Mixing
When is subadditivity preserved under differentiation?
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Let there be a subadditive function $f : mathbb R^n to mathbb R$, that is
,begin{gather}f(a+b) leq f(a) + f(b) quad forall a, b in mathbb R^n,tag{1}label{eq1}end{gather}
and lets write $f' : mathbb R^n to mathbb R^n$ for the gradient of $f$.
Question 1: Under which circumstances can we expect the subadditivity property also hold for $f'$, where "$leq$" is understood component-wise, that is,
begin{gather}f_i'(a+b) leq f_i'(a) + f_i'(b) enspace quad forall i in [n],enspace a, b in mathbb R^ntext? tag{2}label{eq2}end{gather}
A popular example for subadditive functions is the square root $f(t) = sqrt{t}$, which is only defined on a subset of $mathbb R$, but still preserves the subadditivity property under differentiation.
In my specific case, the function $f$ is additionally convex, hence, we know that $f'_i$ possesses no extrema. It is also defined on the whole $mathbb R^n$. Does this have implications for Eq. eqref{eq2}?
Question 2: Out of interest, can somebody provide a counterexample of a function $f$, which is subadditive on the whole $mathbb R$ (ie, Eq. eqref{eq1} is fulfilled), but contradicts Eq. eqref{eq2}?
analysis convex-analysis
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
$endgroup$
add a comment |
$begingroup$
Let there be a subadditive function $f : mathbb R^n to mathbb R$, that is
,begin{gather}f(a+b) leq f(a) + f(b) quad forall a, b in mathbb R^n,tag{1}label{eq1}end{gather}
and lets write $f' : mathbb R^n to mathbb R^n$ for the gradient of $f$.
Question 1: Under which circumstances can we expect the subadditivity property also hold for $f'$, where "$leq$" is understood component-wise, that is,
begin{gather}f_i'(a+b) leq f_i'(a) + f_i'(b) enspace quad forall i in [n],enspace a, b in mathbb R^ntext? tag{2}label{eq2}end{gather}
A popular example for subadditive functions is the square root $f(t) = sqrt{t}$, which is only defined on a subset of $mathbb R$, but still preserves the subadditivity property under differentiation.
In my specific case, the function $f$ is additionally convex, hence, we know that $f'_i$ possesses no extrema. It is also defined on the whole $mathbb R^n$. Does this have implications for Eq. eqref{eq2}?
Question 2: Out of interest, can somebody provide a counterexample of a function $f$, which is subadditive on the whole $mathbb R$ (ie, Eq. eqref{eq1} is fulfilled), but contradicts Eq. eqref{eq2}?
analysis convex-analysis
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
$endgroup$
$begingroup$
Q2: $f(x)=-x^4$ is concave, and therefore subadditive. But its derivative is $f'(x)=-4x^3,$ which is not going to be subadditive (check $x=-2, y=-1.$)
$endgroup$
– Adrian Keister
Jan 15 at 15:01
$begingroup$
@AdrianKeister $f(x) = -x^4$ is not subadditive on all of $mathbb R$ . For example, $f(0) > f(-1) + f(1)$.
$endgroup$
– Robert Israel
Jan 15 at 15:06
$begingroup$
I'm confused: Is $-x^4$ not concave?
$endgroup$
– Adrian Keister
Jan 15 at 16:13
$begingroup$
Oh, I see the problem. If $f$ is concave, and $f(0)ge 0,$ then $f$ is only subadditive on $[0,infty).$ That makes my example less than useful, unfortunately.
$endgroup$
– Adrian Keister
Jan 15 at 16:27
add a comment |
$begingroup$
Let there be a subadditive function $f : mathbb R^n to mathbb R$, that is
,begin{gather}f(a+b) leq f(a) + f(b) quad forall a, b in mathbb R^n,tag{1}label{eq1}end{gather}
and lets write $f' : mathbb R^n to mathbb R^n$ for the gradient of $f$.
