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Let $a, b$, be positives and $a leq b$. For which $p$ the function $frac{1} {x^a+x^b}$ is $L_p(0,+infty)$?

Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$$1le p lt infty$ and $f_k$ nonnegative increasing. Then $f_kto f$ in $L_p$ iff $sup_k||f_k||_p lt infty$.$mathcal L_p([0,1],S,lambda)$ is separable for $pin(0,infty)$Log-convexity of the p-norms of a fixed functionThe exercise regarding $L_p$ and $l_p$ spacesHow to show that $L_p(mu)_+$ is not solid (i.e., $L_p(mu)_+$ has empty interior)?Confused about using Cauchy sequence $(x_n)_1^{infty} in l_p$ to show the sequence space $l_p$ complete$L_p([1,infty],mu)$ isn't contained in $L_q([1,infty],mu)$ where $mu$ is Lebesgue measure.Prove $left|xright|_2leqsqrt{n}left|xright|_infty$What is the dimension of $l_p$-space, $1 leq p < infty$?

1

1

$begingroup$

Let $a, b$, be positives and $a leq b$. For which $p$ the function $frac{1} {x^a+x^b}$ is $L_p(0,+infty)$?

I have solved without any problem the case $p=infty$ (and it is not $L_p$ in this case) , but I can't solve the other case.

I suppose I have to use some inequalities but I have no idea which ones, and how.

Anyone can help me?

functional-analysis hilbert-spaces normed-spaces

share|cite|improve this question

edited 2 days ago

Brian

36012

asked 2 days ago

user149240user149240

755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No idea at all? If you discuss $L_p$ spaces, I guess you know of elementary calculus, and then I think we could expect some effort. Do you know for what values of $alpha$ and $beta$ the integrals $int_0^1 x^alpha$ and $int_1^{+infty}x^beta$ are convergent?
  $endgroup$
  – mickep
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes of course but how can I split the main integral into those very simple integrals? There is a fraction!
  $endgroup$
  – user149240
  2 days ago
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Let $a, b$, be positives and $a leq b$. For which $p$ the function $frac{1} {x^a+x^b}$ is $L_p(0,+infty)$?

I have solved without any problem the case $p=infty$ (and it is not $L_p$ in this case) , but I can't solve the other case.

I suppose I have to use some inequalities but I have no idea which ones, and how.

Anyone can help me?

functional-analysis hilbert-spaces normed-spaces

share|cite|improve this question

edited 2 days ago

Brian

36012

asked 2 days ago

user149240user149240

755

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No idea at all? If you discuss $L_p$ spaces, I guess you know of elementary calculus, and then I think we could expect some effort. Do you know for what values of $alpha$ and $beta$ the integrals $int_0^1 x^alpha$ and $int_1^{+infty}x^beta$ are convergent?
  $endgroup$
  – mickep
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes of course but how can I split the main integral into those very simple integrals? There is a fraction!
  $endgroup$
  – user149240
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $a, b$, be positives and $a leq b$. For which $p$ the function $frac{1} {x^a+x^b}$ is $L_p(0,+infty)$?

I have solved without any problem the case $p=infty$ (and it is not $L_p$ in this case) , but I can't solve the other case.

I suppose I have to use some inequalities but I have no idea which ones, and how.

Anyone can help me?

functional-analysis hilbert-spaces normed-spaces

share|cite|improve this question

edited 2 days ago

Brian

36012

asked 2 days ago

user149240user149240

755

$endgroup$

share|cite|improve this question

edited 2 days ago

Brian

36012

asked 2 days ago

user149240user149240

755

share|cite|improve this question

edited 2 days ago

Brian

36012

asked 2 days ago

user149240user149240

755

edited 2 days ago

Brian

36012

edited 2 days ago

Brian

36012

asked 2 days ago

user149240user149240

755

asked 2 days ago

user149240user149240

755

2 Answers
2

active

oldest

votes

0

$begingroup$

Since $bgeq a>0$ you will have
$$
frac{1}{x^a+x^b}approx frac{1}{x^a}quadtext{if }xapprox 0
$$
and
$$
frac{1}{x^a+x^b}approx frac{1}{x^b}quadtext{as }xto+infty.
$$
Now use comparison to make this argument complete.

share|cite|improve this answer

answered 2 days ago

mickepmickep

18.6k12250

$endgroup$

add a comment | 

0

$begingroup$

For $x in (0,1)$ we have
$$x^a le x^a+x^b le 2x^a$$
so
$$int_{(0,1)}frac1{2^px^{ap}} le int_{(0,1)}frac1{(x^a+x^b)^p} le int_{(0,1)}frac1{x^{ap}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(0,1)$ if and only if $ap > 1$, or $p > frac1a$.

Similarly, for $x in (1, infty)$ we have
$$x^b le x^a+x^b le 2x^b$$
so
$$int_{(1, infty)}frac1{2^px^{bp}} le int_{(1, infty)}frac1{(x^a+x^b)^p} le int_{(1, infty)}frac1{x^{bp}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(1, infty)$ if and only if $bp < 1$, or $p < frac1b$.

