-s)^frac1s+1)
$$
which tends to zero for $snearrow 1$.

If you do not like to take this function with pole at zero, you can replace it by some smooth function: Define $f_epsilon(x) = f(x)$ on $(epsilon,1]$,
$f_epsilon(x)le f(x)$ on $(0,epsilon)$, such that $f_epsilon$ is smooth and decreasing. For $epsilon<x_s$, the $sup_x xf(x)$ is unchanged, while $int_0^1 f_epsilon$ will converge to $int_0^1 f$ for $epsilonto0$.

share|cite|improve this answer

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges

$endgroup$

add a comment | 

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4

$begingroup$

The inf-sup is zero. Clearly, the quantity in question is non-negative. Let me show that the infimum is indeed zero..

Let $s<1$, define $f(x) = x^-s-1$. Then $int_0^1 f(x) = frac11-s-1 = frac ss-1$. The function $x f(x)$ has derivative $(1-s)x^-s-1$. Hence the product is minimal at $x_s=(1-s)^frac1s$.

Then
$$
fracsup_x xf(x) int f dt = ( (1-s)^frac1-ss-(1-s)^frac1s) cdot fracs-1s
= frac1s ((1-s)^frac1s-(1-s)^frac1s+1)
$$
which tends to zero for $snearrow 1$.

If you do not like to take this function with pole at zero, you can replace it by some smooth function: Define $f_epsilon(x) = f(x)$ on $(epsilon,1]$,
$f_epsilon(x)le f(x)$ on $(0,epsilon)$, such that $f_epsilon$ is smooth and decreasing. For $epsilon<x_s$, the $sup_x xf(x)$ is unchanged, while $int_0^1 f_epsilon$ will converge to $int_0^1 f$ for $epsilonto0$.

share|cite|improve this answer

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

The inf-sup is zero. Clearly, the quantity in question is non-negative. Let me show that the infimum is indeed zero..

Let $s<1$, define $f(x) = x^-s-1$. Then $int_0^1 f(x) = frac11-s-1 = frac ss-1$. The function $x f(x)$ has derivative $(1-s)x^-s-1$. Hence the product is minimal at $x_s=(1-s)^frac1s$.

Then
$$
fracsup_x xf(x) int f dt = ( (1-s)^frac1-ss-(1-s)^frac1s) cdot fracs-1s
= frac1s ((1-s)^frac1s-(1-s)^frac1s+1)
$$
which tends to zero for $snearrow 1$.

If you do not like to take this function with pole at zero, you can replace it by some smooth function: Define $f_epsilon(x) = f(x)$ on $(epsilon,1]$,
$f_epsilon(x)le f(x)$ on $(0,epsilon)$, such that $f_epsilon$ is smooth and decreasing. For $epsilon<x_s$, the $sup_x xf(x)$ is unchanged, while $int_0^1 f_epsilon$ will converge to $int_0^1 f$ for $epsilonto0$.

share|cite|improve this answer

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges

$endgroup$

add a comment | 

$begingroup$

The inf-sup is zero. Clearly, the quantity in question is non-negative. Let me show that the infimum is indeed zero..

Let $s<1$, define $f(x) = x^-s-1$. Then $int_0^1 f(x) = frac11-s-1 = frac ss-1$. The function $x f(x)$ has derivative $(1-s)x^-s-1$. Hence the product is minimal at $x_s=(1-s)^frac1s$.

Then
$$
fracsup_x xf(x) int f dt = ( (1-s)^frac1-ss-(1-s)^frac1s) cdot fracs-1s
= frac1s ((1-s)^frac1s-(1-s)^frac1s+1)
$$
which tends to zero for $snearrow 1$.

If you do not like to take this function with pole at zero, you can replace it by some smooth function: Define $f_epsilon(x) = f(x)$ on $(epsilon,1]$,
$f_epsilon(x)le f(x)$ on $(0,epsilon)$, such that $f_epsilon$ is smooth and decreasing. For $epsilon<x_s$, the $sup_x xf(x)$ is unchanged, while $int_0^1 f_epsilon$ will converge to $int_0^1 f$ for $epsilonto0$.

share|cite|improve this answer

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges

$endgroup$

share|cite|improve this answer

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges

answered Jul 22 at 10:26

dawdaw

26.7k1818 silver badges4646 bronze badges