Seperating Hyperplanes and Hahn-Banach
$begingroup$
Let $f:X rightarrow mathbb{R}$ be convex, $x in X$. Assume further that $f$ is continuous and finite in $x.$ Then it says that by Hahn-Banach there is $x^* in X^*$ with
begin{equation}
langle y - x, x^* rangle_{X times X^*} + f(x) leq f(y) forall y in X.
end{equation}
I'm trying to understand why this is true. As far as I know, by Hahn-Banach we can seperate the convex set given by the points above the graph of $f$ from the point $(x,f(x))$ by some hyperplane. More generally, for $A,B subset X$, $Acap B= emptyset$ convex sets with $A$ open there is $x^* in X^*$ and $gamma in mathbb{R}$ such that
begin{equation}
langle a,x^* rangle _{X times X^*} leq gamma leq langle b,x^* rangle _{X times X^*} forall ain A, b in B. (1)
end{equation}
Intuitively, the above estimate makes sense. The desired hyperplane is an affine function below the graph of $f$ which intersects with the graph in the point $(x,f(x))$. But I would like to understand how this follows from the formal statement $(1)$ in the Hahn-Banach Theorem. Thank you for any help.
functional-analysis convex-analysis
$endgroup$
add a comment |
$begingroup$
Let $f:X rightarrow mathbb{R}$ be convex, $x in X$. Assume further that $f$ is continuous and finite in $x.$ Then it says that by Hahn-Banach there is $x^* in X^*$ with
begin{equation}
langle y - x, x^* rangle_{X times X^*} + f(x) leq f(y) forall y in X.
end{equation}
I'm trying to understand why this is true. As far as I know, by Hahn-Banach we can seperate the convex set given by the points above the graph of $f$ from the point $(x,f(x))$ by some hyperplane. More generally, for $A,B subset X$, $Acap B= emptyset$ convex sets with $A$ open there is $x^* in X^*$ and $gamma in mathbb{R}$ such that
begin{equation}
langle a,x^* rangle _{X times X^*} leq gamma leq langle b,x^* rangle _{X times X^*} forall ain A, b in B. (1)
end{equation}
Intuitively, the above estimate makes sense. The desired hyperplane is an affine function below the graph of $f$ which intersects with the graph in the point $(x,f(x))$. But I would like to understand how this follows from the formal statement $(1)$ in the Hahn-Banach Theorem. Thank you for any help.
functional-analysis convex-analysis
$endgroup$
2
$begingroup$
What does it mean for $f:XtoBbb R^n$ to be convex? Did you intend to write $f:XtoBbb R$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 3:47
1
$begingroup$
There are no conditions in your post that ensure that your set $A$ is open, or that the epigraph of $f$ contains interior points.
$endgroup$
– daw
Dec 13 '18 at 7:49
$begingroup$
I'm sorry, you're right. $n$ should be $1$ and I forgot to add that $f$ is assumed to be continuous and finite in $x$, which should guarantee the existence of interior points, I corrected my post
$endgroup$
– jason paper
Dec 14 '18 at 2:15
add a comment |
$begingroup$
Let $f:X rightarrow mathbb{R}$ be convex, $x in X$. Assume further that $f$ is continuous and finite in $x.$ Then it says that by Hahn-Banach there is $x^* in X^*$ with
begin{equation}
langle y - x, x^* rangle_{X times X^*} + f(x) leq f(y) forall y in X.
end{equation}
I'm trying to understand why this is true. As far as I know, by Hahn-Banach we can seperate the convex set given by the points above the graph of $f$ from the point $(x,f(x))$ by some hyperplane. More generally, for $A,B subset X$, $Acap B= emptyset$ convex sets with $A$ open there is $x^* in X^*$ and $gamma in mathbb{R}$ such that
begin{equation}
langle a,x^* rangle _{X times X^*} leq gamma leq langle b,x^* rangle _{X times X^*} forall ain A, b in B. (1)
end{equation}
Intuitively, the above estimate makes sense. The desired hyperplane is an affine function below the graph of $f$ which intersects with the graph in the point $(x,f(x))$. But I would like to understand how this follows from the formal statement $(1)$ in the Hahn-Banach Theorem. Thank you for any help.
functional-analysis convex-analysis
$endgroup$
Let $f:X rightarrow mathbb{R}$ be convex, $x in X$. Assume further that $f$ is continuous and finite in $x.$ Then it says that by Hahn-Banach there is $x^* in X^*$ with
begin{equation}
langle y - x, x^* rangle_{X times X^*} + f(x) leq f(y) forall y in X.
