Mixing
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The behavior of $sqrtff''/f'$ for a strictly convex function
Clash Royale CLAN TAG#URR8PPP
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Let us consider $f:mathbbRlongrightarrow [0,+infty)$ is of class $mathrmC^2$ and is strictly convex with
$$min_xinmathbbR f(x) = f(0) = 0$$
If $f''(x) geq alpha > 0$ for some $alpha >0$ then $f$ is called "strongly convex". Then
$$ f(x) geq f(y) + f'(y)(x-y) + fracalpha2(x-y)^2$$
and as a consequence (see https://rkganti.wordpress.com/2015/08/19/lipschitz-functions-and-convexity/) we have
$$ f(x) leq frac12alpha f'(x)^2$$
for all $xin mathbbR$, which implies that
beginequationlabelxxx
limsup_xlongrightarrow 0 left|fracsqrtf(x)f'(x)right| < infty.tag$star$
endequation
In fact, all we need to show eqrefxxx is $f''(0) > 0$. I am trying to show that eqrefxxx is also true if $f''(0) = 0$. To be precise, I am trying to show that
Given the condition above: $f:mathbbRlongrightarrow
[0,+infty)$ is of class $mathrmC^2$ and is
strictly convex with $$min_xinmathbbR f(x) = f(0) = 0$$ then
beginequationlabelxxxx limsup_xlongrightarrow 0
left|fracf''(x)f'(x)sqrtf(x)right| < infty.tag$starstar$
endequation
real-analysis convex-analysis
edited Aug 2 at 1:30
user357151
13.6k31140
asked Aug 1 at 23:36
Sean
434312
add a comment |Â
up vote
2
down vote
favorite
Let us consider $f:mathbbRlongrightarrow [0,+infty)$ is of class $mathrmC^2$ and is strictly convex with
$$min_xinmathbbR f(x) = f(0) = 0$$
If $f''(x) geq alpha > 0$ for some $alpha >0$ then $f$ is called "strongly convex". Then
$$ f(x) geq f(y) + f'(y)(x-y) + fracalpha2(x-y)^2$$
and as a consequence (see https://rkganti.wordpress.com/2015/08/19/lipschitz-functions-and-convexity/) we have
$$ f(x) leq frac12alpha f'(x)^2$$
for all $xin mathbbR$, which implies that
beginequationlabelxxx
limsup_xlongrightarrow 0 left|fracsqrtf(x)f'(x)right| < infty.tag$star$
endequation
In fact, all we need to show eqrefxxx is $f''(0) > 0$. I am trying to show that eqrefxxx is also true if $f''(0) = 0$. To be precise, I am trying to show that
Given the condition above: $f:mathbbRlongrightarrow
[0,+infty)$ is of class $mathrmC^2$ and is
strictly convex with $$min_xinmathbbR f(x) = f(0) = 0$$ then
beginequationlabelxxxx limsup_xlongrightarrow 0
left|fracf''(x)f'(x)sqrtf(x)right| < infty.tag$starstar$
endequation
real-analysis convex-analysis
edited Aug 2 at 1:30
user357151
13.6k31140
asked Aug 1 at 23:36
Sean
434312
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let us consider $f:mathbbRlongrightarrow [0,+infty)$ is of class $mathrmC^2$ and is strictly convex with
$$min_xinmathbbR f(x) = f(0) = 0$$
If $f''(x) geq alpha > 0$ for some $alpha >0$ then $f$ is called "strongly convex". Then
$$ f(x) geq f(y) + f'(y)(x-y) + fracalpha2(x-y)^2$$
and as a consequence (see https://rkganti.wordpress.com/2015/08/19/lipschitz-functions-and-convexity/) we have
$$ f(x) leq frac12alpha f'(x)^2$$
for all $xin mathbbR$, which implies that
beginequationlabelxxx
limsup_xlongrightarrow 0 left|fracsqrtf(x)f'(x)right| < infty.tag$star$
endequation
In fact, all we need to show eqrefxxx is $f''(0) > 0$. I am trying to show that eqrefxxx is also true if $f''(0) = 0$. To be precise, I am trying to show that
Given the condition above: $f:mathbbRlongrightarrow
[0,+infty)$ is of class $mathrmC^2$ and is
strictly convex with $$min_xinmathbbR f(x) = f(0) = 0$$ then
beginequationlabelxxxx limsup_xlongrightarrow 0
left|fracf''(x)f'(x)sqrtf(x)right| < infty.tag$starstar$
endequation
real-analysis convex-analysis
edited Aug 2 at 1:30
user357151
13.6k31140
asked Aug 1 at 23:36
Sean
434312
Let us consider $f:mathbbRlongrightarrow [0,+infty)$ is of class $mathrmC^2$ and is strictly convex with
$$min_xinmathbbR f(x) = f(0) = 0$$
If $f''(x) geq alpha > 0$ for some $alpha >0$ then $f$ is called "strongly convex". Then
