Question regarding Gateaux differentiability
$begingroup$
Let $E$ be a normed space and $Omega subset E$ be an open convex subset. Let $ain Omega$ and $f:Omegalongrightarrow mathbb{R}$, we say $f$ is differentiable in the direction $v$ at $a$ if the following limit exists and finite
$$ f'(a,v):=lim_{tlongrightarrow 0} frac{f(a+tv) - f(a)}{t}.$$
My question is, which of the following should be the correct definition of Gateaux differentiability at $a$?
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear.
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear and continuous.
The question I have in mind is the following
Let $Omegasubset E$ be open, convex and
$f:Omegalongrightarrowmathbb{R}$ be convex. Let $ain Omega$, if
$f'(a,v)$ exists and finite for all $vin E$ then $f$ is Gateaux
differentiable at $a$.
If I use the first definition then it is obvious, but for the second definition then I don't see how it is true. Since we need $f$ to be bounded or continuous (to get $f$ is Lipschitz near $a$) in order to show $vlongmapsto f'(a,v)$ is continuous.
Proof for $vlongmapsto f'(a,v)$ is linear given that $f$ is convex and $f'(a,v)$ exists for all $v$.
First of all given this hypothesis I can show that
$$ f'_+(a,v):=lim_{tlongrightarrow 0^+} frac{f(a+tv) - f(a)}{t}.$$
exists and finite everywhere. Indeed, the function $varphi(s) = f(a+sv)$ for $sin (-varepsilon,varepsilon)$ is convex from a subset of $mathbb{R}longrightarrow mathbb{R}$, hence it is continuous and locally Lipschitz. By the convexity $slongmapstofrac{f(a+sv) - f(a)}{s}$ is decreasing as $slongrightarrow 0^+$. Together with being bounded from
$$frac{f(a+sv) - f(a)}{s} = frac{varphi(s) - varphi(0)}{s} geq -frac{Cs}{s} = -C.$$
we conclude that $f'_+(a,v)$ exists and finite for all $v$. It is clear that
$$f'_+(a,lambda v) = lambda f'_+(a,v).$$
Let $p(v) = f'_+(a,v)$, it is easy to see that $p:Elongrightarrow mathbb{R}$ is sublinear since for $u,vin E$, we have
begin{align*}
p(u+v) &= lim_{tlongrightarrow 0^+} frac{f(a+t(u+v))-f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)+f(a+2tv) -f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)-f(a)}{2t} + lim_{tlongrightarrow 0^+} frac{f(a+2tv)-f(a)}{2t} = p(u) + p(v).
end{align*}
and $0leq p(0) leq p(-v)+p(v)$, thus $-p(-v)leq p(v)$ for $vin E$.
Now let us define
$$V = {vin E: -p(-v) = p(v)} = {vin E: f'(a,v);text{exists and finite}}.$$
It is easy to see that $V$ is a linear subspace of $E$. For $u,vin V$ we have
begin{equation*}
-p(-u)-p(-v)leq -p(-u-v) leq p(u+v)leq p(u+v) = -p(-u)+p(-v).
end{equation*}
Therefore $u+vin V$. For $lambda in mathbb{R}$ and $vin V$, if $lambda >0$ then clearly $lambda vin V$, while if $lambda < 0$ then $-lambda v > 0$, thus
$$ p(-lambda v) = -lambda p(v) qquad Longrightarrow qquad -p(-lambda v) = p(lambda v)$$
Hence $lambda v in V$. From that definition it is obvious that $p|_V$ is linear.
If $f'(a,v)$ exists for all $vin E$ then $V = E$, hence $vlongmapsto f'(a,v) = p(v)$ is linear on $E$.
Can we show $vlongmapsto f'(a,v)$ is indeed continuous?
convex-analysis gateaux-derivative
$endgroup$
add a comment |
$begingroup$
Let $E$ be a normed space and $Omega subset E$ be an open convex subset. Let $ain Omega$ and $f:Omegalongrightarrow mathbb{R}$, we say $f$ is differentiable in the direction $v$ at $a$ if the following limit exists and finite
$$ f'(a,v):=lim_{tlongrightarrow 0} frac{f(a+tv) - f(a)}{t}.$$
My question is, which of the following should be the correct definition of Gateaux differentiability at $a$?
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear.
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear and continuous.
The question I have in mind is the following
Let $Omegasubset E$ be open, convex and
$f:Omegalongrightarrowmathbb{R}$ be convex. Let $ain Omega$, if
$f'(a,v)$ exists and finite for all $vin E$ then $f$ is Gateaux
differentiable at $a$.