Question 1: Under which circumstances can we expect the subadditivity property also hold for $f'$, where "$leq$" is understood component-wise, that is,
begin{gather}f_i'(a+b) leq f_i'(a) + f_i'(b) enspace quad forall i in [n],enspace a, b in mathbb R^ntext? tag{2}label{eq2}end{gather}
A popular example for subadditive functions is the square root $f(t) = sqrt{t}$, which is only defined on a subset of $mathbb R$, but still preserves the subadditivity property under differentiation.
In my specific case, the function $f$ is additionally convex, hence, we know that $f'_i$ possesses no extrema. It is also defined on the whole $mathbb R^n$. Does this have implications for Eq. eqref{eq2}?
Question 2: Out of interest, can somebody provide a counterexample of a function $f$, which is subadditive on the whole $mathbb R$ (ie, Eq. eqref{eq1} is fulfilled), but contradicts Eq. eqref{eq2}?
analysis convex-analysis
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
$endgroup$
Let there be a subadditive function $f : mathbb R^n to mathbb R$, that is
,begin{gather}f(a+b) leq f(a) + f(b) quad forall a, b in mathbb R^n,tag{1}label{eq1}end{gather}
and lets write $f' : mathbb R^n to mathbb R^n$ for the gradient of $f$.
Question 1: Under which circumstances can we expect the subadditivity property also hold for $f'$, where "$leq$" is understood component-wise, that is,
begin{gather}f_i'(a+b) leq f_i'(a) + f_i'(b) enspace quad forall i in [n],enspace a, b in mathbb R^ntext? tag{2}label{eq2}end{gather}
A popular example for subadditive functions is the square root $f(t) = sqrt{t}$, which is only defined on a subset of $mathbb R$, but still preserves the subadditivity property under differentiation.
In my specific case, the function $f$ is additionally convex, hence, we know that $f'_i$ possesses no extrema. It is also defined on the whole $mathbb R^n$. Does this have implications for Eq. eqref{eq2}?
Question 2: Out of interest, can somebody provide a counterexample of a function $f$, which is subadditive on the whole $mathbb R$ (ie, Eq. eqref{eq1} is fulfilled), but contradicts Eq. eqref{eq2}?
analysis convex-analysis
analysis convex-analysis
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
asked Jan 15 at 14:44
theV0IDtheV0ID
190112
190112
$begingroup$
Q2: $f(x)=-x^4$ is concave, and therefore subadditive. But its derivative is $f'(x)=-4x^3,$ which is not going to be subadditive (check $x=-2, y=-1.$)
$endgroup$
– Adrian Keister
Jan 15 at 15:01
$begingroup$
@AdrianKeister $f(x) = -x^4$ is not subadditive on all of $mathbb R$ . For example, $f(0) > f(-1) + f(1)$.
$endgroup$
– Robert Israel
Jan 15 at 15:06
$begingroup$
I'm confused: Is $-x^4$ not concave?
$endgroup$
– Adrian Keister
Jan 15 at 16:13
$begingroup$
Oh, I see the problem. If $f$ is concave, and $f(0)ge 0,$ then $f$ is only subadditive on $[0,infty).$ That makes my example less than useful, unfortunately.
$endgroup$
– Adrian Keister
Jan 15 at 16:27
add a comment |
$begingroup$
Q2: $f(x)=-x^4$ is concave, and therefore subadditive. But its derivative is $f'(x)=-4x^3,$ which is not going to be subadditive (check $x=-2, y=-1.$)
$endgroup$
– Adrian Keister
Jan 15 at 15:01
$begingroup$
@AdrianKeister $f(x) = -x^4$ is not subadditive on all of $mathbb R$ . For example, $f(0) > f(-1) + f(1)$.
$endgroup$
– Robert Israel
Jan 15 at 15:06
$begingroup$
I'm confused: Is $-x^4$ not concave?
$endgroup$
– Adrian Keister
Jan 15 at 16:13
$begingroup$
Oh, I see the problem. If $f$ is concave, and $f(0)ge 0,$ then $f$ is only subadditive on $[0,infty).$ That makes my example less than useful, unfortunately.
