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Smooth function touching an upper semicontinuous one from above at a maximum point

Definition of smoothness “up to boundary”Show that the upper semicontinuous has a maximumContinuous approximation of upper semicontinuous indicator functionA problem about the special case of the Vitali-Caratheodory theoremBoundedness of an operator in $L^p$ spaceMaximum of a upper semicontinuous functionUpper semicontinuous function attains its supremumHow to prove a function is upper semicontinuous?Show that oscillation function is upper semicontinuousupper semicontinuous function and bounded sequenceShow that the function $x mapsto dim(mathrmIm(D_xf))$ is locally increasing and upper semicontinuous

0

$begingroup$

Let $Omega$ be an open bounded set. Let $u$ be an upper semicontinuous function on $bar Omega$ such that $u = 0$ on $partial Omega$ with a global maximum point at $bar x$.

How can one find a function $u_epsilon in C^infty(bar Omega)$ such that the following holds?

• 
  $bar x$ is a global maximum point of $u_epsilon$,


• 
  $u(bar x) = u_epsilon(bar x)$,


• 
  $u_epsilon ge u$ for every $x in bar Omega$,


• 
  $u_epsilon = 0$ on $partial Omega$,


• 
  $Vert u_epsilon - u Vert_L^infty(bar Omega) le epsilon$.

real-analysis calculus functional-analysis multivariable-calculus

share|cite|improve this question

edited Mar 11 at 19:56

Riku

asked Mar 11 at 19:51

RikuRiku

345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think that this is possible. Take $Omega = (0,1)$ and $u(x) := sqrtx , (1-x)$. Then, you cannot find $v in C^infty(barOmega)$ with $v ge u$ and $v(0) = u(0)$.
  $endgroup$
  – gerw
  Mar 12 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gerw Why do you think not?
  $endgroup$
  – Riku
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $v$ cannot be differentiable in $0$. The difference quotient will diverge.
  $endgroup$
  – gerw
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gerw It only needs to be differentiable in the interior. math.stackexchange.com/questions/421627/…
  $endgroup$
  – Riku
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the answer to this question says that both notions are equivalent. Otherwise, you are invited to construct such a $v$.
  $endgroup$
  – gerw
  18 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $Omega$ be an open bounded set. Let $u$ be an upper semicontinuous function on $bar Omega$ such that $u = 0$ on $partial Omega$ with a global maximum point at $bar x$.

How can one find a function $u_epsilon in C^infty(bar Omega)$ such that the following holds?

• 
  $bar x$ is a global maximum point of $u_epsilon$,


• 
  $u(bar x) = u_epsilon(bar x)$,


• 
  $u_epsilon ge u$ for every $x in bar Omega$,


• 
  $u_epsilon = 0$ on $partial Omega$,


• 
  $Vert u_epsilon - u Vert_L^infty(bar Omega) le epsilon$.

real-analysis calculus functional-analysis multivariable-calculus

share|cite|improve this question

edited Mar 11 at 19:56

Riku

asked Mar 11 at 19:51

RikuRiku

345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think that this is possible. Take $Omega = (0,1)$ and $u(x) := sqrtx , (1-x)$. Then, you cannot find $v in C^infty(barOmega)$ with $v ge u$ and $v(0) = u(0)$.
  $endgroup$
  – gerw
  Mar 12 at 8:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gerw Why do you think not?
  $endgroup$
  – Riku
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $v$ cannot be differentiable in $0$. The difference quotient will diverge.
  $endgroup$
  – gerw
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @gerw It only needs to be differentiable in the interior. math.stackexchange.com/questions/421627/…
  $endgroup$
  – Riku
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  the answer to this question says that both notions are equivalent. Otherwise, you are invited to construct such a $v$.
  $endgroup$
  – gerw
  18 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $Omega$ be an open bounded set. Let $u$ be an upper semicontinuous function on $bar Omega$ such that $u = 0$ on $partial Omega$ with a global maximum point at $bar x$.

How can one find a function $u_epsilon in C^infty(bar Omega)$ such that the following holds?