end{equation}
I'm trying to understand why this is true. As far as I know, by Hahn-Banach we can seperate the convex set given by the points above the graph of $f$ from the point $(x,f(x))$ by some hyperplane. More generally, for $A,B subset X$, $Acap B= emptyset$ convex sets with $A$ open there is $x^* in X^*$ and $gamma in mathbb{R}$ such that
begin{equation}
langle a,x^* rangle _{X times X^*} leq gamma leq langle b,x^* rangle _{X times X^*} forall ain A, b in B. (1)
end{equation}
Intuitively, the above estimate makes sense. The desired hyperplane is an affine function below the graph of $f$ which intersects with the graph in the point $(x,f(x))$. But I would like to understand how this follows from the formal statement $(1)$ in the Hahn-Banach Theorem. Thank you for any help.
functional-analysis convex-analysis
functional-analysis convex-analysis
edited Dec 14 '18 at 2:13
jason paper
asked Dec 13 '18 at 1:30
jason paperjason paper
13519
13519
2
$begingroup$
What does it mean for $f:XtoBbb R^n$ to be convex? Did you intend to write $f:XtoBbb R$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 3:47
1
$begingroup$
There are no conditions in your post that ensure that your set $A$ is open, or that the epigraph of $f$ contains interior points.
$endgroup$
– daw
Dec 13 '18 at 7:49
$begingroup$
I'm sorry, you're right. $n$ should be $1$ and I forgot to add that $f$ is assumed to be continuous and finite in $x$, which should guarantee the existence of interior points, I corrected my post
$endgroup$
– jason paper
Dec 14 '18 at 2:15
add a comment |
2
$begingroup$
What does it mean for $f:XtoBbb R^n$ to be convex? Did you intend to write $f:XtoBbb R$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 3:47
1
$begingroup$
There are no conditions in your post that ensure that your set $A$ is open, or that the epigraph of $f$ contains interior points.
$endgroup$
– daw
Dec 13 '18 at 7:49
$begingroup$
I'm sorry, you're right. $n$ should be $1$ and I forgot to add that $f$ is assumed to be continuous and finite in $x$, which should guarantee the existence of interior points, I corrected my post
$endgroup$
– jason paper
Dec 14 '18 at 2:15
2
2
$begingroup$
What does it mean for $f:XtoBbb R^n$ to be convex? Did you intend to write $f:XtoBbb R$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 3:47
$begingroup$
What does it mean for $f:XtoBbb R^n$ to be convex? Did you intend to write $f:XtoBbb R$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 3:47
1
1
$begingroup$
There are no conditions in your post that ensure that your set $A$ is open, or that the epigraph of $f$ contains interior points.
$endgroup$
– daw
Dec 13 '18 at 7:49
$begingroup$
There are no conditions in your post that ensure that your set $A$ is open, or that the epigraph of $f$ contains interior points.
$endgroup$
– daw
Dec 13 '18 at 7:49
$begingroup$
I'm sorry, you're right. $n$ should be $1$ and I forgot to add that $f$ is assumed to be continuous and finite in $x$, which should guarantee the existence of interior points, I corrected my post
$endgroup$
– jason paper
Dec 14 '18 at 2:15
$begingroup$
I'm sorry, you're right. $n$ should be $1$ and I forgot to add that $f$ is assumed to be continuous and finite in $x$, which should guarantee the existence of interior points, I corrected my post
$endgroup$
– jason paper
Dec 14 '18 at 2:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let me add another answer to address your modified question.
What we want to do is to use the Hahn-Banach separation theorem with the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$. Continuity at $x$ guarantee that $(text{epi} f)^circ$ is not empty, which is not the case if $f$ is a discontinuous linear functional.
Recall that the epigraph $text{epi} f:= {(y,r)in XtimesBbb R: rge f(y)}$ is convex, and that $(Xtimes Bbb R)^*simeq X^*times Bbb R$. Note also that for a convex set $C$ with nonempty interior, we have $overline C = overline{(C^circ)}$.
Let $x_0$ be a point in $X$, take a small $varepsilon>0$ then by continuity of $f$ we can find a neighborhood $V$ of $x_0$ such that $f(y)<f(x_0)+varepsilon/2$ for all $yin V$. This means that
$$
Vtimes (f(x_0)+varepsilon,+ infty) subset text{epi} f
$$
and hence $(text{epi} f)^circne emptyset$.
Now, let's apply the Hahn-Banach separation theorem to the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$ to get $(x^*,lambda)in X^*times Bbb R$ such that
$$
langle x_0,x^* rangle + lambda f(x_0) > langle y,x^* rangle + lambda r
$$
for all $(y,r)in (text{epi} f)^circ$.