If I use the first definition then it is obvious, but for the second definition then I don't see how it is true. Since we need $f$ to be bounded or continuous (to get $f$ is Lipschitz near $a$) in order to show $vlongmapsto f'(a,v)$ is continuous.
Proof for $vlongmapsto f'(a,v)$ is linear given that $f$ is convex and $f'(a,v)$ exists for all $v$.
First of all given this hypothesis I can show that
$$ f'_+(a,v):=lim_{tlongrightarrow 0^+} frac{f(a+tv) - f(a)}{t}.$$
exists and finite everywhere. Indeed, the function $varphi(s) = f(a+sv)$ for $sin (-varepsilon,varepsilon)$ is convex from a subset of $mathbb{R}longrightarrow mathbb{R}$, hence it is continuous and locally Lipschitz. By the convexity $slongmapstofrac{f(a+sv) - f(a)}{s}$ is decreasing as $slongrightarrow 0^+$. Together with being bounded from
$$frac{f(a+sv) - f(a)}{s} = frac{varphi(s) - varphi(0)}{s} geq -frac{Cs}{s} = -C.$$
we conclude that $f'_+(a,v)$ exists and finite for all $v$. It is clear that
$$f'_+(a,lambda v) = lambda f'_+(a,v).$$
Let $p(v) = f'_+(a,v)$, it is easy to see that $p:Elongrightarrow mathbb{R}$ is sublinear since for $u,vin E$, we have
begin{align*}
p(u+v) &= lim_{tlongrightarrow 0^+} frac{f(a+t(u+v))-f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)+f(a+2tv) -f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)-f(a)}{2t} + lim_{tlongrightarrow 0^+} frac{f(a+2tv)-f(a)}{2t} = p(u) + p(v).
end{align*}
and $0leq p(0) leq p(-v)+p(v)$, thus $-p(-v)leq p(v)$ for $vin E$.
Now let us define
$$V = {vin E: -p(-v) = p(v)} = {vin E: f'(a,v);text{exists and finite}}.$$
It is easy to see that $V$ is a linear subspace of $E$. For $u,vin V$ we have
begin{equation*}
-p(-u)-p(-v)leq -p(-u-v) leq p(u+v)leq p(u+v) = -p(-u)+p(-v).
end{equation*}
Therefore $u+vin V$. For $lambda in mathbb{R}$ and $vin V$, if $lambda >0$ then clearly $lambda vin V$, while if $lambda < 0$ then $-lambda v > 0$, thus
$$ p(-lambda v) = -lambda p(v) qquad Longrightarrow qquad -p(-lambda v) = p(lambda v)$$
Hence $lambda v in V$. From that definition it is obvious that $p|_V$ is linear.
If $f'(a,v)$ exists for all $vin E$ then $V = E$, hence $vlongmapsto f'(a,v) = p(v)$ is linear on $E$.
Can we show $vlongmapsto f'(a,v)$ is indeed continuous?
convex-analysis gateaux-derivative
$endgroup$
add a comment |
$begingroup$
Let $E$ be a normed space and $Omega subset E$ be an open convex subset. Let $ain Omega$ and $f:Omegalongrightarrow mathbb{R}$, we say $f$ is differentiable in the direction $v$ at $a$ if the following limit exists and finite
$$ f'(a,v):=lim_{tlongrightarrow 0} frac{f(a+tv) - f(a)}{t}.$$
My question is, which of the following should be the correct definition of Gateaux differentiability at $a$?
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear.
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear and continuous.
The question I have in mind is the following
Let $Omegasubset E$ be open, convex and
$f:Omegalongrightarrowmathbb{R}$ be convex. Let $ain Omega$, if
$f'(a,v)$ exists and finite for all $vin E$ then $f$ is Gateaux
differentiable at $a$.
If I use the first definition then it is obvious, but for the second definition then I don't see how it is true. Since we need $f$ to be bounded or continuous (to get $f$ is Lipschitz near $a$) in order to show $vlongmapsto f'(a,v)$ is continuous.
Proof for $vlongmapsto f'(a,v)$ is linear given that $f$ is convex and $f'(a,v)$ exists for all $v$.
