Sufficient condition for $L^infty$ distance bound
0
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Suppose for two functions $f, gin L^{infty}([0,1])$ and $$||f-g||_{L_1([0,1])}leq epsilon$$ Under what condition can we get $||f-g||_{L^infty([0,1])}leq K(epsilon)$ where $K(cdot)$ is some function (e.g., logarithm function)? What is the weakest condition to ensure $K(epsilon)=tilde{K}cdotepsilon$ for some constant $tilde{K}$?
real-analysis functional-analysis lebesgue-measure lp-spaces
asked Dec 8 '18 at 0:29
user348207user348207
182
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add a comment |
0
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Suppose for two functions $f, gin L^{infty}([0,1])$ and $$||f-g||_{L_1([0,1])}leq epsilon$$ Under what condition can we get $||f-g||_{L^infty([0,1])}leq K(epsilon)$ where $K(cdot)$ is some function (e.g., logarithm function)? What is the weakest condition to ensure $K(epsilon)=tilde{K}cdotepsilon$ for some constant $tilde{K}$?
real-analysis functional-analysis lebesgue-measure lp-spaces
asked Dec 8 '18 at 0:29
user348207user348207
182
$endgroup$
$begingroup$
We can add conditions like $f,g$ satisfy certain regularity conditions, I just want to figure out what could be the weakest condition we need to impose.
$endgroup$
– user348207
Dec 8 '18 at 0:31
$begingroup$
For convenience I'll set $g = 0$ (this doesn't matter at all for what I'm going to say next). The prototypical example of a function with small $L_1$ but large $L_infty$ norm is a very tall function on a very narrow area. To exclude these, you'd like to control how much $f$ can grow in a short interval. One common condition for this is Hoelder continuity, which leads to a bound of $O(C^{1/alpha} epsilon^{ alpha/(alpha + 1)} )$. Note that these conditions imply uniform continuity, so may be too strong for your taste.
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– stochasticboy321
Dec 8 '18 at 5:44
add a comment |
0
0
0
$begingroup$
Suppose for two functions $f, gin L^{infty}([0,1])$ and $$||f-g||_{L_1([0,1])}leq epsilon$$ Under what condition can we get $||f-g||_{L^infty([0,1])}leq K(epsilon)$ where $K(cdot)$ is some function (e.g., logarithm function)? What is the weakest condition to ensure $K(epsilon)=tilde{K}cdotepsilon$ for some constant $tilde{K}$?
real-analysis functional-analysis lebesgue-measure lp-spaces
asked Dec 8 '18 at 0:29
user348207user348207
182
$endgroup$
Suppose for two functions $f, gin L^{infty}([0,1])$ and $$||f-g||_{L_1([0,1])}leq epsilon$$ Under what condition can we get $||f-g||_{L^infty([0,1])}leq K(epsilon)$ where $K(cdot)$ is some function (e.g., logarithm function)? What is the weakest condition to ensure $K(epsilon)=tilde{K}cdotepsilon$ for some constant $tilde{K}$?
real-analysis functional-analysis lebesgue-measure lp-spaces
real-analysis functional-analysis lebesgue-measure lp-spaces
asked Dec 8 '18 at 0:29
user348207user348207
182
asked Dec 8 '18 at 0:29
user348207user348207
182
asked Dec 8 '18 at 0:29
user348207user348207
182
asked Dec 8 '18 at 0:29
user348207user348207
182
asked Dec 8 '18 at 0:29
user348207user348207
182
182
$begingroup$
We can add conditions like $f,g$ satisfy certain regularity conditions, I just want to figure out what could be the weakest condition we need to impose.
$endgroup$
– user348207
Dec 8 '18 at 0:31
$begingroup$
For convenience I'll set $g = 0$ (this doesn't matter at all for what I'm going to say next). The prototypical example of a function with small $L_1$ but large $L_infty$ norm is a very tall function on a very narrow area. To exclude these, you'd like to control how much $f$ can grow in a short interval. One common condition for this is Hoelder continuity, which leads to a bound of $O(C^{1/alpha} epsilon^{ alpha/(alpha + 1)} )$. Note that these conditions imply uniform continuity, so may be too strong for your taste.
$endgroup$
– stochasticboy321
Dec 8 '18 at 5:44
add a comment |
$begingroup$
We can add conditions like $f,g$ satisfy certain regularity conditions, I just want to figure out what could be the weakest condition we need to impose.
$endgroup$
– user348207
Dec 8 '18 at 0:31
$begingroup$
For convenience I'll set $g = 0$ (this doesn't matter at all for what I'm going to say next). The prototypical example of a function with small $L_1$ but large $L_infty$ norm is a very tall function on a very narrow area. To exclude these, you'd like to control how much $f$ can grow in a short interval. One common condition for this is Hoelder continuity, which leads to a bound of $O(C^{1/alpha} epsilon^{ alpha/(alpha + 1)} )$. Note that these conditions imply uniform continuity, so may be too strong for your taste.
$endgroup$
– stochasticboy321
Dec 8 '18 at 5:44
$begingroup$
We can add conditions like $f,g$ satisfy certain regularity conditions, I just want to figure out what could be the weakest condition we need to impose.
$endgroup$
– user348207
Dec 8 '18 at 0:31
$begingroup$
We can add conditions like $f,g$ satisfy certain regularity conditions, I just want to figure out what could be the weakest condition we need to impose.
$endgroup$
– user348207
Dec 8 '18 at 0:31
$begingroup$
For convenience I'll set $g = 0$ (this doesn't matter at all for what I'm going to say next). The prototypical example of a function with small $L_1$ but large $L_infty$ norm is a very tall function on a very narrow area. To exclude these, you'd like to control how much $f$ can grow in a short interval. One common condition for this is Hoelder continuity, which leads to a bound of $O(C^{1/alpha} epsilon^{ alpha/(alpha + 1)} )$. Note that these conditions imply uniform continuity, so may be too strong for your taste.
$endgroup$
– stochasticboy321
Dec 8 '18 at 5:44
$begingroup$
For convenience I'll set $g = 0$ (this doesn't matter at all for what I'm going to say next). The prototypical example of a function with small $L_1$ but large $L_infty$ norm is a very tall function on a very narrow area. To exclude these, you'd like to control how much $f$ can grow in a short interval. One common condition for this is Hoelder continuity, which leads to a bound of $O(C^{1/alpha} epsilon^{ alpha/(alpha + 1)} )$. Note that these conditions imply uniform continuity, so may be too strong for your taste.
$endgroup$
– stochasticboy321
Dec 8 '18 at 5:44
add a comment |
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$begingroup$
We can add conditions like $f,g$ satisfy certain regularity conditions, I just want to figure out what could be the weakest condition we need to impose.
$endgroup$
– user348207
Dec 8 '18 at 0:31
$begingroup$
For convenience I'll set $g = 0$ (this doesn't matter at all for what I'm going to say next). The prototypical example of a function with small $L_1$ but large $L_infty$ norm is a very tall function on a very narrow area. To exclude these, you'd like to control how much $f$ can grow in a short interval. One common condition for this is Hoelder continuity, which leads to a bound of $O(C^{1/alpha} epsilon^{ alpha/(alpha + 1)} )$. Note that these conditions imply uniform continuity, so may be too strong for your taste.
$endgroup$
– stochasticboy321
Dec 8 '18 at 5:44