The above argument with small enough $varepsilon$ shows that $(x_0,r)in (text{epi} f)^circ$ for all $r>f(x_0)$, substitute this into $(y,r)$ and rearrange the terms to get
$$
lambda(f(x_0)-r) > 0
$$
which implies that $lambda<0$. We may thus rescale $(x^*,lambda)$ so that $lambda=-1$.
Since $text{epi} f subset overline{text{epi} f} =overline{(text{epi} f^circ)} $ and that $(y,f(y))in text{epi}f$,
$$
langle x_0,x^* rangle - f(x_0) ge langle y,x^* rangle - f(y)
$$
for all $yin X$ (we take limit $rto f(y)$ so the strict $>$ becomes $ge$). Hence the (rescaled) functional $x^*$ is a subgradient of $f$ at $x_0$, i.e. $x^*in partial f(x_0)$.
$endgroup$
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
add a comment |
$begingroup$
In general, the statement is not true.
Consider a discontinuous linear functional $Lambda:Xto Bbb R$ (which can be constructed from a Hamel basis), since it is a linear functional it is automatically convex. However, at any point $xin X$ its subdifferential $partialLambda(x)$, i.e. the set of all $x^*in X^*$ such that
$$langle y - x, x^* rangle+ f(x) leq f(y) forall y in X,
$$
is empty.
$endgroup$
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let me add another answer to address your modified question.
What we want to do is to use the Hahn-Banach separation theorem with the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$. Continuity at $x$ guarantee that $(text{epi} f)^circ$ is not empty, which is not the case if $f$ is a discontinuous linear functional.
Recall that the epigraph $text{epi} f:= {(y,r)in XtimesBbb R: rge f(y)}$ is convex, and that $(Xtimes Bbb R)^*simeq X^*times Bbb R$. Note also that for a convex set $C$ with nonempty interior, we have $overline C = overline{(C^circ)}$.
Let $x_0$ be a point in $X$, take a small $varepsilon>0$ then by continuity of $f$ we can find a neighborhood $V$ of $x_0$ such that $f(y)<f(x_0)+varepsilon/2$ for all $yin V$. This means that
$$
Vtimes (f(x_0)+varepsilon,+ infty) subset text{epi} f
$$
and hence $(text{epi} f)^circne emptyset$.
Now, let's apply the Hahn-Banach separation theorem to the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$ to get $(x^*,lambda)in X^*times Bbb R$ such that
$$
langle x_0,x^* rangle + lambda f(x_0) > langle y,x^* rangle + lambda r
$$
for all $(y,r)in (text{epi} f)^circ$.
The above argument with small enough $varepsilon$ shows that $(x_0,r)in (text{epi} f)^circ$ for all $r>f(x_0)$, substitute this into $(y,r)$ and rearrange the terms to get
$$
lambda(f(x_0)-r) > 0
$$
which implies that $lambda<0$. We may thus rescale $(x^*,lambda)$ so that $lambda=-1$.
Since $text{epi} f subset overline{text{epi} f} =overline{(text{epi} f^circ)} $ and that $(y,f(y))in text{epi}f$,
$$
langle x_0,x^* rangle - f(x_0) ge langle y,x^* rangle - f(y)
$$
for all $yin X$ (we take limit $rto f(y)$ so the strict $>$ becomes $ge$). Hence the (rescaled) functional $x^*$ is a subgradient of $f$ at $x_0$, i.e. $x^*in partial f(x_0)$.
$endgroup$
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
add a comment |
$begingroup$
Let me add another answer to address your modified question.
What we want to do is to use the Hahn-Banach separation theorem with the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$. Continuity at $x$ guarantee that $(text{epi} f)^circ$ is not empty, which is not the case if $f$ is a discontinuous linear functional.
Recall that the epigraph $text{epi} f:= {(y,r)in XtimesBbb R: rge f(y)}$ is convex, and that $(Xtimes Bbb R)^*simeq X^*times Bbb R$. Note also that for a convex set $C$ with nonempty interior, we have $overline C = overline{(C^circ)}$.
Let $x_0$ be a point in $X$, take a small $varepsilon>0$ then by continuity of $f$ we can find a neighborhood $V$ of $x_0$ such that $f(y)<f(x_0)+varepsilon/2$ for all $yin V$. This means that
$$
Vtimes (f(x_0)+varepsilon,+ infty) subset text{epi} f
$$
and hence $(text{epi} f)^circne emptyset$.