First of all given this hypothesis I can show that
$$ f'_+(a,v):=lim_{tlongrightarrow 0^+} frac{f(a+tv) - f(a)}{t}.$$
exists and finite everywhere. Indeed, the function $varphi(s) = f(a+sv)$ for $sin (-varepsilon,varepsilon)$ is convex from a subset of $mathbb{R}longrightarrow mathbb{R}$, hence it is continuous and locally Lipschitz. By the convexity $slongmapstofrac{f(a+sv) - f(a)}{s}$ is decreasing as $slongrightarrow 0^+$. Together with being bounded from
$$frac{f(a+sv) - f(a)}{s} = frac{varphi(s) - varphi(0)}{s} geq -frac{Cs}{s} = -C.$$
we conclude that $f'_+(a,v)$ exists and finite for all $v$. It is clear that
$$f'_+(a,lambda v) = lambda f'_+(a,v).$$
Let $p(v) = f'_+(a,v)$, it is easy to see that $p:Elongrightarrow mathbb{R}$ is sublinear since for $u,vin E$, we have
begin{align*}
p(u+v) &= lim_{tlongrightarrow 0^+} frac{f(a+t(u+v))-f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)+f(a+2tv) -f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)-f(a)}{2t} + lim_{tlongrightarrow 0^+} frac{f(a+2tv)-f(a)}{2t} = p(u) + p(v).
end{align*}
and $0leq p(0) leq p(-v)+p(v)$, thus $-p(-v)leq p(v)$ for $vin E$.
Now let us define
$$V = {vin E: -p(-v) = p(v)} = {vin E: f'(a,v);text{exists and finite}}.$$
It is easy to see that $V$ is a linear subspace of $E$. For $u,vin V$ we have
begin{equation*}
-p(-u)-p(-v)leq -p(-u-v) leq p(u+v)leq p(u+v) = -p(-u)+p(-v).
end{equation*}
Therefore $u+vin V$. For $lambda in mathbb{R}$ and $vin V$, if $lambda >0$ then clearly $lambda vin V$, while if $lambda < 0$ then $-lambda v > 0$, thus
$$ p(-lambda v) = -lambda p(v) qquad Longrightarrow qquad -p(-lambda v) = p(lambda v)$$
Hence $lambda v in V$. From that definition it is obvious that $p|_V$ is linear.
If $f'(a,v)$ exists for all $vin E$ then $V = E$, hence $vlongmapsto f'(a,v) = p(v)$ is linear on $E$.
Can we show $vlongmapsto f'(a,v)$ is indeed continuous?
convex-analysis gateaux-derivative
$endgroup$
Let $E$ be a normed space and $Omega subset E$ be an open convex subset. Let $ain Omega$ and $f:Omegalongrightarrow mathbb{R}$, we say $f$ is differentiable in the direction $v$ at $a$ if the following limit exists and finite
$$ f'(a,v):=lim_{tlongrightarrow 0} frac{f(a+tv) - f(a)}{t}.$$
My question is, which of the following should be the correct definition of Gateaux differentiability at $a$?
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear.
$f'(a,v)$ exists and finite for all $vin E$ and $vlongmapsto f'(a,v)$ is linear and continuous.
The question I have in mind is the following
Let $Omegasubset E$ be open, convex and
$f:Omegalongrightarrowmathbb{R}$ be convex. Let $ain Omega$, if
$f'(a,v)$ exists and finite for all $vin E$ then $f$ is Gateaux
differentiable at $a$.
If I use the first definition then it is obvious, but for the second definition then I don't see how it is true. Since we need $f$ to be bounded or continuous (to get $f$ is Lipschitz near $a$) in order to show $vlongmapsto f'(a,v)$ is continuous.
Proof for $vlongmapsto f'(a,v)$ is linear given that $f$ is convex and $f'(a,v)$ exists for all $v$.
First of all given this hypothesis I can show that
$$ f'_+(a,v):=lim_{tlongrightarrow 0^+} frac{f(a+tv) - f(a)}{t}.$$
exists and finite everywhere. Indeed, the function $varphi(s) = f(a+sv)$ for $sin (-varepsilon,varepsilon)$ is convex from a subset of $mathbb{R}longrightarrow mathbb{R}$, hence it is continuous and locally Lipschitz. By the convexity $slongmapstofrac{f(a+sv) - f(a)}{s}$ is decreasing as $slongrightarrow 0^+$. Together with being bounded from
$$frac{f(a+sv) - f(a)}{s} = frac{varphi(s) - varphi(0)}{s} geq -frac{Cs}{s} = -C.$$
we conclude that $f'_+(a,v)$ exists and finite for all $v$. It is clear that
$$f'_+(a,lambda v) = lambda f'_+(a,v).$$
Let $p(v) = f'_+(a,v)$, it is easy to see that $p:Elongrightarrow mathbb{R}$ is sublinear since for $u,vin E$, we have
begin{align*}
p(u+v) &= lim_{tlongrightarrow 0^+} frac{f(a+t(u+v))-f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)+f(a+2tv) -f(a)}{t}\
&leq lim_{tlongrightarrow 0^+} frac{f(a+2tu)-f(a)}{2t} + lim_{tlongrightarrow 0^+} frac{f(a+2tv)-f(a)}{2t} = p(u) + p(v).
end{align*}
and $0leq p(0) leq p(-v)+p(v)$, thus $-p(-v)leq p(v)$ for $vin E$.
