Question about strong convexity
$begingroup$
I don't really know how to begin. I tried substituting $y$ for $x + h$ and taking the Taylor approx. of $f(x + h)$ around $f(x)$. The RHS becomes
$h^T nabla f(x) + phi(x)$
Where $phi$ is our remainder function. Since $f$ is only $C^1$ smooth, the remainder function isn't well conditioned and this is sorta where I hit a wall. My friends in similar classes recommended finding something to integrate, but I don't really know what.
Furthermore, this sorta fails some sanity checks. Like if you take a convex function and negate it, the result shouldn't be convex; however, negating a function doesn't affect whether it satisfies $|nabla f(y) - nabla f(x)|^2 geq l |f(y) - f(x)|$.
Any tips would be appreciated :)
real-analysis convex-analysis nonlinear-optimization
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
$endgroup$
add a comment |
$begingroup$
I don't really know how to begin. I tried substituting $y$ for $x + h$ and taking the Taylor approx. of $f(x + h)$ around $f(x)$. The RHS becomes
$h^T nabla f(x) + phi(x)$
Where $phi$ is our remainder function. Since $f$ is only $C^1$ smooth, the remainder function isn't well conditioned and this is sorta where I hit a wall. My friends in similar classes recommended finding something to integrate, but I don't really know what.
Furthermore, this sorta fails some sanity checks. Like if you take a convex function and negate it, the result shouldn't be convex; however, negating a function doesn't affect whether it satisfies $|nabla f(y) - nabla f(x)|^2 geq l |f(y) - f(x)|$.
Any tips would be appreciated :)
real-analysis convex-analysis nonlinear-optimization
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
$endgroup$
add a comment |
$begingroup$
I don't really know how to begin. I tried substituting $y$ for $x + h$ and taking the Taylor approx. of $f(x + h)$ around $f(x)$. The RHS becomes
$h^T nabla f(x) + phi(x)$
Where $phi$ is our remainder function. Since $f$ is only $C^1$ smooth, the remainder function isn't well conditioned and this is sorta where I hit a wall. My friends in similar classes recommended finding something to integrate, but I don't really know what.
Furthermore, this sorta fails some sanity checks. Like if you take a convex function and negate it, the result shouldn't be convex; however, negating a function doesn't affect whether it satisfies $|nabla f(y) - nabla f(x)|^2 geq l |f(y) - f(x)|$.
Any tips would be appreciated :)
real-analysis convex-analysis nonlinear-optimization
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
$endgroup$
I don't really know how to begin. I tried substituting $y$ for $x + h$ and taking the Taylor approx. of $f(x + h)$ around $f(x)$. The RHS becomes
$h^T nabla f(x) + phi(x)$
Where $phi$ is our remainder function. Since $f$ is only $C^1$ smooth, the remainder function isn't well conditioned and this is sorta where I hit a wall. My friends in similar classes recommended finding something to integrate, but I don't really know what.
Furthermore, this sorta fails some sanity checks. Like if you take a convex function and negate it, the result shouldn't be convex; however, negating a function doesn't affect whether it satisfies $|nabla f(y) - nabla f(x)|^2 geq l |f(y) - f(x)|$.
Any tips would be appreciated :)
real-analysis convex-analysis nonlinear-optimization
real-analysis convex-analysis nonlinear-optimization
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
asked Dec 11 '18 at 2:09
Louis CastricatoLouis Castricato
285
285
add a comment |
add a comment |
1 Answer
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$begingroup$
The claim as stated is false: $f=0$ is an obvious counter-example.
And, as you already mentioned, if $f$ satisfies the inequality then also $-f$.
In addition, every smooth function satisfying the inequality has to be a constant:
If $f$ satisfies the inequality and $nabla f$ is Lipschitz, then necessarily $f$ is constant:
$$
|f(x+h)-f(x)|le l |nabla f(x+h)-nabla f(x)|^2 le l cdot L cdot |h|^2.
$$
Dividing by $|h|$ and letting $hto 0$ proves $nabla f(x)=0$.
Hence, the claimed statement is not even wrong.
Edit: Also every convex $C^1$ function satisfying the inequality is constant. I found a result of Rockafellar ('Second-order convex analysis', Journal of Nonlinear and Convex Analysis 1 (1999), 1-16, available at https://sites.math.washington.edu/~rtr/papers/rtr176-SecondOrderConvexAnalysis.pdf), which could be of use here:
Let $f$ be $C^1$ and convex. Then for almost all $x$ the map $nabla f$ satisfies a Lipschitz estimate. Then all such functions $f$ satisfying the inequality above have $nabla f=0$ almost everywhere. Hence, almost all points in $mathbb R^n$ are global minimizers of $f$, so $f$ has to be constant.
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
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add a comment |
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$begingroup$
The claim as stated is false: $f=0$ is an obvious counter-example.
And, as you already mentioned, if $f$ satisfies the inequality then also $-f$.