$$ f(x) geq f(y) + f'(y)(x-y) + fracalpha2(x-y)^2$$
and as a consequence (see https://rkganti.wordpress.com/2015/08/19/lipschitz-functions-and-convexity/) we have
$$ f(x) leq frac12alpha f'(x)^2$$
for all $xin mathbbR$, which implies that
beginequationlabelxxx
limsup_xlongrightarrow 0 left|fracsqrtf(x)f'(x)right| < infty.tag$star$
endequation
In fact, all we need to show eqrefxxx is $f''(0) > 0$. I am trying to show that eqrefxxx is also true if $f''(0) = 0$. To be precise, I am trying to show that
Given the condition above: $f:mathbbRlongrightarrow
[0,+infty)$ is of class $mathrmC^2$ and is
strictly convex with $$min_xinmathbbR f(x) = f(0) = 0$$ then
beginequationlabelxxxx limsup_xlongrightarrow 0
left|fracf''(x)f'(x)sqrtf(x)right| < infty.tag$starstar$
endequation
real-analysis convex-analysis
edited Aug 2 at 1:30
user357151
13.6k31140
asked Aug 1 at 23:36
Sean
434312
edited Aug 2 at 1:30
user357151
13.6k31140
edited Aug 2 at 1:30
user357151
13.6k31140
edited Aug 2 at 1:30
user357151
13.6k31140
13.6k31140
asked Aug 1 at 23:36
Sean
434312
asked Aug 1 at 23:36
Sean
434312
asked Aug 1 at 23:36
Sean
434312
434312
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
This is not true. Let $g$ be a continuous function such that $x^6 le g(x)le x^6+x^2$ for all $xinmathbbR$, and more precisely $g(x)=x^6$ except in the intervals $(2^-n, 2^-n + 2^-8n)$ in which $g$ reaches $x^6+x^2$ somewhere. Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$. Integration yields
$|f'(x)|le |x|^7/7 + sum_2^-n le 2^-2n2^-8n$. The geometric sum is dominated by its largest term, which is at most $|x|^10$, negligible compared to $|x|^7/7$. Thus, $|f'(x)| le |x|^7$ for all sufficiently small $|x|$.
$ f(x) ge x^8/56$ by comparison of the second derivatives of both sides.
Thus, at the points where $g(x) = x^6+x^2$ we have
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge fracx^2x^7fracx^4sqrt56 = frac1xsqrt56
$$
which is unbounded as $xto 0$.
Let's generalize the above to get a $C^2,alpha$ counterexample for any $alphain (0, 1)$. Let $a, b, c$ be positive numbers to be chosen later. Define
$$
g(x) = |x|^a + sum_n=1^infty 2^-bnmax(0, 1- 2^cn|x-2^-n|)
$$
where the sum is a formal way of saying: add a triangle of height $2^-bn$ with the base $[2^-n-2^-cn, 2^-n+2^-cn]$.
We will choose $b$ large, $c = b+2$, and $a=2b+1$. Note that $g$ is Hölder continuous with the exponent $b/c = b/(b+2)$ which is close to $1$ when $b$ is large.
Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$.
Integration yields $|f'(x)| le |x|^a+1/(a+1) + 2sum_2^-n le 2^-bn2^-cn$. The geometric sum is dominated by its largest term, which is of size $x^b+c = x^a+1$. Thus, $|f'(x)| = O(|x|^a+1)$ as $xto 0$.
$f(x) ge x^a+2/((a+1)(a+2))$ by comparison of the second derivatives of both sides.
At the points $x=2^-n$ we have $f''(x) ge x^b$, which implies, for some constant $nu>0$,
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge nu fracx^bx^a+1 x^(a+2)/2 = nu x^b - a/2 = nu x^-1/2
$$
which is unbounded as $xto 0$.
answered Aug 2 at 1:29
user357151
13.6k31140
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
This is not true. Let $g$ be a continuous function such that $x^6 le g(x)le x^6+x^2$ for all $xinmathbbR$, and more precisely $g(x)=x^6$ except in the intervals $(2^-n, 2^-n + 2^-8n)$ in which $g$ reaches $x^6+x^2$ somewhere. Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$. Integration yields
$|f'(x)|le |x|^7/7 + sum_2^-n le 2^-2n2^-8n$. The geometric sum is dominated by its largest term, which is at most $|x|^10$, negligible compared to $|x|^7/7$. Thus, $|f'(x)| le |x|^7$ for all sufficiently small $|x|$.
$ f(x) ge x^8/56$ by comparison of the second derivatives of both sides.
Thus, at the points where $g(x) = x^6+x^2$ we have
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge fracx^2x^7fracx^4sqrt56 = frac1xsqrt56
$$
which is unbounded as $xto 0$.