$endgroup$
– Adrian Keister
Jan 15 at 16:27
$begingroup$
Q2: $f(x)=-x^4$ is concave, and therefore subadditive. But its derivative is $f'(x)=-4x^3,$ which is not going to be subadditive (check $x=-2, y=-1.$)
$endgroup$
– Adrian Keister
Jan 15 at 15:01
$begingroup$
Q2: $f(x)=-x^4$ is concave, and therefore subadditive. But its derivative is $f'(x)=-4x^3,$ which is not going to be subadditive (check $x=-2, y=-1.$)
$endgroup$
– Adrian Keister
Jan 15 at 15:01
$begingroup$
@AdrianKeister $f(x) = -x^4$ is not subadditive on all of $mathbb R$ . For example, $f(0) > f(-1) + f(1)$.
$endgroup$
– Robert Israel
Jan 15 at 15:06
$begingroup$
@AdrianKeister $f(x) = -x^4$ is not subadditive on all of $mathbb R$ . For example, $f(0) > f(-1) + f(1)$.
$endgroup$
– Robert Israel
Jan 15 at 15:06
$begingroup$
I'm confused: Is $-x^4$ not concave?
$endgroup$
– Adrian Keister
Jan 15 at 16:13
$begingroup$
I'm confused: Is $-x^4$ not concave?
$endgroup$
– Adrian Keister
Jan 15 at 16:13
$begingroup$
Oh, I see the problem. If $f$ is concave, and $f(0)ge 0,$ then $f$ is only subadditive on $[0,infty).$ That makes my example less than useful, unfortunately.
$endgroup$
– Adrian Keister
Jan 15 at 16:27
$begingroup$
Oh, I see the problem. If $f$ is concave, and $f(0)ge 0,$ then $f$ is only subadditive on $[0,infty).$ That makes my example less than useful, unfortunately.
$endgroup$
– Adrian Keister
Jan 15 at 16:27
add a comment |
1 Answer
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$begingroup$
Consider $f(x) = |x|$ which is subadditive on $mathbb R$. Its derivative is not defined at $0$, but even where it is defined it fails to be subadditive, e.g.
$f'(1) > f'(-1) + f'(2)$.
For an example that is differentiable on all of $mathbb R$, try
$$ f(x) = cases{2|x| & if $|x|ge 1$cr x^2+1 & otherwise}$$
EDIT: Somewhat more generally, the only cases where an even
differentiable function $f$ and its derivative $f'$ are both subadditive on all of $mathbb R$ are when $f$ is constant.
Indeed, suppose $f$ is subadditive, differentiable and even, so $f'$ is odd. But then if $f'$ is subadditive it must be additive (and thus linear), since if $f'(x + y) < f'(x) + f'(y)$ we would have $f'(-x-y) = -f'(x+y) > -f'(x) - f'(y) = f'(-x) + f'(-y)$. And since a quadratic is never subadditive on $mathbb R$, this can only happen if $f$ is constant.
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
add a comment |
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$begingroup$
Consider $f(x) = |x|$ which is subadditive on $mathbb R$. Its derivative is not defined at $0$, but even where it is defined it fails to be subadditive, e.g.
$f'(1) > f'(-1) + f'(2)$.
For an example that is differentiable on all of $mathbb R$, try
$$ f(x) = cases{2|x| & if $|x|ge 1$cr x^2+1 & otherwise}$$
EDIT: Somewhat more generally, the only cases where an even
differentiable function $f$ and its derivative $f'$ are both subadditive on all of $mathbb R$ are when $f$ is constant.
Indeed, suppose $f$ is subadditive, differentiable and even, so $f'$ is odd. But then if $f'$ is subadditive it must be additive (and thus linear), since if $f'(x + y) < f'(x) + f'(y)$ we would have $f'(-x-y) = -f'(x+y) > -f'(x) - f'(y) = f'(-x) + f'(-y)$. And since a quadratic is never subadditive on $mathbb R$, this can only happen if $f$ is constant.
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
add a comment |
$begingroup$
Consider $f(x) = |x|$ which is subadditive on $mathbb R$. Its derivative is not defined at $0$, but even where it is defined it fails to be subadditive, e.g.
$f'(1) > f'(-1) + f'(2)$.