Now let us define
$$V = {vin E: -p(-v) = p(v)} = {vin E: f'(a,v);text{exists and finite}}.$$
It is easy to see that $V$ is a linear subspace of $E$. For $u,vin V$ we have
begin{equation*}
-p(-u)-p(-v)leq -p(-u-v) leq p(u+v)leq p(u+v) = -p(-u)+p(-v).
end{equation*}
Therefore $u+vin V$. For $lambda in mathbb{R}$ and $vin V$, if $lambda >0$ then clearly $lambda vin V$, while if $lambda < 0$ then $-lambda v > 0$, thus
$$ p(-lambda v) = -lambda p(v) qquad Longrightarrow qquad -p(-lambda v) = p(lambda v)$$
Hence $lambda v in V$. From that definition it is obvious that $p|_V$ is linear.
If $f'(a,v)$ exists for all $vin E$ then $V = E$, hence $vlongmapsto f'(a,v) = p(v)$ is linear on $E$.
Can we show $vlongmapsto f'(a,v)$ is indeed continuous?
convex-analysis gateaux-derivative
convex-analysis gateaux-derivative
edited Dec 31 '18 at 1:34
Sean
asked Dec 31 '18 at 0:46
SeanSean
532513
532513
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The usual definition of Gateaux derivative (at $a$ in the direction $v$) is identical to the definition of directional derivative you gave. Gateaux derivative does not need to exist in all directions, let alone be linear with respect to the directional vector. The quoted passage is then simply defining what is meant by Gateaux differentiability at $a$. It is similar for instance, to defining continuity in an interval via continuity at each point of the interval.
$endgroup$
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
1
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
add a comment |
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1 Answer
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$begingroup$
The usual definition of Gateaux derivative (at $a$ in the direction $v$) is identical to the definition of directional derivative you gave. Gateaux derivative does not need to exist in all directions, let alone be linear with respect to the directional vector. The quoted passage is then simply defining what is meant by Gateaux differentiability at $a$. It is similar for instance, to defining continuity in an interval via continuity at each point of the interval.
$endgroup$
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
1
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
add a comment |
$begingroup$
The usual definition of Gateaux derivative (at $a$ in the direction $v$) is identical to the definition of directional derivative you gave. Gateaux derivative does not need to exist in all directions, let alone be linear with respect to the directional vector. The quoted passage is then simply defining what is meant by Gateaux differentiability at $a$. It is similar for instance, to defining continuity in an interval via continuity at each point of the interval.
$endgroup$
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
1
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
add a comment |
$begingroup$
The usual definition of Gateaux derivative (at $a$ in the direction $v$) is identical to the definition of directional derivative you gave. Gateaux derivative does not need to exist in all directions, let alone be linear with respect to the directional vector. The quoted passage is then simply defining what is meant by Gateaux differentiability at $a$. It is similar for instance, to defining continuity in an interval via continuity at each point of the interval.
$endgroup$
The usual definition of Gateaux derivative (at $a$ in the direction $v$) is identical to the definition of directional derivative you gave. Gateaux derivative does not need to exist in all directions, let alone be linear with respect to the directional vector. The quoted passage is then simply defining what is meant by Gateaux differentiability at $a$. It is similar for instance, to defining continuity in an interval via continuity at each point of the interval.
answered Dec 31 '18 at 1:05
timurtimur
12.2k2144
12.2k2144
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
1
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
add a comment |
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
1
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I mean, the question really is, given the hypothesis of the quote passage, can I show $vlongmapsto f'(a,v)$ is bounded? (it is linear already).
$endgroup$
– Sean
Dec 31 '18 at 1:09
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
I followed the definition of Gateaux derivative in this paper, which requires the map to be linear bounded. m-hikari.com/ams/ams-password-2008/ams-password17-20-2008/…
$endgroup$
– Sean
Dec 31 '18 at 1:12
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
$begingroup$
So are saying that convexity of $f$ implies linearity of the Gateaux derivative?
$endgroup$
– timur
Dec 31 '18 at 1:18
1
1
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
$begingroup$
Yes, unless I am wrong, I am going to put the proof for that into the post now.
$endgroup$
– Sean
Dec 31 '18 at 1:19
add a comment |
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