Mixing
A strictly increasing convex function with zero right-derivative at a single point.
$begingroup$
Let $f: mathbb{R} to [0, infty)$ be strictly increasing and convex. Since $f$ is convex its right derivative at, say, zero exists and is finite.
Is it possible to construct an $f$ such that its right derivative is zero at some point?
Intuitively, this should not be possible since it would constitute a point where the tangent at $f$ (from the right) is flat and $f$ can not become smaller when we move to the left, i.e. we would either loose convexity or strictly increasing"ness". However, I cannot find a proper proof or counterexample.
derivatives convex-analysis
asked Jan 29 at 14:59
NiUNiU
753413
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R} to [0, infty)$ be strictly increasing and convex. Since $f$ is convex its right derivative at, say, zero exists and is finite.
Is it possible to construct an $f$ such that its right derivative is zero at some point?
Intuitively, this should not be possible since it would constitute a point where the tangent at $f$ (from the right) is flat and $f$ can not become smaller when we move to the left, i.e. we would either loose convexity or strictly increasing"ness". However, I cannot find a proper proof or counterexample.
derivatives convex-analysis
asked Jan 29 at 14:59
NiUNiU
753413
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R} to [0, infty)$ be strictly increasing and convex. Since $f$ is convex its right derivative at, say, zero exists and is finite.
Is it possible to construct an $f$ such that its right derivative is zero at some point?
Intuitively, this should not be possible since it would constitute a point where the tangent at $f$ (from the right) is flat and $f$ can not become smaller when we move to the left, i.e. we would either loose convexity or strictly increasing"ness". However, I cannot find a proper proof or counterexample.
derivatives convex-analysis
asked Jan 29 at 14:59
NiUNiU
753413
$endgroup$
Let $f: mathbb{R} to [0, infty)$ be strictly increasing and convex. Since $f$ is convex its right derivative at, say, zero exists and is finite.
Is it possible to construct an $f$ such that its right derivative is zero at some point?
Intuitively, this should not be possible since it would constitute a point where the tangent at $f$ (from the right) is flat and $f$ can not become smaller when we move to the left, i.e. we would either loose convexity or strictly increasing"ness". However, I cannot find a proper proof or counterexample.
derivatives convex-analysis
derivatives convex-analysis
asked Jan 29 at 14:59
NiUNiU
753413
asked Jan 29 at 14:59
NiUNiU
753413
asked Jan 29 at 14:59
NiUNiU
753413
asked Jan 29 at 14:59
NiUNiU
753413
asked Jan 29 at 14:59
NiUNiU
753413
753413
add a comment |
add a comment |
1 Answer
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$begingroup$
Say $D_Rf$ is the right derivative. $f$ convex implies that $D_Rf$ is non-decreasing, while $f$ strictly increasing implies that $D_Rfge0$. So if $D_Rf(0)=0$ then $D_Rf(x)=0$ for all $x<0$, hence $f$ is not strictly increasing.
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Say $D_Rf$ is the right derivative. $f$ convex implies that $D_Rf$ is non-decreasing, while $f$ strictly increasing implies that $D_Rfge0$. So if $D_Rf(0)=0$ then $D_Rf(x)=0$ for all $x<0$, hence $f$ is not strictly increasing.
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
$endgroup$
add a comment |
$begingroup$
Say $D_Rf$ is the right derivative. $f$ convex implies that $D_Rf$ is non-decreasing, while $f$ strictly increasing implies that $D_Rfge0$. So if $D_Rf(0)=0$ then $D_Rf(x)=0$ for all $x<0$, hence $f$ is not strictly increasing.
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
$endgroup$
add a comment |
$begingroup$
Say $D_Rf$ is the right derivative. $f$ convex implies that $D_Rf$ is non-decreasing, while $f$ strictly increasing implies that $D_Rfge0$. So if $D_Rf(0)=0$ then $D_Rf(x)=0$ for all $x<0$, hence $f$ is not strictly increasing.
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
$endgroup$
Say $D_Rf$ is the right derivative. $f$ convex implies that $D_Rf$ is non-decreasing, while $f$ strictly increasing implies that $D_Rfge0$. So if $D_Rf(0)=0$ then $D_Rf(x)=0$ for all $x<0$, hence $f$ is not strictly increasing.
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
answered Jan 29 at 15:25
David C. UllrichDavid C. Ullrich
61.7k43995
61.7k43995
add a comment |
add a comment |
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