Can a function be neither convex nor concave everywhere?
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For a simpliest example, define continuous $f:mathbb Rtomathbb R$ to be locally convex in neighborhood $Usubsetmathbb R$ if ${y>f(x)|xin U}$ is a convex set.
$f:mathbb Rtomathbb R$ to be locally concave in $Usubsetmathbb R$ if ${y<f(x)|xin U}$ is a convex set.
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all neighborhood set $Usubsetmathbb R$ and $|U|>3$. Is it possible? (I guess yes)
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all measure non zero neighborhood set $Usubsetmathbb R$. Is it possible? (I guess no)
real-analysis functions continuity convex-analysis
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
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add a comment |
$begingroup$
For a simpliest example, define continuous $f:mathbb Rtomathbb R$ to be locally convex in neighborhood $Usubsetmathbb R$ if ${y>f(x)|xin U}$ is a convex set.
$f:mathbb Rtomathbb R$ to be locally concave in $Usubsetmathbb R$ if ${y<f(x)|xin U}$ is a convex set.
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all neighborhood set $Usubsetmathbb R$ and $|U|>3$. Is it possible? (I guess yes)
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all measure non zero neighborhood set $Usubsetmathbb R$. Is it possible? (I guess no)
real-analysis functions continuity convex-analysis
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
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1
$begingroup$
Yes in both cases. Take a look at this for an idea.
$endgroup$
– metamorphy
Dec 20 '18 at 10:50
add a comment |
$begingroup$
For a simpliest example, define continuous $f:mathbb Rtomathbb R$ to be locally convex in neighborhood $Usubsetmathbb R$ if ${y>f(x)|xin U}$ is a convex set.
$f:mathbb Rtomathbb R$ to be locally concave in $Usubsetmathbb R$ if ${y<f(x)|xin U}$ is a convex set.
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all neighborhood set $Usubsetmathbb R$ and $|U|>3$. Is it possible? (I guess yes)
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all measure non zero neighborhood set $Usubsetmathbb R$. Is it possible? (I guess no)
real-analysis functions continuity convex-analysis
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
$endgroup$
For a simpliest example, define continuous $f:mathbb Rtomathbb R$ to be locally convex in neighborhood $Usubsetmathbb R$ if ${y>f(x)|xin U}$ is a convex set.
$f:mathbb Rtomathbb R$ to be locally concave in $Usubsetmathbb R$ if ${y<f(x)|xin U}$ is a convex set.
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all neighborhood set $Usubsetmathbb R$ and $|U|>3$. Is it possible? (I guess yes)
Say $f$ is neither convex nor concave everywhere if $f$ is neither locally concave nor convex in all measure non zero neighborhood set $Usubsetmathbb R$. Is it possible? (I guess no)
real-analysis functions continuity convex-analysis
real-analysis functions continuity convex-analysis
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
edited Dec 26 '18 at 20:29
High GPA
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
asked Dec 20 '18 at 10:36
High GPAHigh GPA
899420
899420
1
$begingroup$
Yes in both cases. Take a look at this for an idea.
$endgroup$
– metamorphy
Dec 20 '18 at 10:50
add a comment |
1
$begingroup$
Yes in both cases. Take a look at this for an idea.
$endgroup$
– metamorphy
Dec 20 '18 at 10:50
1
1
$begingroup$
Yes in both cases. Take a look at this for an idea.
$endgroup$
– metamorphy
Dec 20 '18 at 10:50
$begingroup$
Yes in both cases. Take a look at this for an idea.
$endgroup$
– metamorphy
Dec 20 '18 at 10:50
add a comment |
1 Answer
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The Weierstrass function provides an example of a function that is continuous but not convex or concave in any open neighborhood.
The reason being that it has infinitely fine oscillation everywhere (since it is a fractal). It is defined by the infinite series
$$
f(x):=sum_{n=0}^infty a^n cos(b^n pi x)
$$
where $a,bin Bbb R$ satisfy certain assumptions. You can read more about it in the wikipedia link I provided.
To see why this function has such properties, we recall that if $f$ is convex/concave, then it is (locally) Lipschitz. However, Rademacher's theorem says that a Lipschitz continuous function is differentiable almost everywhere, which contradicts the fact that our $f$ is nowhere differentiable.
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
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$begingroup$
The Weierstrass function provides an example of a function that is continuous but not convex or concave in any open neighborhood.
The reason being that it has infinitely fine oscillation everywhere (since it is a fractal). It is defined by the infinite series
$$
f(x):=sum_{n=0}^infty a^n cos(b^n pi x)
$$
where $a,bin Bbb R$ satisfy certain assumptions. You can read more about it in the wikipedia link I provided.
To see why this function has such properties, we recall that if $f$ is convex/concave, then it is (locally) Lipschitz. However, Rademacher's theorem says that a Lipschitz continuous function is differentiable almost everywhere, which contradicts the fact that our $f$ is nowhere differentiable.
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
$endgroup$
add a comment |
$begingroup$
The Weierstrass function provides an example of a function that is continuous but not convex or concave in any open neighborhood.
The reason being that it has infinitely fine oscillation everywhere (since it is a fractal). It is defined by the infinite series
$$
f(x):=sum_{n=0}^infty a^n cos(b^n pi x)
$$
where $a,bin Bbb R$ satisfy certain assumptions. You can read more about it in the wikipedia link I provided.
To see why this function has such properties, we recall that if $f$ is convex/concave, then it is (locally) Lipschitz. However, Rademacher's theorem says that a Lipschitz continuous function is differentiable almost everywhere, which contradicts the fact that our $f$ is nowhere differentiable.
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
$endgroup$
add a comment |
$begingroup$
The Weierstrass function provides an example of a function that is continuous but not convex or concave in any open neighborhood.
The reason being that it has infinitely fine oscillation everywhere (since it is a fractal). It is defined by the infinite series
$$
f(x):=sum_{n=0}^infty a^n cos(b^n pi x)
$$
where $a,bin Bbb R$ satisfy certain assumptions. You can read more about it in the wikipedia link I provided.
To see why this function has such properties, we recall that if $f$ is convex/concave, then it is (locally) Lipschitz. However, Rademacher's theorem says that a Lipschitz continuous function is differentiable almost everywhere, which contradicts the fact that our $f$ is nowhere differentiable.
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
$endgroup$
The Weierstrass function provides an example of a function that is continuous but not convex or concave in any open neighborhood.
The reason being that it has infinitely fine oscillation everywhere (since it is a fractal). It is defined by the infinite series
$$
f(x):=sum_{n=0}^infty a^n cos(b^n pi x)
$$
where $a,bin Bbb R$ satisfy certain assumptions. You can read more about it in the wikipedia link I provided.
To see why this function has such properties, we recall that if $f$ is convex/concave, then it is (locally) Lipschitz. However, Rademacher's theorem says that a Lipschitz continuous function is differentiable almost everywhere, which contradicts the fact that our $f$ is nowhere differentiable.
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
edited Dec 20 '18 at 15:20
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
answered Dec 20 '18 at 10:57
BigbearZzzBigbearZzz
8,80421652
8,80421652
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Yes in both cases. Take a look at this for an idea.
$endgroup$
– metamorphy
Dec 20 '18 at 10:50