In addition, every smooth function satisfying the inequality has to be a constant:
If $f$ satisfies the inequality and $nabla f$ is Lipschitz, then necessarily $f$ is constant:
$$
|f(x+h)-f(x)|le l |nabla f(x+h)-nabla f(x)|^2 le l cdot L cdot |h|^2.
$$
Dividing by $|h|$ and letting $hto 0$ proves $nabla f(x)=0$.
Hence, the claimed statement is not even wrong.
Edit: Also every convex $C^1$ function satisfying the inequality is constant. I found a result of Rockafellar ('Second-order convex analysis', Journal of Nonlinear and Convex Analysis 1 (1999), 1-16, available at https://sites.math.washington.edu/~rtr/papers/rtr176-SecondOrderConvexAnalysis.pdf), which could be of use here:
Let $f$ be $C^1$ and convex. Then for almost all $x$ the map $nabla f$ satisfies a Lipschitz estimate. Then all such functions $f$ satisfying the inequality above have $nabla f=0$ almost everywhere. Hence, almost all points in $mathbb R^n$ are global minimizers of $f$, so $f$ has to be constant.
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
$endgroup$
add a comment |
$begingroup$
The claim as stated is false: $f=0$ is an obvious counter-example.
And, as you already mentioned, if $f$ satisfies the inequality then also $-f$.
In addition, every smooth function satisfying the inequality has to be a constant:
If $f$ satisfies the inequality and $nabla f$ is Lipschitz, then necessarily $f$ is constant:
$$
|f(x+h)-f(x)|le l |nabla f(x+h)-nabla f(x)|^2 le l cdot L cdot |h|^2.
$$
Dividing by $|h|$ and letting $hto 0$ proves $nabla f(x)=0$.
Hence, the claimed statement is not even wrong.
Edit: Also every convex $C^1$ function satisfying the inequality is constant. I found a result of Rockafellar ('Second-order convex analysis', Journal of Nonlinear and Convex Analysis 1 (1999), 1-16, available at https://sites.math.washington.edu/~rtr/papers/rtr176-SecondOrderConvexAnalysis.pdf), which could be of use here:
Let $f$ be $C^1$ and convex. Then for almost all $x$ the map $nabla f$ satisfies a Lipschitz estimate. Then all such functions $f$ satisfying the inequality above have $nabla f=0$ almost everywhere. Hence, almost all points in $mathbb R^n$ are global minimizers of $f$, so $f$ has to be constant.
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
$endgroup$
add a comment |
$begingroup$
The claim as stated is false: $f=0$ is an obvious counter-example.
And, as you already mentioned, if $f$ satisfies the inequality then also $-f$.
In addition, every smooth function satisfying the inequality has to be a constant:
If $f$ satisfies the inequality and $nabla f$ is Lipschitz, then necessarily $f$ is constant:
$$
|f(x+h)-f(x)|le l |nabla f(x+h)-nabla f(x)|^2 le l cdot L cdot |h|^2.
$$
Dividing by $|h|$ and letting $hto 0$ proves $nabla f(x)=0$.
Hence, the claimed statement is not even wrong.
Edit: Also every convex $C^1$ function satisfying the inequality is constant. I found a result of Rockafellar ('Second-order convex analysis', Journal of Nonlinear and Convex Analysis 1 (1999), 1-16, available at https://sites.math.washington.edu/~rtr/papers/rtr176-SecondOrderConvexAnalysis.pdf), which could be of use here:
Let $f$ be $C^1$ and convex. Then for almost all $x$ the map $nabla f$ satisfies a Lipschitz estimate. Then all such functions $f$ satisfying the inequality above have $nabla f=0$ almost everywhere. Hence, almost all points in $mathbb R^n$ are global minimizers of $f$, so $f$ has to be constant.
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
$endgroup$
The claim as stated is false: $f=0$ is an obvious counter-example.
And, as you already mentioned, if $f$ satisfies the inequality then also $-f$.
In addition, every smooth function satisfying the inequality has to be a constant:
If $f$ satisfies the inequality and $nabla f$ is Lipschitz, then necessarily $f$ is constant:
$$
|f(x+h)-f(x)|le l |nabla f(x+h)-nabla f(x)|^2 le l cdot L cdot |h|^2.
$$
Dividing by $|h|$ and letting $hto 0$ proves $nabla f(x)=0$.
Hence, the claimed statement is not even wrong.
Edit: Also every convex $C^1$ function satisfying the inequality is constant. I found a result of Rockafellar ('Second-order convex analysis', Journal of Nonlinear and Convex Analysis 1 (1999), 1-16, available at https://sites.math.washington.edu/~rtr/papers/rtr176-SecondOrderConvexAnalysis.pdf), which could be of use here:
Let $f$ be $C^1$ and convex. Then for almost all $x$ the map $nabla f$ satisfies a Lipschitz estimate. Then all such functions $f$ satisfying the inequality above have $nabla f=0$ almost everywhere. Hence, almost all points in $mathbb R^n$ are global minimizers of $f$, so $f$ has to be constant.
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
edited Dec 11 '18 at 8:46
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
answered Dec 11 '18 at 8:14
dawdaw
24.5k1645
24.5k1645
add a comment |
add a comment |
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