Let's generalize the above to get a $C^2,alpha$ counterexample for any $alphain (0, 1)$. Let $a, b, c$ be positive numbers to be chosen later. Define
$$
g(x) = |x|^a + sum_n=1^infty 2^-bnmax(0, 1- 2^cn|x-2^-n|)
$$
where the sum is a formal way of saying: add a triangle of height $2^-bn$ with the base $[2^-n-2^-cn, 2^-n+2^-cn]$.
We will choose $b$ large, $c = b+2$, and $a=2b+1$. Note that $g$ is Hölder continuous with the exponent $b/c = b/(b+2)$ which is close to $1$ when $b$ is large.
Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$.
Integration yields $|f'(x)| le |x|^a+1/(a+1) + 2sum_2^-n le 2^-bn2^-cn$. The geometric sum is dominated by its largest term, which is of size $x^b+c = x^a+1$. Thus, $|f'(x)| = O(|x|^a+1)$ as $xto 0$.
$f(x) ge x^a+2/((a+1)(a+2))$ by comparison of the second derivatives of both sides.
At the points $x=2^-n$ we have $f''(x) ge x^b$, which implies, for some constant $nu>0$,
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge nu fracx^bx^a+1 x^(a+2)/2 = nu x^b - a/2 = nu x^-1/2
$$
which is unbounded as $xto 0$.
answered Aug 2 at 1:29
user357151
13.6k31140
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
add a comment |Â
up vote
2
down vote
This is not true. Let $g$ be a continuous function such that $x^6 le g(x)le x^6+x^2$ for all $xinmathbbR$, and more precisely $g(x)=x^6$ except in the intervals $(2^-n, 2^-n + 2^-8n)$ in which $g$ reaches $x^6+x^2$ somewhere. Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$. Integration yields
$|f'(x)|le |x|^7/7 + sum_2^-n le 2^-2n2^-8n$. The geometric sum is dominated by its largest term, which is at most $|x|^10$, negligible compared to $|x|^7/7$. Thus, $|f'(x)| le |x|^7$ for all sufficiently small $|x|$.
$ f(x) ge x^8/56$ by comparison of the second derivatives of both sides.
Thus, at the points where $g(x) = x^6+x^2$ we have
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge fracx^2x^7fracx^4sqrt56 = frac1xsqrt56
$$
which is unbounded as $xto 0$.
Let's generalize the above to get a $C^2,alpha$ counterexample for any $alphain (0, 1)$. Let $a, b, c$ be positive numbers to be chosen later. Define
$$
g(x) = |x|^a + sum_n=1^infty 2^-bnmax(0, 1- 2^cn|x-2^-n|)
$$
where the sum is a formal way of saying: add a triangle of height $2^-bn$ with the base $[2^-n-2^-cn, 2^-n+2^-cn]$.
We will choose $b$ large, $c = b+2$, and $a=2b+1$. Note that $g$ is Hölder continuous with the exponent $b/c = b/(b+2)$ which is close to $1$ when $b$ is large.
Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$.
Integration yields $|f'(x)| le |x|^a+1/(a+1) + 2sum_2^-n le 2^-bn2^-cn$. The geometric sum is dominated by its largest term, which is of size $x^b+c = x^a+1$. Thus, $|f'(x)| = O(|x|^a+1)$ as $xto 0$.
$f(x) ge x^a+2/((a+1)(a+2))$ by comparison of the second derivatives of both sides.
At the points $x=2^-n$ we have $f''(x) ge x^b$, which implies, for some constant $nu>0$,
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge nu fracx^bx^a+1 x^(a+2)/2 = nu x^b - a/2 = nu x^-1/2
$$
which is unbounded as $xto 0$.
answered Aug 2 at 1:29
user357151
13.6k31140
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is not true. Let $g$ be a continuous function such that $x^6 le g(x)le x^6+x^2$ for all $xinmathbbR$, and more precisely $g(x)=x^6$ except in the intervals $(2^-n, 2^-n + 2^-8n)$ in which $g$ reaches $x^6+x^2$ somewhere. Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$. Integration yields
$|f'(x)|le |x|^7/7 + sum_2^-n le 2^-2n2^-8n$. The geometric sum is dominated by its largest term, which is at most $|x|^10$, negligible compared to $|x|^7/7$. Thus, $|f'(x)| le |x|^7$ for all sufficiently small $|x|$.
$ f(x) ge x^8/56$ by comparison of the second derivatives of both sides.
Thus, at the points where $g(x) = x^6+x^2$ we have
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge fracx^2x^7fracx^4sqrt56 = frac1xsqrt56
$$
which is unbounded as $xto 0$.