• 
  $bar x$ is a global maximum point of $u_epsilon$,


• 
  $u(bar x) = u_epsilon(bar x)$,


• 
  $u_epsilon ge u$ for every $x in bar Omega$,


• 
  $u_epsilon = 0$ on $partial Omega$,


• 
  $Vert u_epsilon - u Vert_L^infty(bar Omega) le epsilon$.

real-analysis calculus functional-analysis multivariable-calculus

share|cite|improve this question

edited Mar 11 at 19:56

Riku

asked Mar 11 at 19:51

RikuRiku

345

$endgroup$

share|cite|improve this question

edited Mar 11 at 19:56

Riku

asked Mar 11 at 19:51

RikuRiku

345

share|cite|improve this question

edited Mar 11 at 19:56

Riku

asked Mar 11 at 19:51

RikuRiku

345

asked Mar 11 at 19:51

RikuRiku

345

asked Mar 11 at 19:51

RikuRiku

345

1 Answer
1

active

oldest

votes

1

$begingroup$

If you rephrase everything in terms of lower semicontinuity and minimums you can use the Moreau Envelope (also called the Yosida regularization) and get something close to what you need.

edit: with your assumptions we have that the function $-u$ is lower semicontinuous on $barOmega$ with $-u=0$ on $partial Omega$ and a global minimum at $barx$. Consider the function:
$$
(-u)^(beta)(x) = inflimits_yleft-u(y) + frac12beta
$$

Then, $barx$ is a global minimum of $(-u)^(beta)$, $(-u)^(beta)leq (-u)$ for all $xinbarOmega$ and $(-u)^(beta)(barx)=-u(barx)$.

share|cite|improve this answer

edited Mar 11 at 20:03

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you add more details on this?
  $endgroup$
  – Riku
  Mar 11 at 19:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well now that you added the last part about the norm it's no longer true...
  $endgroup$
  – Tony S.F.
  Mar 11 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If it fits the other, it'd still be interesting to me.
  $endgroup$
  – Riku
  Mar 11 at 20:00
  

  
  
  
  

add a comment | 

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1

$begingroup$

If you rephrase everything in terms of lower semicontinuity and minimums you can use the Moreau Envelope (also called the Yosida regularization) and get something close to what you need.

edit: with your assumptions we have that the function $-u$ is lower semicontinuous on $barOmega$ with $-u=0$ on $partial Omega$ and a global minimum at $barx$. Consider the function:
$$
(-u)^(beta)(x) = inflimits_yleft-u(y) + frac12beta
$$

Then, $barx$ is a global minimum of $(-u)^(beta)$, $(-u)^(beta)leq (-u)$ for all $xinbarOmega$ and $(-u)^(beta)(barx)=-u(barx)$.

share|cite|improve this answer

edited Mar 11 at 20:03

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you add more details on this?
  $endgroup$
  – Riku
  Mar 11 at 19:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well now that you added the last part about the norm it's no longer true...
  $endgroup$
  – Tony S.F.
  Mar 11 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If it fits the other, it'd still be interesting to me.
  $endgroup$
  – Riku
  Mar 11 at 20:00
  

  
  
  
  

add a comment | 

1

$begingroup$

If you rephrase everything in terms of lower semicontinuity and minimums you can use the Moreau Envelope (also called the Yosida regularization) and get something close to what you need.

edit: with your assumptions we have that the function $-u$ is lower semicontinuous on $barOmega$ with $-u=0$ on $partial Omega$ and a global minimum at $barx$. Consider the function:
$$
(-u)^(beta)(x) = inflimits_yleft-u(y) + frac12beta
$$

Then, $barx$ is a global minimum of $(-u)^(beta)$, $(-u)^(beta)leq (-u)$ for all $xinbarOmega$ and $(-u)^(beta)(barx)=-u(barx)$.

share|cite|improve this answer

edited Mar 11 at 20:03

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Could you add more details on this?
  $endgroup$
  – Riku
  Mar 11 at 19:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well now that you added the last part about the norm it's no longer true...
  $endgroup$
  – Tony S.F.
  Mar 11 at 20:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If it fits the other, it'd still be interesting to me.
  $endgroup$
  – Riku
  Mar 11 at 20:00
  

  
  
  
  

add a comment | 

$begingroup$

If you rephrase everything in terms of lower semicontinuity and minimums you can use the Moreau Envelope (also called the Yosida regularization) and get something close to what you need.

edit: with your assumptions we have that the function $-u$ is lower semicontinuous on $barOmega$ with $-u=0$ on $partial Omega$ and a global minimum at $barx$. Consider the function:
$$
(-u)^(beta)(x) = inflimits_yleft-u(y) + frac12beta
$$

Then, $barx$ is a global minimum of $(-u)^(beta)$, $(-u)^(beta)leq (-u)$ for all $xinbarOmega$ and $(-u)^(beta)(barx)=-u(barx)$.

share|cite|improve this answer

edited Mar 11 at 20:03

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028

$endgroup$

share|cite|improve this answer

edited Mar 11 at 20:03

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028

answered Mar 11 at 19:59

Tony S.F.Tony S.F.

3,33321028