For an example that is differentiable on all of $mathbb R$, try
$$ f(x) = cases{2|x| & if $|x|ge 1$cr x^2+1 & otherwise}$$
EDIT: Somewhat more generally, the only cases where an even
differentiable function $f$ and its derivative $f'$ are both subadditive on all of $mathbb R$ are when $f$ is constant.
Indeed, suppose $f$ is subadditive, differentiable and even, so $f'$ is odd. But then if $f'$ is subadditive it must be additive (and thus linear), since if $f'(x + y) < f'(x) + f'(y)$ we would have $f'(-x-y) = -f'(x+y) > -f'(x) - f'(y) = f'(-x) + f'(-y)$. And since a quadratic is never subadditive on $mathbb R$, this can only happen if $f$ is constant.
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
$endgroup$
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
add a comment |
$begingroup$
Consider $f(x) = |x|$ which is subadditive on $mathbb R$. Its derivative is not defined at $0$, but even where it is defined it fails to be subadditive, e.g.
$f'(1) > f'(-1) + f'(2)$.
For an example that is differentiable on all of $mathbb R$, try
$$ f(x) = cases{2|x| & if $|x|ge 1$cr x^2+1 & otherwise}$$
EDIT: Somewhat more generally, the only cases where an even
differentiable function $f$ and its derivative $f'$ are both subadditive on all of $mathbb R$ are when $f$ is constant.
Indeed, suppose $f$ is subadditive, differentiable and even, so $f'$ is odd. But then if $f'$ is subadditive it must be additive (and thus linear), since if $f'(x + y) < f'(x) + f'(y)$ we would have $f'(-x-y) = -f'(x+y) > -f'(x) - f'(y) = f'(-x) + f'(-y)$. And since a quadratic is never subadditive on $mathbb R$, this can only happen if $f$ is constant.
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
$endgroup$
Consider $f(x) = |x|$ which is subadditive on $mathbb R$. Its derivative is not defined at $0$, but even where it is defined it fails to be subadditive, e.g.
$f'(1) > f'(-1) + f'(2)$.
For an example that is differentiable on all of $mathbb R$, try
$$ f(x) = cases{2|x| & if $|x|ge 1$cr x^2+1 & otherwise}$$
EDIT: Somewhat more generally, the only cases where an even
differentiable function $f$ and its derivative $f'$ are both subadditive on all of $mathbb R$ are when $f$ is constant.
Indeed, suppose $f$ is subadditive, differentiable and even, so $f'$ is odd. But then if $f'$ is subadditive it must be additive (and thus linear), since if $f'(x + y) < f'(x) + f'(y)$ we would have $f'(-x-y) = -f'(x+y) > -f'(x) - f'(y) = f'(-x) + f'(-y)$. And since a quadratic is never subadditive on $mathbb R$, this can only happen if $f$ is constant.
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
edited Jan 16 at 17:38
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
answered Jan 15 at 15:10
Robert IsraelRobert Israel
324k23214468
324k23214468
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
add a comment |
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
$begingroup$
I will accept this answer if you could elaborate on implications for my first question in particular.
$endgroup$
– theV0ID
Jan 16 at 13:53
add a comment |
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$begingroup$
Q2: $f(x)=-x^4$ is concave, and therefore subadditive. But its derivative is $f'(x)=-4x^3,$ which is not going to be subadditive (check $x=-2, y=-1.$)
$endgroup$
– Adrian Keister
Jan 15 at 15:01
$begingroup$
@AdrianKeister $f(x) = -x^4$ is not subadditive on all of $mathbb R$ . For example, $f(0) > f(-1) + f(1)$.
$endgroup$
– Robert Israel
Jan 15 at 15:06
$begingroup$
I'm confused: Is $-x^4$ not concave?
$endgroup$
– Adrian Keister
Jan 15 at 16:13
$begingroup$
Oh, I see the problem. If $f$ is concave, and $f(0)ge 0,$ then $f$ is only subadditive on $[0,infty).$ That makes my example less than useful, unfortunately.
$endgroup$
– Adrian Keister
Jan 15 at 16:27