Smooth function touching an upper semicontinuous one from above at a maximum pointDefinition 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
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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
$begingroup$
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
asked Mar 11 at 19:51
RikuRiku
345
$endgroup$
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
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
asked Mar 11 at 19:51
RikuRiku
345
$endgroup$
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
real-analysis calculus functional-analysis multivariable-calculus
asked Mar 11 at 19:51
RikuRiku
345
asked Mar 11 at 19:51
RikuRiku
345
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
345
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@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
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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 $(-u)^(beta)(barx)=-u(barx)$.
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 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $(-u)^(beta)(barx)=-u(barx)$.
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 $(-u)^(beta)(barx)=-u(barx)$.
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 $(-u)^(beta)(barx)=-u(barx)$.
answered Mar 11 at 19:59
Tony S.F.Tony S.F.
3,33321028
$endgroup$
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 $(-u)^(beta)(barx)=-u(barx)$.
answered Mar 11 at 19:59
Tony S.F.Tony S.F.
3,33321028
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
3,33321028
$begingroup$
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 |
$begingroup$
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
$begingroup$
Could you add more details on this?
$endgroup$
– Riku
Mar 11 at 19:59
$begingroup$
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
$begingroup$
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
$begingroup$
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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$begingroup$
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