Let's generalize the above to get a $C^2,alpha$ counterexample for any $alphain (0, 1)$. Let $a, b, c$ be positive numbers to be chosen later. Define
$$
g(x) = |x|^a + sum_n=1^infty 2^-bnmax(0, 1- 2^cn|x-2^-n|)
$$
where the sum is a formal way of saying: add a triangle of height $2^-bn$ with the base $[2^-n-2^-cn, 2^-n+2^-cn]$.
We will choose $b$ large, $c = b+2$, and $a=2b+1$. Note that $g$ is Hölder continuous with the exponent $b/c = b/(b+2)$ which is close to $1$ when $b$ is large.
Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$.
Integration yields $|f'(x)| le |x|^a+1/(a+1) + 2sum_2^-n le 2^-bn2^-cn$. The geometric sum is dominated by its largest term, which is of size $x^b+c = x^a+1$. Thus, $|f'(x)| = O(|x|^a+1)$ as $xto 0$.
$f(x) ge x^a+2/((a+1)(a+2))$ by comparison of the second derivatives of both sides.
At the points $x=2^-n$ we have $f''(x) ge x^b$, which implies, for some constant $nu>0$,
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge nu fracx^bx^a+1 x^(a+2)/2 = nu x^b - a/2 = nu x^-1/2
$$
which is unbounded as $xto 0$.
answered Aug 2 at 1:29
user357151
13.6k31140
This is not true. Let $g$ be a continuous function such that $x^6 le g(x)le x^6+x^2$ for all $xinmathbbR$, and more precisely $g(x)=x^6$ except in the intervals $(2^-n, 2^-n + 2^-8n)$ in which $g$ reaches $x^6+x^2$ somewhere. Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$. Integration yields
$|f'(x)|le |x|^7/7 + sum_2^-n le 2^-2n2^-8n$. The geometric sum is dominated by its largest term, which is at most $|x|^10$, negligible compared to $|x|^7/7$. Thus, $|f'(x)| le |x|^7$ for all sufficiently small $|x|$.
$ f(x) ge x^8/56$ by comparison of the second derivatives of both sides.
Thus, at the points where $g(x) = x^6+x^2$ we have
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge fracx^2x^7fracx^4sqrt56 = frac1xsqrt56
$$
which is unbounded as $xto 0$.
Let's generalize the above to get a $C^2,alpha$ counterexample for any $alphain (0, 1)$. Let $a, b, c$ be positive numbers to be chosen later. Define
$$
g(x) = |x|^a + sum_n=1^infty 2^-bnmax(0, 1- 2^cn|x-2^-n|)
$$
where the sum is a formal way of saying: add a triangle of height $2^-bn$ with the base $[2^-n-2^-cn, 2^-n+2^-cn]$.
We will choose $b$ large, $c = b+2$, and $a=2b+1$. Note that $g$ is Hölder continuous with the exponent $b/c = b/(b+2)$ which is close to $1$ when $b$ is large.
Let $f$ be such that $f''=g$, with $f(0)=f'(0)=0$.
Integration yields $|f'(x)| le |x|^a+1/(a+1) + 2sum_2^-n le 2^-bn2^-cn$. The geometric sum is dominated by its largest term, which is of size $x^b+c = x^a+1$. Thus, $|f'(x)| = O(|x|^a+1)$ as $xto 0$.
$f(x) ge x^a+2/((a+1)(a+2))$ by comparison of the second derivatives of both sides.
At the points $x=2^-n$ we have $f''(x) ge x^b$, which implies, for some constant $nu>0$,
$$
left|fracf''(x)f'(x)sqrtf(x)right| ge nu fracx^bx^a+1 x^(a+2)/2 = nu x^b - a/2 = nu x^-1/2
$$
which is unbounded as $xto 0$.
answered Aug 2 at 1:29
user357151
13.6k31140
edited Aug 3 at 20:29
answered Aug 2 at 1:29
user357151
13.6k31140
answered Aug 2 at 1:29
user357151
13.6k31140
answered Aug 2 at 1:29
user357151
13.6k31140
13.6k31140
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
add a comment |Â
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
Thanks! Is it correct if I assume $fin C^2,alpha$ for $alphain (0,1)$? Since it is correct for $fin C^3$.
– Sean
Aug 3 at 19:48
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
No, because the function $g$ here can be Hölder continuous. Its main feature is spikes of size $2^-4n$ on an interval of size $2^-8n$, which allows for $alpha=1/2$. And the example isn't really optimal.
– user357151
Aug 3 at 19:52
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
So this is correct if $fin C^2,alpha$ for $alpha>1/2$?
– Sean
Aug 3 at 20:16
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
No. As I said, the original example wasn't optimized for Hölder continuity. I added a general version, which works for any $C^2,alpha$ with $alphain (0,1 )$.
– user357151
Aug 3 at 20:30
add a comment |Â
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