Now, let's apply the Hahn-Banach separation theorem to the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$ to get $(x^*,lambda)in X^*times Bbb R$ such that
$$
langle x_0,x^* rangle + lambda f(x_0) > langle y,x^* rangle + lambda r
$$
for all $(y,r)in (text{epi} f)^circ$.
The above argument with small enough $varepsilon$ shows that $(x_0,r)in (text{epi} f)^circ$ for all $r>f(x_0)$, substitute this into $(y,r)$ and rearrange the terms to get
$$
lambda(f(x_0)-r) > 0
$$
which implies that $lambda<0$. We may thus rescale $(x^*,lambda)$ so that $lambda=-1$.
Since $text{epi} f subset overline{text{epi} f} =overline{(text{epi} f^circ)} $ and that $(y,f(y))in text{epi}f$,
$$
langle x_0,x^* rangle - f(x_0) ge langle y,x^* rangle - f(y)
$$
for all $yin X$ (we take limit $rto f(y)$ so the strict $>$ becomes $ge$). Hence the (rescaled) functional $x^*$ is a subgradient of $f$ at $x_0$, i.e. $x^*in partial f(x_0)$.
$endgroup$
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
add a comment |
$begingroup$
Let me add another answer to address your modified question.
What we want to do is to use the Hahn-Banach separation theorem with the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$. Continuity at $x$ guarantee that $(text{epi} f)^circ$ is not empty, which is not the case if $f$ is a discontinuous linear functional.
Recall that the epigraph $text{epi} f:= {(y,r)in XtimesBbb R: rge f(y)}$ is convex, and that $(Xtimes Bbb R)^*simeq X^*times Bbb R$. Note also that for a convex set $C$ with nonempty interior, we have $overline C = overline{(C^circ)}$.
Let $x_0$ be a point in $X$, take a small $varepsilon>0$ then by continuity of $f$ we can find a neighborhood $V$ of $x_0$ such that $f(y)<f(x_0)+varepsilon/2$ for all $yin V$. This means that
$$
Vtimes (f(x_0)+varepsilon,+ infty) subset text{epi} f
$$
and hence $(text{epi} f)^circne emptyset$.
Now, let's apply the Hahn-Banach separation theorem to the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$ to get $(x^*,lambda)in X^*times Bbb R$ such that
$$
langle x_0,x^* rangle + lambda f(x_0) > langle y,x^* rangle + lambda r
$$
for all $(y,r)in (text{epi} f)^circ$.
The above argument with small enough $varepsilon$ shows that $(x_0,r)in (text{epi} f)^circ$ for all $r>f(x_0)$, substitute this into $(y,r)$ and rearrange the terms to get
$$
lambda(f(x_0)-r) > 0
$$
which implies that $lambda<0$. We may thus rescale $(x^*,lambda)$ so that $lambda=-1$.
Since $text{epi} f subset overline{text{epi} f} =overline{(text{epi} f^circ)} $ and that $(y,f(y))in text{epi}f$,
$$
langle x_0,x^* rangle - f(x_0) ge langle y,x^* rangle - f(y)
$$
for all $yin X$ (we take limit $rto f(y)$ so the strict $>$ becomes $ge$). Hence the (rescaled) functional $x^*$ is a subgradient of $f$ at $x_0$, i.e. $x^*in partial f(x_0)$.
$endgroup$
Let me add another answer to address your modified question.
What we want to do is to use the Hahn-Banach separation theorem with the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$. Continuity at $x$ guarantee that $(text{epi} f)^circ$ is not empty, which is not the case if $f$ is a discontinuous linear functional.
Recall that the epigraph $text{epi} f:= {(y,r)in XtimesBbb R: rge f(y)}$ is convex, and that $(Xtimes Bbb R)^*simeq X^*times Bbb R$. Note also that for a convex set $C$ with nonempty interior, we have $overline C = overline{(C^circ)}$.
Let $x_0$ be a point in $X$, take a small $varepsilon>0$ then by continuity of $f$ we can find a neighborhood $V$ of $x_0$ such that $f(y)<f(x_0)+varepsilon/2$ for all $yin V$. This means that
$$
Vtimes (f(x_0)+varepsilon,+ infty) subset text{epi} f
$$
and hence $(text{epi} f)^circne emptyset$.
Now, let's apply the Hahn-Banach separation theorem to the open set $(text{epi} f)^circ$ and the compact set ${(x_0,f(x_0)}$ to get $(x^*,lambda)in X^*times Bbb R$ such that
$$
langle x_0,x^* rangle + lambda f(x_0) > langle y,x^* rangle + lambda r
$$
for all $(y,r)in (text{epi} f)^circ$.