Conclusion: $frac1{x^a+x^b}$ is in $L^p(0, infty)$ if and only if
$$p in left(frac1a, frac1bright)$$

share|cite|improve this answer

answered 2 days ago

mechanodroidmechanodroid

28.5k62548

$endgroup$

add a comment | 

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0

$begingroup$

Since $bgeq a>0$ you will have
$$
frac{1}{x^a+x^b}approx frac{1}{x^a}quadtext{if }xapprox 0
$$
and
$$
frac{1}{x^a+x^b}approx frac{1}{x^b}quadtext{as }xto+infty.
$$
Now use comparison to make this argument complete.

share|cite|improve this answer

answered 2 days ago

mickepmickep

18.6k12250

$endgroup$

add a comment | 

0

$begingroup$

Since $bgeq a>0$ you will have
$$
frac{1}{x^a+x^b}approx frac{1}{x^a}quadtext{if }xapprox 0
$$
and
$$
frac{1}{x^a+x^b}approx frac{1}{x^b}quadtext{as }xto+infty.
$$
Now use comparison to make this argument complete.

share|cite|improve this answer

answered 2 days ago

mickepmickep

18.6k12250

$endgroup$

add a comment | 

$begingroup$

Since $bgeq a>0$ you will have
$$
frac{1}{x^a+x^b}approx frac{1}{x^a}quadtext{if }xapprox 0
$$
and
$$
frac{1}{x^a+x^b}approx frac{1}{x^b}quadtext{as }xto+infty.
$$
Now use comparison to make this argument complete.

share|cite|improve this answer

answered 2 days ago

mickepmickep

18.6k12250

$endgroup$

share|cite|improve this answer

answered 2 days ago

mickepmickep

18.6k12250

answered 2 days ago

mickepmickep

18.6k12250

answered 2 days ago

mickepmickep

18.6k12250

0

$begingroup$

For $x in (0,1)$ we have
$$x^a le x^a+x^b le 2x^a$$
so
$$int_{(0,1)}frac1{2^px^{ap}} le int_{(0,1)}frac1{(x^a+x^b)^p} le int_{(0,1)}frac1{x^{ap}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(0,1)$ if and only if $ap > 1$, or $p > frac1a$.

Similarly, for $x in (1, infty)$ we have
$$x^b le x^a+x^b le 2x^b$$
so
$$int_{(1, infty)}frac1{2^px^{bp}} le int_{(1, infty)}frac1{(x^a+x^b)^p} le int_{(1, infty)}frac1{x^{bp}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(1, infty)$ if and only if $bp < 1$, or $p < frac1b$.

Conclusion: $frac1{x^a+x^b}$ is in $L^p(0, infty)$ if and only if
$$p in left(frac1a, frac1bright)$$

share|cite|improve this answer

answered 2 days ago

mechanodroidmechanodroid

28.5k62548

$endgroup$

add a comment | 

0

$begingroup$

For $x in (0,1)$ we have
$$x^a le x^a+x^b le 2x^a$$
so
$$int_{(0,1)}frac1{2^px^{ap}} le int_{(0,1)}frac1{(x^a+x^b)^p} le int_{(0,1)}frac1{x^{ap}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(0,1)$ if and only if $ap > 1$, or $p > frac1a$.

Similarly, for $x in (1, infty)$ we have
$$x^b le x^a+x^b le 2x^b$$
so
$$int_{(1, infty)}frac1{2^px^{bp}} le int_{(1, infty)}frac1{(x^a+x^b)^p} le int_{(1, infty)}frac1{x^{bp}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(1, infty)$ if and only if $bp < 1$, or $p < frac1b$.

Conclusion: $frac1{x^a+x^b}$ is in $L^p(0, infty)$ if and only if
$$p in left(frac1a, frac1bright)$$

share|cite|improve this answer

answered 2 days ago

mechanodroidmechanodroid

28.5k62548

$endgroup$

add a comment | 

$begingroup$

For $x in (0,1)$ we have
$$x^a le x^a+x^b le 2x^a$$
so
$$int_{(0,1)}frac1{2^px^{ap}} le int_{(0,1)}frac1{(x^a+x^b)^p} le int_{(0,1)}frac1{x^{ap}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(0,1)$ if and only if $ap > 1$, or $p > frac1a$.

Similarly, for $x in (1, infty)$ we have
$$x^b le x^a+x^b le 2x^b$$
so
$$int_{(1, infty)}frac1{2^px^{bp}} le int_{(1, infty)}frac1{(x^a+x^b)^p} le int_{(1, infty)}frac1{x^{bp}}$$
Hence $frac1{x^a+x^b}$ is in $L^p(1, infty)$ if and only if $bp < 1$, or $p < frac1b$.

Conclusion: $frac1{x^a+x^b}$ is in $L^p(0, infty)$ if and only if
$$p in left(frac1a, frac1bright)$$

share|cite|improve this answer

answered 2 days ago

mechanodroidmechanodroid

28.5k62548

$endgroup$

share|cite|improve this answer

answered 2 days ago

mechanodroidmechanodroid

28.5k62548

answered 2 days ago

mechanodroidmechanodroid

28.5k62548

answered 2 days ago

mechanodroidmechanodroid

28.5k62548