Cfrgtkky

In Bertsekas, Convex Optimization Algorithms The following Proposition is proved.

Let $Phi: mathbb{R}^{n} to mathbb{R}$ be a convex function. For every $x in mathbb{R}^{n}$, we have

(a) The subgradient $partial Phi(x)$ is a nonempty, convex and compact set, and we have
begin{equation}
label{eq-quotient}
Phi'(x;d):= lim limits_{alpha to 0} dfrac{Phi(x+alpha d)-Phi(x)}{alpha} =max_{g in partial f(x)} g^{intercal} d quad forall d in mathbb{R^{n}}
end{equation}
(b) If $Phi$ is differentiable at $x$ with gradient $nabla Phi(x)$, then $nabla Phi(x)$ is its unique subgradient at $x$, and we have $Phi'(x;d)= nabla Phi(x)^{intercal}$

What I want to show: I want to generalize the nonemptyness, closedness and compactness of the subgradient to convex functions, defined on arbitrary open convex subsets of $mathbb{R}^{n}$. My Proof goes as follows: Let $Phi : X to mathbb{R}^{n}$ be a convex function with $X subseteq mathbb{R}^{n}$ a convex open set.

We can (I think) extend $Phi$ to a convex function $Phi _{text{ext}}: mathbb{R}^{n} to mathbb{R}$ by defining

begin{equation}
Phi_{text{ext}}(x)=Phi(x) text{ if } x in X text{ and } Phi_{text{ext}}(x)= infty text{ if } x notin X
end{equation}
. By the proposition stated above, we know that for $p in X$, $partial Phi_{text{ext}}(p)$ is non-empty, closed and compact. We also know that $partial Phi_{text{ext}}(p)= partial Phi(p)$ holds, since
begin{gather}
partial Phi_{text{ext}}(p)={ y in mathbb{R}^{n} | langle y, q-p rangle leq Phi_{text{ext}}(q) - Phi_{text{ext}}(p) forall q in mathbb{R}^{n} } \
= { y in mathbb{R}^{n} | langle y, q-p rangle leq Phi_{text{ext}}(q) - Phi(p) forall q in X } cap { y in mathbb{R^{n}} | langle y, rangle leq Phi_{text{ext}}(q) - Phi(p) forall q in mathbb{R}^{n} setminus X } \
={ y in mathbb{R}^{n} | langle y, q-p rangle leq Phi_{text{ext}}(q) - Phi(p) forall q in X } cap mathbb{R}^{n} \
={ y in mathbb{R}^{n} | langle y, q-p rangle leq Phi(q) - Phi(p) forall q in X } \
= partial Phi(p)
end{gather}

In particular, the fact that $partial Phi(p)$ is non empty, closed and compact follows from the fact that $partial Phi_{text{ext}}(p)$ fulfills the property.

Question: Is my proof correct?

Edit: As pointed out by littleO, the proof in bertsekas assumes that the convex function has finite values. Hence my modified question is then: Given that $Phi$ is finite, is my proof correct if we would replace the definition of $Phi_{text{ext}}$ with

begin{equation}
Phi_{text{ext}}(x)=Phi(x) text{ if } x in X text{ and } Phi_{text{ext}}(x)= sup_{s in X} Phi(s) text{ if } x notin X
end{equation}

Your proof is incorrect because $Phi$ might not be extendable to a convex function on entire $Bbb{R^n}$. For example take $Phi$ the function whose graph is the lower half of the unit circle.

Now how to proof what you claimed: Subdifferentials and directional derivatives is a local feature of function. So first prove that $ y in partial Phi (p) $ if and only if there exist an open neighborhood of $p$, say $X$ such that

$$ langle y, q-p rangle leq Phi(q) - Phi(p) quad forall q in X $$

Hint: For right to left define the function $ f(x)= Phi(x) - Phi(p) - langle y, x-p rangle$ observe that $f$ is convex on the whole $Bbb{R^n}$ and take a local minimum at $x =p$, so it has to be global minimum too.

Now mimic the Bertsekas' proof, for the local version of subdifferential .