The above argument with small enough $varepsilon$ shows that $(x_0,r)in (text{epi} f)^circ$ for all $r>f(x_0)$, substitute this into $(y,r)$ and rearrange the terms to get
$$
lambda(f(x_0)-r) > 0
$$
which implies that $lambda<0$. We may thus rescale $(x^*,lambda)$ so that $lambda=-1$.
Since $text{epi} f subset overline{text{epi} f} =overline{(text{epi} f^circ)} $ and that $(y,f(y))in text{epi}f$,
$$
langle x_0,x^* rangle - f(x_0) ge langle y,x^* rangle - f(y)
$$
for all $yin X$ (we take limit $rto f(y)$ so the strict $>$ becomes $ge$). Hence the (rescaled) functional $x^*$ is a subgradient of $f$ at $x_0$, i.e. $x^*in partial f(x_0)$.
edited Dec 14 '18 at 3:51
answered Dec 14 '18 at 3:45
BigbearZzzBigbearZzz
8,85921652
8,85921652
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
add a comment |
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
$begingroup$
Thank you very much, great answer! =) I think my biggest problem was not seeing that we can combine the dual pairing of $X$ and the Euclidean product as some pairing on $X times mathbb{R}$. It should be clear to me now, so I won't need the above counter example anymore, I just thought this could clearify the relation. Thanks!
$endgroup$
– jason paper
Dec 15 '18 at 1:40
add a comment |
$begingroup$
In general, the statement is not true.
Consider a discontinuous linear functional $Lambda:Xto Bbb R$ (which can be constructed from a Hamel basis), since it is a linear functional it is automatically convex. However, at any point $xin X$ its subdifferential $partialLambda(x)$, i.e. the set of all $x^*in X^*$ such that
$$langle y - x, x^* rangle+ f(x) leq f(y) forall y in X,
$$
is empty.
$endgroup$
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
add a comment |
$begingroup$
In general, the statement is not true.
Consider a discontinuous linear functional $Lambda:Xto Bbb R$ (which can be constructed from a Hamel basis), since it is a linear functional it is automatically convex. However, at any point $xin X$ its subdifferential $partialLambda(x)$, i.e. the set of all $x^*in X^*$ such that
$$langle y - x, x^* rangle+ f(x) leq f(y) forall y in X,
$$
is empty.
$endgroup$
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
add a comment |
$begingroup$
In general, the statement is not true.
Consider a discontinuous linear functional $Lambda:Xto Bbb R$ (which can be constructed from a Hamel basis), since it is a linear functional it is automatically convex. However, at any point $xin X$ its subdifferential $partialLambda(x)$, i.e. the set of all $x^*in X^*$ such that
$$langle y - x, x^* rangle+ f(x) leq f(y) forall y in X,
$$
is empty.
$endgroup$
In general, the statement is not true.
Consider a discontinuous linear functional $Lambda:Xto Bbb R$ (which can be constructed from a Hamel basis), since it is a linear functional it is automatically convex. However, at any point $xin X$ its subdifferential $partialLambda(x)$, i.e. the set of all $x^*in X^*$ such that
$$langle y - x, x^* rangle+ f(x) leq f(y) forall y in X,
$$
is empty.
answered Dec 13 '18 at 4:53
BigbearZzzBigbearZzz
8,85921652
8,85921652
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
add a comment |
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
Thanks for the example, unfortunately as mentioned above I forgot some assumptions. What I don't get is how we see that the subdifferential is empty in this case (or non-empty under the correctend assumptions).. of course, this corresponds directly to what I'm trying to see above, but I don't understand how we get it, or how it is related to the Hahn-Banach Theorem.
$endgroup$
– jason paper
Dec 14 '18 at 2:26
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
$begingroup$
@jasonpaper I've added another answer to address your modified question. If you're still interested in knowing why a discontinuous linear map is a counter example you can ask another question and I might come to answer. I think my answer here is already too long :)
$endgroup$
– BigbearZzz
Dec 14 '18 at 3:47
add a comment |
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What does it mean for $f:XtoBbb R^n$ to be convex? Did you intend to write $f:XtoBbb R$?
$endgroup$
– BigbearZzz
Dec 13 '18 at 3:47
1
$begingroup$
There are no conditions in your post that ensure that your set $A$ is open, or that the epigraph of $f$ contains interior points.
$endgroup$
– daw
Dec 13 '18 at 7:49
$begingroup$
I'm sorry, you're right. $n$ should be $1$ and I forgot to add that $f$ is assumed to be continuous and finite in $x$, which should guarantee the existence of interior points, I corrected my post
$endgroup$
– jason paper
Dec 14 '18 at 2:15